Proof By Remark 1.2, we may suppose without loss of generality that $T = A \oplus B$ , and so we decompose the underlying Hilbert space as $\mathcal H = \mathcal H_A \oplus \mathcal H_B$ , where $A \in \mathcal B(\mathcal H_A)$ , $B \in \mathcal B(\mathcal H_B)$ , and $\sigma (A^2) \cap \sigma (B^2) = \varnothing $ .

1. (a) implies (b). The hypothesis that allows us to choose square-zero operators $M,N \in \mathcal B ( \mathcal H)$ such that $T=MN-NM$ . Routine calculations show that

   $$\begin{align*}T^2 M = M N M N M = M T^2, \end{align*}$$
   and
   $$\begin{align*}T^2 N = N M N M N = N T^2.\end{align*}$$

   With respect to the decomposition $\mathcal H = \mathcal H_A \oplus \mathcal H_B$ , we write

   $$\begin{align*}M=\begin{bmatrix}M_1& M_2 \\ M_3& M_4 \end{bmatrix},~~N=\begin{bmatrix}N_1& N_2 \\ N_3& N_4 \end{bmatrix}.\end{align*}$$

   By comparing the $(1,2)$ and the $(2,1)$ entries of $T^2M$ and $MT^2$ , we conclude that

   $$\begin{align*}A^2 M_2=M_2 B^2,~~B^2 M_3= M_3 A^2.\end{align*}$$

   Since $\sigma (A^2) \cap \sigma (B^2) =\varnothing $ , by Theorem 1.6, the corresponding Rosenblum operators $\tau _{A^2, B^2}$ and $\tau _{B^2, A^2}$ are injective. From this, we conclude that $M_2 = 0 = M_3$ , so that $M = M_1 \oplus M_4$ . Similarly, $N = N_1 \oplus N_4$ . Since $M^2 = 0 = N^2$ , we conclude that $M_1, M_4, N_1$ , and $N_4$ are all nilpotent of order at most two.

   But $T = [M, N]$ , and thus, $A = [M_1, N_1]$ and $B = [M_4, N_4]$ lie in .

2. (b) implies (a). If $A = [M_1, N_1]$ and $B = [M_2, N_2]$ , where $M_i^2 = 0 = N_i^2$ , $i = 1, 2$ , then $(M_1 \oplus M_2)^2 = 0 = (N_1 \oplus N_2)^2$ and

   $$\begin{align*}T = [(M_1 \oplus M_2), (N_1 \oplus N_2)].\\[-34pt] \end{align*}$$

Proof

1. (a) implies (b). Suppose that . Let $B = [M, N]$ , where $M, N \in \mathcal B ( \mathcal H)$ satisfy $M^2 = 0 = N^2$ .

   Decompose $\mathcal H = \mathcal H_1 \oplus \mathcal H_2$ , where $\mathcal H_1 = \ker \, M$ . Relative to this decomposition, we write

   $$\begin{align*}M=\begin{bmatrix} 0 & A \\ 0 & 0 \end{bmatrix}, \ \ \ \ \ N = \begin{bmatrix} W & X \\ Y & Z \end{bmatrix}. \end{align*}$$

   It follows that

   $$\begin{align*}B = [M, N] = \begin{bmatrix} A Y & A Z - W A \\ 0 & - Y A \end{bmatrix}. \end{align*}$$

   If $\dim \, \mathcal H < \infty $ , then the invertibility of B implies that of both $A Y$ and $-Y A$ , which in turn implies (by considering the ranks of $AY$ and of $-YA$ ) that $\dim \, \mathcal H_1 = \dim \, \mathcal H_2$ and thus $\dim \, \mathcal H$ is even.

   If $\dim \, \mathcal H = \infty $ , then the invertibility of B implies that M has infinite rank. Since $M^2 = 0$ , it clearly has infinite nullity, and thus $\dim \, \mathcal H_1 = \dim \, \mathcal H_2$ . (The same argument shows that the rank and nullity of N are also infinite.)

   In either the finite-dimensional or the infinite-dimensional setting, it follows from the invertibility of B that $A Y$ is left-invertible and that $Y A$ is right-invertible. In particular, $\ker \, AY = \{ 0\}$ and $ {\mathrm {ran}}\, Y A = \mathcal H_2$ . From this, it follows that $\ker \, Y = \{ 0\}$ and $ {\mathrm {ran}}\, Y = \mathcal H_2$ , so that Y is a (continuous) bijection. By the Open Mapping Theorem, Y is invertible.

   Since $\dim \, \mathcal H_1 = \dim \, \mathcal H_2$ and therefore $\mathcal H_1$ is isomorphic to $\mathcal H_2$ , we may identify elements of $\mathcal B(\mathcal H_1 \oplus \mathcal H_2)$ with elements of $\mathbb M_2(\mathcal B(\mathcal H_1))$ , and view each of A, W, X, Y, and Z as sitting in $\mathcal B(\mathcal H_1)$ .

   Let $S := I \oplus Y$ and consider $M_0 := S^{-1} M S$ , $N_0 := S^{-1} N S$ . Then, for appropriate choices of $A_0, W_0, X_0$ , and $Z_0$ , $M_0$ and $N_0$ are of the form

   $$\begin{align*}M_0 = \begin{bmatrix} 0 & A_0 \\ 0 & 0 \end{bmatrix}, \ \ \ \ \ N_0 = \begin{bmatrix} W_0 & X_0 \\ I & Z_0 \end{bmatrix}. \end{align*}$$

   Since $N^2 = 0$ , we see that $Z_0 = -W_0$ and that $X_0 = -Z_0^2$ . Thus, we may write

   $$\begin{align*}N_0 = \begin{bmatrix} W_0 & - W_0^2 \\ I & - W_0 \end{bmatrix}. \end{align*}$$

   Next, note that

   $$\begin{align*}S^{-1} B S = S^{-1} [M, N] S = [M_0, N_0] = \begin{bmatrix} A_0 & -A_0 W_0 - W_0 A_0 \\ 0 & -A_0 \end{bmatrix}. \end{align*}$$

   Set $R = \begin {bmatrix} W_0 & I \\ I & 0 \end {bmatrix}$ , so that R is invertible with inverse $R^{-1} = \begin {bmatrix} 0 & I \\ I & -W_0 \end {bmatrix}$ . A routine calculation shows that

   $$\begin{align*}(SR)^{-1} B (SR) = R^{-1} (S^{-1} B S) R = (-A_0) \oplus A_0. \end{align*}$$

   The last statement is clear.

2. (b) implies (a). This was demonstrated in Paragraph 1.3.

Proof This follows immediately from Proposition 2.4.

Proof First note that as T and $-T$ are both normal, if T is similar to $-T$ , then T is unitarily equivalent to $-T$ by [Reference Halmos17, Problem 192].

The next part of the argument is almost identical to that of [Reference Drnovšek, Radjavi and Rosenthal12, Proposition 3]. Let E be the spectral measure of the normal operator T, and let

$$\begin{align*}\Omega_1=\{z\in \mathbb{C}: z\neq 0, 0\leq \text{arg}z<\pi\},~~\Omega_2=\{z\in \mathbb{C}: z\neq 0, \pi\leq \text{arg}z<2\pi\}.\end{align*}$$

Since T is unitarily equivalent to $-T$ , $T|_{E(\Omega _2)}$ is (unitarily) similar to $-T|_{E(\Omega _1)}$ . Thus, without loss of generality, one can assume that T has the form $A\oplus -A \oplus 0$ , with respect to the decomposition $\mathcal H=\text {Ran }E(\Omega _1)\oplus \text {Ran }E(\Omega _2)\oplus \text {Ran }E(\{0\})$ . As seen in Paragraph 1.3, it easily follows that .

Proof Note that $NM$ commutes with T. Since T is normal, this implies that

$$\begin{align*}NM=\begin{bmatrix}0 & WA\\ 0& YA\\ \end{bmatrix}~~\text{and}~~T^*=\begin{bmatrix}(AY)^* & 0\\ (AZ-WA)^* & -(YA)^*\\ \end{bmatrix}\end{align*}$$

commute. In particular,

$$\begin{align*}YA(AZ-WA)^*=0,\end{align*}$$

so that $(AZ-WA)^*=0$ by hypothesis.

Proof

1. (a) implies (b). This is trivial, since each unitary operator is invertible.

2. (b) implies (c). This is Proposition 2.9.

3. (c) implies (a). By Zorn’s lemma, there is a maximal subspace $\mathcal {L}$ invariant under both M and N on which the respective restrictions of M and N commute. (It is possible that $\mathcal L = \{ 0\}$ .) Relative to the decomposition $\mathcal H = \mathcal L \oplus \mathcal L^\perp $ , M and N have the forms

   $$\begin{align*}M= \begin{bmatrix}M_1 & *\\ 0& M_2\\ \end{bmatrix} ,~~\text{and}~~N=\begin{bmatrix}N_1 & *\\ 0& N_2\\ \end{bmatrix}.\end{align*}$$

   We may then apply the same argument to the adjoints of the nilpotents $M_2$ and $N_2$ . This leads to a decomposition of $\mathcal L^\perp $ as $\mathcal L^\perp = \mathcal M \oplus \mathcal N$ where, with respect to the decomposition $\mathcal H = \mathcal L \oplus \mathcal M \oplus \mathcal N$ , we have

   $$\begin{align*}M=\begin{bmatrix} M_1&*&*\\ 0&M_0&*\\ 0&0&M_3 \\ \end{bmatrix}~~\text{and}~~ N=\begin{bmatrix} N_1&*&*\\ 0&N_0&*\\ 0& 0& N_3\\ \end{bmatrix}. \end{align*}$$

   Recall that the kernel and the range of a normal operator are reducing for that operator. Since T is normal and $T=MN-NM$ , it follows that

   $$\begin{align*}T=\begin{bmatrix} 0&0&0\\ 0&T_0&0\\ 0& 0&0\\ \end{bmatrix},\end{align*}$$

   where $T_0$ is the commutator of $M_0$ and $N_0$ . Note that $M_0$ and $N_0$ cannot have common kernel (by the maximality of $\mathcal L$ and $\mathcal N$ ). Neither can their adjoints.

   We now write

   $$\begin{align*}M_0=\begin{bmatrix}0 & A\\ 0& 0\\ \end{bmatrix},~~\text{and}~~N_0=\begin{bmatrix}W & X\\ Y& Z\\ \end{bmatrix},\end{align*}$$

   relative to the decomposition $\mathcal {M}=\mathcal {M}_1\oplus (\mathcal {M}\ominus \mathcal {M}_1)$ with $\mathcal {M}_1=\ker M_0$ . Then A is injective, and with respect to the decomposition $\mathcal M = \mathcal M_1 \oplus (\mathcal M \ominus \mathcal M_1)$ , $T_0$ has the form

   $$\begin{align*}\begin{bmatrix}AY&AZ-WA\\ 0&-YA\\ \end{bmatrix}.\end{align*}$$

   We claim that Y is injective. Otherwise, Y has nonzero kernel $\mathcal {M}_2$ contained in $\mathcal {M}_1$ . Note that $N_0^2=0$ implies $YW+ZY=0$ , and thus $\mathcal {M}_2$ is invariant under W. Then with respect to the decomposition $\mathcal {M}=\mathcal {M}_2\oplus (\mathcal {M}_1\ominus \mathcal {M}_2)\oplus (\mathcal {M}\ominus \mathcal {M}_1)$ , we have

   $$\begin{align*}M_0=\begin{bmatrix} 0&0&A_1\\ 0&0&A_2\\ 0&0&0 \\ \end{bmatrix}~~\text{and}~~ N_0=\begin{bmatrix} W_1&W_2&X_1\\ 0&W_3&X_2\\ 0& Y_1& Z\\ \end{bmatrix}.\end{align*}$$

   Note that $\mathcal {L}\oplus \mathcal {M}_2$ is an invariant subspace under both M and N on which the respective restrictions of M and N commute. This contradicts the maximality of $\mathcal {L}$ . Thus, Y is injective.

   By Lemma 2.10, $AZ-WA=0$ . Then

   $$\begin{align*}T_0=\begin{bmatrix}AY&0\\ 0&-YA\\ \end{bmatrix}.\end{align*}$$

   Note that $Y(AY)=(YA)Y$ , $A(YA)=(AY)A$ , and $A,Y$ are injective. Furthermore, since $T_0$ is normal, so are $AY$ and $YA$ . Thus, by [Reference Chen, Herrero and Wu9, Lemma 4.1], $AY$ is unitarily equivalent to $YA$ . Therefore, $T_0$ is unitarily equivalent to $-T_0$ , and hence, T is unitarily equivalent to $-T$ .

Proof $\bullet $ Step one.

Let $T = [M, N] = MN - NM$ , where $M^2 = 0 = N^2$ . Let $\mathcal H_1 := \ker \, M$ and $\mathcal H_2 = \mathcal H_1^\perp $ , so that relative to the decomposition $\mathcal H = \mathcal H_1 \oplus \mathcal H_2$ , we may write $M = \begin {bmatrix} 0 & A \\ 0 & 0 \end {bmatrix}$ and $N = \begin {bmatrix} W & X \\ Y & Z \end {bmatrix}$ . Set $L := MN$ and $R := NM$ , and observe that

$$\begin{align*}LR = (MN)(NM) = M N^2 M = 0 \end{align*}$$

and $RL = 0$ as well.

Now, $L = \begin {bmatrix} AY & AZ \\ 0 & 0 \end {bmatrix}$ and $R = \begin {bmatrix} 0 & WA \\ 0 & YA \end {bmatrix}$ . If $0 \ne \alpha \in \sigma _p(L)$ , then $\alpha \in \sigma _p(AY)$ . Similarly, if $0 \ne \alpha \in \sigma _p(R)$ , then $\alpha \in \sigma _p(YA)$ .

$\bullet $ Step two.

Let $k \in \mathbb N$ and $0 \ne \alpha \in \mathbb C$ .

1. (II.1) Note that if $0 \ne w = \begin {bmatrix} w_1 \\ w_2 \end {bmatrix} \in \ker \, (L - \alpha I)^k$ , then (for an appropriate operator $L_{k, \alpha , 2}$ )

   $$\begin{align*}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = (L - \alpha I)^k w = \begin{bmatrix} (AY - \alpha I)^k w_1 + L_{k, \alpha, 2} w_2 \\ (-\alpha)^k \ w_2 \end{bmatrix}, \end{align*}$$
   implying that $w_2 = 0$ .

2. (II.2) Similarly, if $0 \ne w \in \ker \, (R+\alpha I)^k$ , then (for an appropriate operator $R_{k, \alpha , 2}$ )

   $$\begin{align*}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = (R+ \alpha I)^k w = \begin{bmatrix} \alpha^k w_1 + R_{k, \alpha, 2}\ w_2 \\ (YA + \alpha)^k w_2 \end{bmatrix}. \end{align*}$$

   It follows that $w_2 \ne 0$ , for otherwise, $\alpha ^k w_1 = 0$ , and hence, $w_1 = 0$ , a contradiction.

$\bullet $ Step three.

Now let $0 \ne \alpha \in \sigma _p(T)$ , and let $z \in \ker \, (T-\alpha I)^k$ . Then (for an appropriate operator $T_{k, \alpha , 2}$ )

$$\begin{align*}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = (T - \alpha I)^k z = \begin{bmatrix} AY - \alpha I & A Z - W A \\ 0 & - YA - \alpha I \end{bmatrix}^k \ \begin{bmatrix} z_1 \\ z_2 \end{bmatrix} = \begin{bmatrix} (AY - \alpha I)^k z_1 + T_{k, \alpha, 2}\ z_2 \\ (-YA-\alpha I)^k z_2 \end{bmatrix}. \end{align*}$$

This leaves us with two possibilities. Either

1. (a) $z_2 \ne 0$ , and thus, $z_2 \in \ker \, (YA + \alpha I)^k$ ; or

2. (b) $z_2 = 0$ , in which case $z_1 \ne 0$ and $(AY - \alpha I)^k z_1 = 0$ .

$\bullet $ Step four.

Conversely, suppose that condition (a) above holds: that is, $0 \ne y_2 \in \mathcal H_2$ and that $(Y A + \alpha I)^k y_2 = 0$ .

Let $y_1 := -(\alpha ^{-k}) R_{k, \alpha , 2} \ y_2$ , where $R_{k, \alpha , 2}$ is defined as above, and set $y = \begin {bmatrix} y_1 \\ y_2 \end {bmatrix}$ . Clearly, $y \ne 0$ as $y_2 \ne 0$ . Then

$$\begin{align*}(R + \alpha I)^k \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} =\begin{bmatrix} \alpha^k y_1 + R_{k, \alpha, 2} \ y_2 \\ (YA + \alpha)^k y_2 \end{bmatrix} = \begin{bmatrix} \alpha^k y_1 + R_{k, \alpha, 2} \ y_2 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{align*}$$

That is, $(R + \alpha I)^k y = 0$ . Note that $(R + \alpha I)^k L y = L (R + \alpha I)^k y = L 0 = 0$ . Hence, $L y \in \ker \, (R+\alpha I)^k$ . But then

$$\begin{align*}L y = \begin{bmatrix} A Y & A Z \\ 0 & 0 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} (AY) y_1 + (AZ) y_2 \\ 0 \end{bmatrix}. \end{align*}$$

By (II.2) of Step two above, we must have $Ly = 0$ .

A tedious but routine calculation shows that

$$ \begin{align*} (T-\alpha I)^k &= (L - (R + \alpha I))^k \\ &= L^k - \binom{k}{1} \alpha L^{k-1} + \binom{k}{2} \alpha^2 L^{k-2} + \cdots \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots + (-1)^{k-1} \binom{k}{k-1} \alpha^{k-1} L^{1} + (-1)^k (R+\alpha I)^k. \end{align*} $$

From this, it follows that

$$\begin{align*}(T - \alpha I)^k y = (L - (R + \alpha I))^k y = 0, \end{align*}$$

so that $y \in \ker \, (T - \alpha I)^k$ .

More generally, if $y_2^{(1)}, y_2^{(2)}, \ldots , y_2^{(m)}$ is a linearly independent set of vectors in the kernel of $(YA + \alpha I)^k$ , then with $y_1^{(j)} := (-\alpha )^{-k} R_{k, \alpha , 2} \ y_2^{(j)}$ , $1 \le j \le m$ , we see that

$$\begin{align*}y^{(j)} := \begin{bmatrix} y_1^{(j)} \\ y_2^{(j)} \end{bmatrix}, \ \ \ 1 \le j \le m \end{align*}$$

is a linearly independent family of vectors in $\ker \, (T - \alpha I)^k$ .

$\bullet $ Step five.

Suppose that condition (b) holds: that is, $0 \ne x_1 \in \mathcal H_1$ and that $(AY - \alpha I)^k x_1 = 0$ .

Let $x := \begin {bmatrix} x_1 \\ 0 \end {bmatrix}$ . Then

$$\begin{align*}(L - \alpha I)^k x = 0. \end{align*}$$

A routine calculation similar to the one above shows that

$$ \begin{align*} (T-\alpha I)^k &= ( (L-\alpha I) - R)^k \\ &= (L-\alpha I)^ k \\ &\ \ \ + (-1)^k \left ( \binom{k}{1} \alpha^{k-1} R + \binom{k}{2} \alpha^{k-2} R^2 + \cdots + \binom{k}{k-1} \alpha^1 R^{k-1} + R^k \right). \end{align*} $$

From this and the fact that $R x = 0$ , it follows that

$$\begin{align*}(T - \alpha I)^k x = (L-\alpha I)^k x + (-1)^k \sum_{j=1}^k \binom{k}{j} \alpha^{k-j} R^j x = \begin{bmatrix} 0 \\ 0 \end{bmatrix}. \end{align*}$$

Hence, $x \in \ker \, (T-\alpha I)^k$ . More generally, if $x_1^{(1)}, x_1^{(2)}, \ldots , x_1^{(\ell )}$ is a linearly independent set of vectors in $\ker \, (AY - \alpha I)^k$ , then

$$\begin{align*}x^{(j)} := \begin{bmatrix} x_1^{(j)} \\ 0 \end{bmatrix}, \ \ \ 1 \le j \le \ell \end{align*}$$

is a linearly independent family of vectors in $\ker \, (T-\alpha I)^k$ .

$\bullet $ Step six.

We also observe that $\{ x^{(1)}, x^{(2)}, \ldots , x^{(\ell )}, y^{(1)}, y^{(2)}, \ldots , y^{(m)} \}$ is linearly independent. Indeed, if $\eta _i, \zeta _j \in \mathbb C$ , $1 \le i \le \ell , 1 \le j \le m$ and

$$\begin{align*}\sum_{i=1}^\ell \eta_i x^{(i)} + \sum_{j=1}^m \zeta_j y^{(j)} = 0, \end{align*}$$

then by considering the second coordinate of these sums, we see that $\zeta _j = 0$ , $1 \le j \le m$ , and hence by considering the remaining first coordinates, $\eta _i = 0$ , $1 \le i \le \ell $ .

By considering the case where $\{ x^{(1)}, x^{(2)}, \ldots , x^{(\ell )}\}$ is a basis for $\ker \, (AY -\alpha I)^k$ and $\{ y^{(1)}, y^{(2)}, \ldots , y^{(m)} \}$ is a basis for $\ker \, (YA + \alpha I)^k$ , we conclude that

$$\begin{align*}{\mathrm {nul}}\, (T-\alpha I)^k \ge {\mathrm {nul}}\, (AY - \alpha I)^k + {\mathrm {nul}}\, (YA + \alpha I)^k. \end{align*}$$

The reverse inequality follows easily from the form of $(T-\alpha I)^k$ exhibited in Step three.

$\bullet $ Step seven.

Consider the following equation from Horn and Johnson [Reference Horn and Johnson21, Theorem 1.3.20]:

$$\begin{align*}\begin{bmatrix} I & - A \\ 0 & I \end{bmatrix} \ \begin{bmatrix} A Y & 0 \\ Y & 0 \end{bmatrix} \ \begin{bmatrix} I & A \\ 0 & I \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ Y & YA \end{bmatrix}. \end{align*}$$

Since $\begin {bmatrix} I & - A \\ 0 & I \end {bmatrix}$ is invertible with inverse $\begin {bmatrix} I & A \\ 0 & I \end {bmatrix}$ , it routinely follows that

$$\begin{align*}\begin{bmatrix} A Y - \alpha I & 0 \\ Y & -\alpha I\end{bmatrix}^k \quad \text{and} \quad \begin{bmatrix} -\alpha I & 0 \\ Y & Y A - \alpha I \end{bmatrix}^k \end{align*}$$

are similar.

An argument similar to that employed in Step two shows that

$$\begin{align*}{\mathrm {nul}}\, \begin{bmatrix} A Y - \alpha I & 0 \\ Y & -\alpha I\end{bmatrix}^k = {\mathrm {nul}} (AY - \alpha I)^k, \end{align*}$$

and that

$$\begin{align*}{\mathrm {nul}}\, \begin{bmatrix} -\alpha I & 0 \\ Y & Y A - \alpha I \end{bmatrix}^k = {\mathrm {nul}} \, (Y A - \alpha I)^k. \end{align*}$$

It follows that

$$\begin{align*}{\mathrm {nul}}\, (AY - \alpha I)^k = {\mathrm {nul}}\, (YA -\alpha I)^k, \end{align*}$$

and therefore,

$$ \begin{align*} {\mathrm {nul}} \, (T + \alpha I)^k &= {\mathrm {nul}}(A Y + \alpha I)^k + {\mathrm {nul}} (Y A - \alpha I)^k \\ &= {\mathrm {nul}}(Y A + \alpha I)^k + {\mathrm {nul}} (A Y - \alpha I)^k \\ &= {\mathrm {nul}}(T - \alpha I)^k.\\[-34pt] \end{align*} $$

Proof Let $T = [M, N]$ , where $M^2 = 0 = N^2$ , and write $M = \begin {bmatrix} 0 & A \\ 0 & 0 \end {bmatrix}$ and $N =\begin {bmatrix} W & X \\ Y & Z \end {bmatrix}$ relative to $\ker \, M \oplus (\ker \, M)^\perp $ . Then $T =\begin {bmatrix} AY & AZ-WA\\ 0 & -YA \end {bmatrix}$ . We claim that

$$\begin{align*}\sigma_{ap}(T)\setminus\{0\}=(\sigma_{ap}(AY)\setminus\{0\})\cup (\sigma_{ap}(-YA)\setminus\{0\}). \end{align*}$$

$\bullet $ Step one.

Let $\lambda \in \sigma _{ap}(T)\setminus \{0\}$ . Then there exists a sequence $\left ( \begin {bmatrix} x_n\\ y_n \\ \end {bmatrix} \right )_{n=1}^{\infty }$ of unit vectors such that

$$\begin{align*}\lim_n \begin{bmatrix} AY-\lambda & AZ-WA\\ 0 & -YA-\lambda \end{bmatrix}\begin{bmatrix} x_n\\ y_n \\ \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ \end{bmatrix}.\end{align*}$$

If $\underset {n\in \mathbb {N}}{\inf }\|y_n\|>0$ , it is easy to see that $\lambda \in \sigma _{ap}(-YA)$ ; otherwise, $\underset {n\in \mathbb {N}}{\inf }\|y_n\|=0$ , and it is then easy to show that $\lambda \in \sigma _{ap}(AY)$ . Therefore,

$$\begin{align*}\sigma_{ap}(T)\setminus\{0\}\subset(\sigma_{ap}(AY)\setminus\{0\})\cup (\sigma_{ap}(-YA)\setminus\{0\}).\end{align*}$$

$\bullet $ Step two.

• • Let $0\neq \alpha \in \sigma _{ap}(AY)$ , and $(x_n)_n \subset \ker M$ be a sequence of unit vectors such that $\lim _n \|(AY-\alpha )x_n\| = 0$ . Then

  $$\begin{align*}\lim_n \begin{bmatrix} AY-\alpha & AZ-WA\\ 0 & -YA-\alpha \end{bmatrix}\begin{bmatrix} x_n\\ 0 \\ \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ \end{bmatrix},\end{align*}$$
  which in turn implies that $\alpha \in \sigma _{ap}(T)$ .

• • Let $0\neq \beta \in \sigma _{ap}(-YA)$ , and $\{y_n\}\subset (\ker M)^\perp $ be a sequence of unit vectors such that $\lim _n \|(-YA-\beta )y_n\| = 0$ . Let $x_n=-\frac {1}{\beta }WAy_n$ , $n\in \mathbb {N}$ . Define $z_n=\begin {bmatrix} x_n\\ y_n \\ \end {bmatrix}$ . Then $\|z_n\|\geq 1$ , $n\in \mathbb {N}$ . We shall require the operators $L := MN$ and $R := NM$ . Then

  $$\begin{align*}\lim_n (R+\beta)z_n= \lim_n \begin{bmatrix} \beta & WA\\ 0 & YA+\beta \end{bmatrix}\begin{bmatrix} x_n\\ y_n \\ \end{bmatrix} = \begin{bmatrix} 0\\ 0 \\ \end{bmatrix}.\end{align*}$$
  As $LR=RL=0$ ,
  $$\begin{align*}\lim_n \beta Lz_n = \lim_n L[(R+\beta)z_n] = \begin{bmatrix} 0\\ 0 \\ \end{bmatrix}.\end{align*}$$
  Since $\beta \neq 0$ , it follows $\lim _n Lz_n = 0$ . Hence,
  $$\begin{align*}\lim_n (T-\beta)z_n = \lim_n (L-R-\beta)z_n = \lim_n Lz_n-(R+\beta)z_n = 0,\end{align*}$$
  and thus, $\beta \in \sigma _{ap}(T)$ .

Therefore,

$$\begin{align*}\sigma_{ap}(T)\setminus\{0\}\supset(\sigma_{ap}(AY)\setminus\{0\})\cup (\sigma_{ap}(-YA)\setminus\{0\}),\end{align*}$$

and the claim that $\sigma _{ap}(T)\setminus \{0\}=(\sigma _{ap}(AY)\setminus \{0\})\cup (\sigma _{ap}(-YA)\setminus \{0\})$ holds.

By Proposition 2.15,

$$ \begin{align*} \sigma_{ap}(T)\setminus\{0\} & =(\sigma_{ap}(AY)\setminus\{0\})\cup (\sigma_{ap}(-YA)\setminus\{0\}) \\ &= (\sigma_{ap}(YA)\setminus\{0\})\cup (\sigma_{ap}(-AY)\setminus\{0\})\\ &= \sigma_{ap}(-T)\setminus\{0\}. \end{align*} $$

This verifies that $\sigma _{ap}(T)=\sigma _{ap}(-T)$ .

Proof Let $T = [M, N]$ , where $M^2 = 0 = N^2$ . Write $M = \begin {bmatrix} 0 & A \\ 0 & 0 \end {bmatrix}$ and $N =\begin {bmatrix} W & X \\ Y & Z \end {bmatrix}$ relative to $\ker \, M \oplus (\ker \, M)^\perp $ , and so $T =\begin {bmatrix} AY & AZ-WA\\ 0 & -YA \end {bmatrix}$ .

1. (a) Let $\Omega $ be a component of $\sigma (T)$ . Then $\Omega \cap \partial \sigma (T)\neq \varnothing $ . Since $\partial \sigma (T)\subset \sigma _{ap}(T)$ ([Reference Conway10, Chapter VII Proposition 6.7], it follows that $\Omega \cap \sigma _{ap}(T)\neq \varnothing $ . By Lemma 2.16, $\Omega \cap \sigma _{ap}(-T)\neq \varnothing $ . In particular, $\Omega \cap \sigma (-T)\neq \varnothing $ .

2. (b) The essential spectrum only makes sense if $\dim \, \mathcal H = \infty $ .

   In this case, the fact that $M^2 = 0$ implies that $\dim \, \ker \, M = \infty $ . If $\dim \, (\ker \, M)^\perp < \infty $ , then both M and (therefore) T have finite rank, whence $\sigma _e(T)=\sigma _e(-T)=\{0\}$ . As such, we may assume without loss of generality that $\dim \ker \, M=\dim (\ker \, M)^\perp =\dim \, \mathcal H = \infty $ .

   As per the comments preceding this Proposition, we think of $A, W, Y$ , and Z as elements of $\mathcal B ( \mathcal H)$ .

   Let $\pi $ be the quotient map $\pi $ from $\mathcal {B}(\mathcal H)$ onto the Calkin algebra $\mathcal {B}(\mathcal H)/\mathcal {K}(\mathcal H)$ . Then

   $$\begin{align*}t:=\pi\otimes \text{id}_{\mathbb{M}_2}(T)=\begin{bmatrix} \pi(A)\pi(Y) & \pi(AZ-WA)\\ 0 & -\pi(Y)\pi(A) \end{bmatrix}.\end{align*}$$

   Let $\rho $ be a unital faithful $C^*$ -representation of $C^*(\pi (A),\pi (Y),\pi (Z),\pi (W))$ to $\mathcal {B}(\mathcal L)$ , where $\mathcal L$ is a separable Hilbert space. Then

   $$\begin{align*}\tilde{T}:=(\rho\otimes \text{id}_{\mathbb{M}_2})(t)=\begin{bmatrix} (\rho\circ\pi)(A)\cdot(\rho\circ\pi)(Y) & (\rho\circ\pi)(AZ-WA)\\ 0 & -(\rho\circ\pi)(Y)\cdot(\rho\circ\pi)(A) \end{bmatrix}.\end{align*}$$

   Moreover, $\sigma _e(T)=\sigma (t)=\sigma (\tilde {T})$ and . The proof is completed by applying (a).

Proof Let us write the Jordan form of T as $B \oplus J$ , where B is invertible and J is nilpotent. By Proposition 2.2, and . Now, by Proposition 2.4 and the comment which follows it, $B \sim -B$ . Since J is nilpotent and acts on a finite-dimensional space, $J \sim - J$ , from which it easily follows that $T \sim -T$ .

Proof Let $J \in \mathcal B(\mathbb C^n)$ be nilpotent, and suppose that , say $J = [M, N]$ , where $M^2 = 0 = N^2$ . Let $L := MN$ and $R := NM$ . Clearly, $L R = 0 = RL$ , from which we deduce that $J^k = (L - R)^k = L^k + (-1)^k R^k$ for all $k \ge 1$ .

Since $J = [M, N]$ is nilpotent, by Theorem 3.3, there exists an orthonormal basis $\{ e_k\}_{k=1}^n$ for $\mathbb C^n$ relative to which the matrices $[M]$ and $[N]$ corresponding to M and N respectively are upper-triangular. In fact, since $M^2 = 0 = N^2$ , $[M]$ and $[N]$ are strictly upper-triangular. But the algebra $\mathfrak T_n^\circ \subseteq \mathbb M_n(\mathbb C)$ of strictly upper-triangular $n \times n$ complex matrices is a nil algebra, and clearly, $A_1 A_2 \cdots A_n = 0$ for all choices of $A_k \in \mathfrak T_n^\circ $ , $1 \le k \le n$ . It follows that $L^k = 0 = R^k$ , and thus $J^k = 0$ , if $k \ge \lfloor \frac {n+1}{2} \rfloor $ .

The second sentence in the statement of the Corollary is an immediate consequence of the first.

Proof First note that $J_2^{(2)} = J_2 \oplus J_2 \sim J_2 \oplus - J_2 \simeq [M, N]$ , where

$$\begin{align*}M = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}, \ \ \ \ \ N = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}. \end{align*}$$

Thus, if $\mu = 2 \kappa $ for some $\kappa \ge 1$ , then

$$\begin{align*}J_2^{(\mu)} = J_2^{(2 \kappa)} \sim (J_2 \oplus - J_2)^{(\kappa)} \simeq [M, N]^{(\kappa)} = [M^{(\kappa)}, N^{(\kappa)}]. \end{align*}$$

Since $M^{(\kappa )}$ and $N^{(\kappa )}$ are nilpotent of order two, we see that when $\mu $ is even.

Next, we will show that if $0 \le \kappa $ is an integer, then $J_2^{(2\kappa +1)}$ is not in . Otherwise, there exist square-zero operators $M,N \in \mathcal B(\mathbb C^{2(2\kappa +1)})$ such that

$$\begin{align*}J_2^{(2 \kappa +1)} \simeq J := \begin{bmatrix}0& I_{2\kappa +1} \\ 0& 0 \end{bmatrix} = [M, N].\end{align*}$$

As $M J = - J M$ and $N J = - J N$ , relative to the same decomposition of $\mathcal H$ , M, and N admit the following decompositions:

$$\begin{align*}M=\begin{bmatrix}M_0& M_1 \\ 0 & -M_0 \end{bmatrix},~~N=\begin{bmatrix}N_0& N_1 \\ 0 & -N_0 \end{bmatrix},\end{align*}$$

where $M_i,N_i\in \mathcal B(\mathbb C^{2\kappa +1})$ , $i=0,1$ .

Since $M^2 = 0$ , it follows that $M_0^2 = 0$ , and so we can find an invertible operator $S \in \mathcal B(\mathbb C^{2 \kappa + 1})$ and $0 \le s \le \kappa $ such that $S^{-1} M_0 S = \begin {bmatrix} 0 & I_s \\ 0 & 0 \end {bmatrix} \oplus 0^{(2 \kappa + 1 - 2 s)}$ . Observe that

$$\begin{align*}(S\oplus S)^{-1} J (S \oplus S) = J, \end{align*}$$

and thus, by replacing M and N by $(S\oplus S)^{-1} M (S \oplus S)$ and $(S \oplus S)^{-1} N (S \oplus S)$ , we may assume without loss of generality that (relative to a decomposition of $\mathbb C^{4 \kappa +2} = \mathcal H_1 \oplus \mathcal H_2 \oplus \mathcal H_3$ , where $\dim \, \mathcal H_1 = s = \dim \,\mathcal H_2$ and $\dim \, \mathcal H_3 = (2 \kappa +1) - 2 s$ )

$$\begin{align*}M_0 = \begin{bmatrix} 0 & I_s & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & I_s \\ 0 & 0 \end{bmatrix} \oplus 0^{(2 \kappa + 1 - 2 s)}. \end{align*}$$

Obviously, $J \not = 0$ implies that $s \ge 1$ . Furthermore, $N^2 = 0$ implies that $N_0^2 = 0$ , and the equations $M^2 = 0$ and $J = [M, N]$ imply that

• • $M_0 M_1=M_1 M_0$ ;

• • $M_0 N_0 = N_0 M_0$ ; and

• • $M_0N_1-M_1N_0-N_0M_1+N_1M_0=I_{2\kappa +1}.$

The first two equations imply that relative to the decomposition $\mathbb C^{4 \kappa + 2} = \mathcal H_1 \oplus \mathcal H_2 \oplus \mathcal H_3$ from above, $M_1$ and $N_0$ have the following forms:

$$\begin{align*}M_1=\begin{bmatrix} A_0& A_1& B\\ 0& A_0& 0\\ 0& C& D\\ \end{bmatrix},~~N_0=\begin{bmatrix} E_0& E_1 & F\\ 0& E_0& 0\\0 & G& K\\ \end{bmatrix}, \end{align*}$$

where $A_i,E_i\in \mathbb {M}_{s}(\mathbb C)$ , $i=0,1$ , and $D, K\in \mathcal B(\mathbb C^{2\kappa +1 - 2s})$ . As $N_0$ is square-zero, so is K.

A straightforward calculation based upon the third equation above shows that the compression of $I_{2 \kappa + 1}$ to $\mathcal H_3$ is nothing more than

$$\begin{align*}I_{2 \kappa + 1 - 2s} = - (D K + K D). \end{align*}$$

Now $K \in \mathcal B(\mathbb C^{2\kappa +1 - 2s })$ is a square-zero operator, and so K is similar to $J_2^{(t)} \oplus 0^{(2 \kappa + 1 - 2s -2t)}$ . Crucially, we note that $2\kappa +1 - 2s$ is an odd number, and therefore, $(2 \kappa + 1) - 2 s - 2 t \ge 1$ . Noting that $I_{2 \kappa + 1 - 2s}$ is invariant under conjugation by invertible elements, we may choose a basis for $\mathbb C^{2 \kappa + 1 - 2 s}$ with respect to which the last column and last row of the corresponding matrix for K are both zero. It follows that the $(2 \kappa + 1 - 2s, 2 \kappa + 1 - 2s)$ entry of $I_{2 \kappa + 1 - 2s} = -(D K + K D)$ is zero, which is clearly false.

This contradiction implies that $J_2^{(2\kappa +1)} \simeq J$ is not in .

Proof Step 1.

The uniqueness of the Jordan canonical form of J (up to permutation of the irreducible summands) allows us to find integers $1 \le d, \mu _1, \mu _2, \ldots , \mu _d$ and $\kappa _1 \ge \kappa _2 \ge \cdots \ge \kappa _d \ge 1$ such that

$$\begin{align*}J := J_{\kappa_1}^{(\mu_1)} \oplus J_{\kappa_2}^{(\mu_2)} \oplus \cdots \oplus J_{\kappa_d}^{(\mu_d)} \end{align*}$$

relative to the decomposition $\mathbb C^n = \mathcal H_1^{(\mu _1)} \oplus \mathcal H_2^{(\mu _2)} \oplus \cdots \oplus \mathcal H_d^{(\mu _d)}$ . Let $\mathcal H_{(i,j)}$ denote the $j^{th}$ copy of $\mathcal H_i \simeq \mathbb C^{\kappa _i}$ in $\mathcal H_i^{(\mu _i)}$ , $1 \le j \le \mu _i$ , and let $\{ e_{(i, j, s)}: 1 \le s \le \kappa _i\}$ denote the orthonormal basis of $\mathcal H_{(i,j)}$ that gives rise to the standard super-diagonal form for the $j^{th}$ copy of $J_{\kappa _i}$ in $J_{\kappa _i}^{(\mu _i)}$ . Also, let $P_{(i,j)}$ denote the orthogonal projection of $\mathbb C^n$ onto $\mathcal H_{(i,j)}$ , $1 \le j \le \mu _i$ , $1 \le i \le d$ .

We have that $J = [M, N]$ , that $M^2 = 0 = N^2$ , and thus $J M = - M J$ , $J N = - N J$ . Observe that

$$\begin{align*}P_{(i_2, j_2)} M P_{(i_1,j_1)} \in \mathcal B(\mathcal H_{(i_1, j_1)}, \mathcal H_{(i_2,j_2)}) \simeq \mathbb M_{\kappa_{i_2} \times \kappa_{i_1}}(\mathbb C). \end{align*}$$

A routine calculation ensuing from the fact that $J M = - M J$ yields

$$\begin{align*}J_{\kappa_{i_2}} (P_{(i_2, j_2)} M P_{(i_1, j_1)}) = - (P_{(i_2, j_2)} M P_{(i_1,j_1)}) J_{\kappa_{i_1}}. \end{align*}$$

Step 2.

Suppose that $1 \le b < d$ and that $\kappa _b - \kappa _{b+1} \ge 2$ . We define four spaces $\mathfrak M_\ell $ , $1 \le \ell \le 4$ .

• • We define $\mathfrak M_3 := \mathrm {span} \{ e_{(i, j, \kappa _i)} : 1 \le i \le b, 1 \le j \le \mu _i \}$ . In other words, we are choosing the last basis vector $e_{(i,j,\kappa _i)}$ from each copy $\mathcal H_{(i,j)}$ of the space upon which each $J_{\kappa _i}$ acts, up to and including all copies of $\mathcal H_{(b, j)}$ , $1 \le j \le \mu _b$ , but no copies occurring after that.

• • We define $\mathfrak M_2 := \mathrm {span} \{ e_{(i, j, \kappa _i - 1)} : 1 \le i \le b, 1 \le j \le \mu _i \}$ . In other words, we are choosing the second to last basis vector $e_{(i,j,\kappa _i -1)}$ from each copy $\mathcal H_{(i,j)}$ of the space upon which each $J_{\kappa _i}$ acts, up to and including all copies of $\mathcal H_{(b, j)}$ , $1 \le j \le \mu _b$ , but no copies occurring after that.

• • We define $\mathfrak M_1 := \mathrm {span} \{ e_{(i,j, s)} : 1 \le i \le b, 1 \le j \le \mu _i, 1 \le s \le \kappa _i -2 \}$ . This is the space

  $$\begin{align*}\left( \mathcal H_1^{(\mu_1)} \oplus \cdots \oplus \mathcal H_b^{(\mu_b)} \right) \ominus (\mathfrak M_2 \oplus \mathfrak M_3). \end{align*}$$

• • We define

  $$\begin{align*}\mathfrak M_4 := \mathbb C^n \ominus (\mathfrak M_1 \oplus \mathfrak M_2 \oplus \mathfrak M_3) = \mathcal H_{b+1}^{(\mu_{b+1})} \oplus \mathcal H_{b+2}^{(\mu_{b+2})} \oplus \cdots \oplus \mathcal H_d^{(\mu_d)}. \end{align*}$$

Step 3.

Write $M = [M_{i,j}]$ relative to the decomposition

$$\begin{align*}\mathbb C^n = \mathfrak M_1 \oplus \mathfrak M_2 \oplus \mathfrak M_3 \oplus \mathfrak M_4, \end{align*}$$

and note that with respect to this decomposition, J is of the form

$$\begin{align*}J = \begin{bmatrix} Q_{1,1} & Q_{1,2} & 0 & 0 \\ 0 & 0 & I_\mu & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & Q_{4,4} \end{bmatrix}, \end{align*}$$

where $\mu := \sum _{i=1}^b \mu _i$ . Our immediate goal is to show that the equation $J M = - M J$ implies that $M_{2,1} = M_{2, 4} = M_{3, 1} = M_{3, 2} = M_{3, 4} = 0$ , so that M is of the form

$$\begin{align*}\begin{bmatrix} M_{1, 1} & M_{1, 2} & M_{1, 3} & M_{1,4} \\ 0 & M_{2, 2} & M_{2,3} & 0 \\ 0 & 0 & M_{3, 3} & 0 \\ M_{4, 1} & M_{4, 2} & M_{4, 3} & M_{4, 4} \end{bmatrix}. \end{align*}$$

(In fact, we can do better than this. For example, it can be shown that $M_{3, 3} = - M_{2, 2}$ , and we can better specify other matrix entries of M, but for our purposes, the stated form of M will suffice.)

To that end, choose an arbitrary basis vector $e_{(i_1, j_1, s_1)}$ for $\mathfrak M_1$ ; that is, $1 \le i_1 \le b$ , $1 \le j_1 \le \mu _{i_1}$ and $1 \le s_1 \le \kappa _{i_1} - 2$ . Next, fix an arbitrary basis vector $e_{(i_2, j_2, \kappa _{i_2} - 1)}$ for $\mathfrak M_2$ , so that $1 \le i_2 \le b$ , $1 \le j_2 \le \mu _{i_2}$ . Note that $e_{(i_2, j_2, \kappa _{i_2})}$ is an arbitrary basis vector for $\mathfrak M_3$ .

Then

$$\begin{align*}\langle M e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa_{i_2}-1)} \rangle = \langle P_{(i_2, j_2)} M P_{(i_1, j_1)} e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa_{i_2}-1)} \rangle, \end{align*}$$

and

$$\begin{align*}\langle M e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa_{i_2})} \rangle = \langle P_{(i_2, j_2)} M P_{(i_1, j_1)} e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa_{i_2})} \rangle. \end{align*}$$

But $P_{(i_2, j_2)} M P_{(i_1, j_1)} \in \mathcal B(\mathcal H_{(i_1, j_1)}, \mathcal H_{(i_2, j_2)}) \simeq \mathbb M_{\kappa _{i_2} \times \kappa _{i_1}}(\mathbb C)$ , and so we may think of $P_{(i_2, j_2)} M P_{(i_1, j_1)}$ as playing the role of D in Lemma 3.10 and Remark 3.11 above. By that Remark,

• • if $\langle (P_{(i_2, j_2)} M P_{(i_1, j_1)}) e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa _{i_2})}\rangle \ne 0$ , then $\kappa _{i_2} \le \kappa _{i_1}$ and $s_1 = \kappa _{i_1}$ . The condition that $\kappa _{i_2} \le \kappa _{i_1}$ implies that $i_1 \le i_2 \le b$ , and so we may conclude that $M_{3, 4} = 0$ . The condition that $s_1 = \kappa _{i_1}$ then implies that $e_{(i_1, j_1, s_1)} = e_{(i_1, j_1, \kappa _{i_1})} \in \mathfrak M_3$ . In other words, $M_{3, 1} = M_{3, 2} = 0$ .

• • Next, suppose that $\langle P_{(i_2, j_2)} M P_{(i_1, j_1)} e_{(i_1, j_1, s_1)}, e_{(i_2, j_2, \kappa _{i_2}-1)} \rangle \ne 0$ . By Remark 3.11(ii), either $\kappa _{i_2} - 1 \le \kappa _{i_1}$ . The hypotheses that $i_2 \le b$ and $\kappa _b - \kappa _{b+1} \ge 2$ then imply that $i_1 \le i_2$ , whence $M_{2,4} = 0$ . The same remark also shows that $s_1 \in \{ \kappa _{i_1} - 1, \kappa _{i_1}\}$ , implying that $e_{(i_1, j_1, s_1)} \in \mathfrak M_2 \oplus \mathfrak M_3$ . In other words, $M_{2,1} = 0$ .

Hence, M has the desired form. Since $M^2 = 0$ , we conclude that $\begin {bmatrix} M_{2, 2} & M_{2, 3} \\ 0 & M_{3, 3} \end {bmatrix}$ is nilpotent of order two.

Since $J N = - N J$ , the same argument implies that

$$\begin{align*}N = \begin{bmatrix} N_{1,1} & N_{1, 2} & N_{1, 3} & N_{1, 4} \\ 0 & N_{2, 2} & N_{2, 3} & 0 \\ 0 & 0 & N_{3, 3} & 0 \\ N_{4, 1} & N_{4, 2} & N_{4, 3} & N_{4, 4} \end{bmatrix}, \end{align*}$$

and as before, since $N^2 = 0$ , we conclude that $\begin {bmatrix} N_{2, 2} & N_{2, 3} \\ 0 & N_{3, 3} \end {bmatrix}$ is nilpotent of order two.

Step 4.

Now $J = [M, N]$ . By comparing the compression of J and of $[M, N]$ to $\mathfrak M_2 \oplus \mathfrak M_3$ , we find that

By Proposition 3.6, $\mu $ is even. However,

$$\begin{align*}\mu = \dim\, \mathfrak M_3 = \dim\, \mathfrak M_2 = \sum_{i = 1}^b \mu_i. \end{align*}$$

In other words, $\mu $ counts exactly how many operators there are up to and including the $\mu _b^{th}$ copy of $J_{\kappa _b}$ .

Step 5.

There remains to prove that if $\kappa _d \ge 2$ , then $\sum _{i=1}^d \mu _i$ is even. The proof is a routine adaptation of the previous case (where $1 \le b < d$ ), noting that when $b = d$ , then $\mathfrak M_4 = \{ 0\}$ , in which case, relative to the decomposition $\mathcal H = \mathfrak M_1 \oplus \mathfrak M_2 \oplus \mathfrak M_3$ ,

$$\begin{align*}J= \begin{bmatrix} Q_{1, 1} & Q_{1, 2} & 0 \\ 0 & 0 & I_\mu \\ 0 & 0 & 0 \end{bmatrix}, \end{align*}$$

while

$$\begin{align*}M = \begin{bmatrix} M_{1, 1} & M_{1, 2} & M_{1, 3} \\ 0 & M_{2, 2} & M_{2, 3} \\ 0 & 0 & M_{3, 3} \end{bmatrix}, \ \ \ \ \ \ \ \ \ \ N = \begin{bmatrix} N_{1, 1} & N_{1, 2} & N_{1, 3} \\ 0 & N_{2, 2} & N_{2, 3} \\ 0 & 0 & N_{3, 3} \end{bmatrix}. \end{align*}$$

Arguing as before, this implies that $\begin {bmatrix} 0 & I_\mu \\ 0 & 0 \end {bmatrix}$ is a commutator of the two square-zero operators

$$\begin{align*}\begin{bmatrix} M_{2, 2} & M_{2, 3} \\ 0 & M_{3, 3} \end{bmatrix}, \ \ \ \ \text{ and } \ \ \ \ \ \begin{bmatrix} N_{2, 2} & N_{2, 3} \\ 0 & N_{3, 3} \end{bmatrix}. \end{align*}$$

By Proposition 3.6, this can happen only if $\mu $ is even. In this case, however, $\mu = \sum _{i=1}^d \mu _i$ .

The result now follows by applying Theorem 3.8, the reformulation of the Theorem of Borwein and Richmond.

Proof Set $T_1 := J_\kappa \oplus J_\kappa $ and $T_2 := J_{\kappa +1} \oplus J_\kappa $ .

In the case of $T_1$ , simply note that $J_\kappa \sim -J_\kappa $ , and thus $T_1 \sim J_\kappa \oplus -J_\kappa $ . Since is invariant under similarity, we then have that

In the case of $T_2$ , set $A = \begin {bmatrix} 0 \\ I_\kappa \end {bmatrix} \in \mathbb M_{\kappa +1, \kappa }(\mathbb C)$ and $Y = \begin {bmatrix} I_\kappa & 0 \end {bmatrix} \in \mathbb M_{\kappa , \kappa +1}(\mathbb C)$ . If we set

$$\begin{align*}M = \begin{bmatrix} 0 & A \\ 0 & 0 \end{bmatrix}, N = \begin{bmatrix} 0 & 0 \\ Y & 0 \end{bmatrix} \in \mathcal B(\mathbb C^{\kappa+1} \oplus \mathbb C^\kappa) \simeq \mathbb M_{2\kappa+1}(\mathbb C), \end{align*}$$

then

$$\begin{align*}[M, N] = AY \oplus - YA = J_{\kappa+1}^* \oplus - J_\kappa^* \sim J_{\kappa+1} \oplus J_\kappa = T_2.\end{align*}$$

Since , .

Proof Define

$$\begin{align*}Q_{0} := \oplus_{j=1}^{\lfloor \frac{d}{2} \rfloor} (J_{\kappa_{2j-1}} \oplus J_{\kappa_{2j}}). \end{align*}$$

By Proposition 3.13, each , from which it easily follows that . If d is even, then $Q = Q_0$ , and we are done. If d is odd, then $Q= Q_{0} \oplus J_1$ , which is clearly in , as $Q_{0}$ and $J_1$ are in .

Finally, because $J \sim Q$ , and is invariant under similarity.

Proof

1. (a) implies (b).

   This is the content of Theorem 3.12.

2. (b) implies (a).

   Choose integers $1 \le d$ and $\kappa _1 \ge \kappa _2 \ge \cdots \ge \kappa _d$ such that $J = \oplus _{i=1}^d J_{\kappa _i}$ . Suppose that J admits a square root. By Theorem 3.7, $\kappa _{2 j -1} - \kappa _{2 j} \le 1$ for all $1 \le j \le \lfloor \frac {d}{2} \rfloor $ , and if d is odd, then $\kappa _d = 1$ , implying that $J_{\kappa _d} = 0$ .

   By Corollary 3.14, .

Proof Since both and the set $\{ Z^2 : Z \in \mathcal B(\mathbb C^n \}$ are invariant under similarity, we see from Remark 1.2 that without loss of generality, we may assume that $T = B \oplus Q$ , where B is invertible and Q is nilpotent. By Proposition 2.2, the hypothesis that implies that and .

Any invertible operator $B \in \mathcal B(\mathcal H)$ (where $\dim \, \mathcal H < \infty $ ) admits a square root X; one can either appeal to [Reference Borwein and Richmond7, Theorem 1], or one may simply apply the functional calculus to B using a branch of the square-root function $f(z) = z^{\frac {1}{2}}$ that avoids the finitely many nonzero eigenvalues of B. Theorem 3.15 above shows that if is nilpotent and acts on a finite-dimensional space, then $Q = Y^2$ for some nilpotent operator Y.

Thus, $T = B \oplus Q = X^2 \oplus Y^2 = (X \oplus Y)^2$ .

Proof

1. (a) implies (b). Suppose that . By Remark 1.2, we may assume without loss of generality that $T = B \oplus Q$ , where B is invertible and Q is nilpotent. By Proposition 2.2, and . (One of B and Q may be absent. We deal with the case where both are present.) If B is present, then by Proposition 2.4, B acts on a space of even dimension, say $2k$ , and there exists $A \in \mathcal B(\mathbb C^k)$ such that $B \sim A \oplus -A$ . By Theorem 3.12, $Q = L^2$ for some nilpotent operator $L \in \mathcal B(\mathbb C^{n-2k})$ . From this, the result immediately follows.

2. (b) implies (a). If $T \sim (A \oplus -A) \oplus L^2$ , where A is invertible and L is nilpotent, then setting $B := (A \oplus -A)$ and $Q = L^2$ , we see from Proposition 2.4 and Theorem 3.12 that B and Q lie in , whence .

Proof Suppose that $T = [M, N]$ , where $M^2 = 0 = N^2$ . By Theorem 3.17, there exist an invertible operator A and a nilpotent operator L such that

$$\begin{align*}T \sim (A \oplus -A) \oplus L^2. \end{align*}$$

It follows directly from this that . That is, .

Since is closed [Reference Marcoux, Radjavi and Zhang28, Proposition 3.3],

Conversely, suppose that , and let $ {\varepsilon }> 0$ . The proof of [Reference Marcoux, Radjavi and Zhang28, Theorem 3.4] shows we can find an operator $X \in \mathcal B ( \mathcal H)$ such that $ {\Vert } X - T {\Vert } < {\varepsilon }$ , and X is similar to a diagonal operator such that for all $1 \le k \le \lfloor \frac {n}{2}\rfloor $ , $\beta _{2k} = -\beta _{2k-1}$ , $\beta _n = 0$ if n is odd, and for $1 \le i, j \le n$ , $\beta _i = \beta _j$ if and only if $i = j$ . In particular,

where $m = \begin {cases} 0 & \text { if } n \text { is even } \\ 1 & \text { if } n \text { is odd.}\end {cases}$

Note that for each $\alpha \in \mathbb C$ , from which we conclude that . Since is closed under similarity, , whence . That is,

This completes the proof of the Proposition.

Proof This follows immediately from [Reference Marcoux, Radjavi and Zhang28, Corollary 3.5].

Proof Suppose to the contrary that there exists a connected component $\Omega \subseteq \sigma (a)$ satisfying

$$\begin{align*}\Omega \cap -\sigma(a) = \Omega \cap \sigma(-a) = \varnothing. \end{align*}$$

Note that $\Omega $ and $\sigma (-a)\cup (\sigma (a)\setminus \Omega )$ are disjoint, compact sets of $\mathbb C$ . Thus, we can find disjoint open sets $G_1, G_2$ of $\mathbb C$ such that $\Omega \subseteq G_1$ , $\sigma (-a)\cup (\sigma (a)\setminus \Omega ) \subseteq G_2$ and $\text {d}(G_1,G_2)=\delta>0$ .

By applying the upper semi-continuity of the spectrum and Newburgh’s Theorem (c.f. [Reference Herrero19, Theorem 1.1]), we find that there exists $\delta> 0$ such that $ {\Vert } y - a {\Vert } < \delta $ implies that

$$\begin{align*}\sigma(y)\subset G_1\cup G_2, \sigma(-y) \subseteq G_2, \text{and}~\sigma(y) \cap G_1 \ne \varnothing.\end{align*}$$

In particular, for $N \in \mathbb N$ sufficiently large, we have that

$$\begin{align*}\sigma(b_N)\subset G_1\cup G_2, \sigma(-b_N) \subseteq G_2, \text{and}~\sigma(b_N) \cap G_1 \ne \varnothing.\end{align*}$$

Let $\beta \in \sigma (b_N) \cap G_1 \ne \varnothing $ . Clearly, the connected component $\Gamma _\beta $ of $\beta $ in $\sigma (b_N)$ also lies in $G_1$ . But then

$$\begin{align*}\Gamma_\beta \cap \sigma(-b_N) \subseteq G_1 \cap G_2 = \varnothing, \end{align*}$$

contradicting our hypothesis. This completes the proof.

Proof Without loss of generality, we may assume that $\alpha \neq 0$ and that

$$\begin{align*}\min\{ \dim\, \mathcal H(\{ \alpha\}; T), \dim\, \mathcal H(\{ -\alpha\}; T)\}<+\infty.\end{align*}$$

We will consider the following two subcases.

1. (a) If $\dim \, \mathcal H(\{ \alpha \}; T)=+\infty $ , then $\alpha \in \sigma _e(T)$ , but $-\alpha \notin \sigma _e(T)$ and hence $\alpha \notin \sigma _e(-T)$ . Then $\{\alpha \}$ , which is a component of $\sigma _e(T)$ , has no intersection with $\sigma _e(-T)$ . This, however, contradicts Proposition 4.8 (b). A similar argument may be applied if $\dim \, \mathcal H(\{-\alpha \}; T) = +\infty $ .

2. (b) Next, suppose that

   $$\begin{align*}\max\{ \dim\, \mathcal H(\{ \alpha\}; T), \dim\, \mathcal H(\{ -\alpha\}; T)\}<+\infty.\end{align*}$$

   Choose $r{\kern-1pt}\in{\kern-1pt} (0, \frac {|\alpha |}{3})$ , set $G_1 {\kern-1pt}={\kern-1pt} B(\alpha ,r) {\kern-1pt}:={\kern-1pt} \{ \lambda \in \mathbb C: |\lambda - \alpha | < r\}$ , and $G_2{\kern-1pt}={\kern-1pt} B(-\alpha ,r)$ . Then $G_1,G_2$ are disjoint open sets. By [Reference Herrero19, Corollary 1.6], there exists $\delta>0$ such that $\|Y-T\|<\delta $ implies that

   • • $\sigma (Y) \cap G_k \ne \varnothing $ , $k = 1, 2$ ,

   • • $\dim \, \mathcal H(\sigma (Y) \cap G_1; Y)= \dim \, \mathcal H(\{ \alpha \}; T)<\infty $ , and

   • • $\dim \, \mathcal H(\sigma (Y) \cap G_2; Y) =\dim \,\mathcal H(\{ - \alpha \}; T)<\infty $ .

   Since , we can choose , such that $\|X-T\|<\delta $ . By Theorem 2.13,

   $$\begin{align*}\dim\, \mathcal H(\sigma(X) \cap G_1; X)=\dim\, \mathcal H(\sigma(X) \cap G_2; X).\end{align*}$$
   Therefore,
   $$\begin{align*}\dim\, \mathcal H(\{ \alpha\}; T) = \dim\, \mathcal H(\{ -\alpha\}; T). \end{align*}$$

Proof By [Reference Marcoux, Radjavi and Zhang27, Theorem 4.11] (a), (b), and (c) are equivalent. By Proposition 4.8 and Proposition 4.9, (d) implies (c).

By examining the proof of (c) implying (a) of [Reference Marcoux, Radjavi and Zhang27, Theorem 3.5] and also [Reference Marcoux, Radjavi and Zhang27, Theorem 4.11], we find that to show that (c) implies (d), it suffices to note that balanced operators acting on a finite-dimensional Hilbert space are in (Corollary 4.3)) and to show that . The last fact is easy to check by factoring $X\oplus -X$ as

$$\begin{align*}\begin{bmatrix}X&\\& -X \end{bmatrix}=\begin{bmatrix} 0& X\\ & 0 \end{bmatrix}\begin{bmatrix} 0& \\I & 0 \end{bmatrix}-\begin{bmatrix} 0& \\I & 0 \end{bmatrix}\begin{bmatrix} 0& X\\ & 0 \end{bmatrix}.\\[-34pt] \end{align*}$$

Proof This follows from [Reference Marcoux, Radjavi and Zhang27, Theorem 2.13] and Theorem 4.10 above.

Proof Since every compact operator is biquasitriangular, this is an immediate consequence of the equivalence of (c) and (d) in Theorem 4.10.

Proof

1. (a) if and only if (b). This follows immediately from Theorem 4.10 above.

2. (b) implies (c). This is Lemma 4.13.

3. (c) implies (d). This follows from the definitions of both sets.

4. (d) implies (e). Let be a compact operator. We shall show that given $ {\varepsilon }> 0$ , there exists a finite-dimensional subspace $\mathcal M \subseteq \mathcal H$ and such that $ {\Vert } K - (F \oplus 0) {\Vert } < \frac { {\varepsilon }}{2}$ . If we can do this, then by applying Theorem 4.2, we can find such that $ {\Vert } B - F {\Vert }< \frac { {\varepsilon }}{2}$ , whence , and $ {\Vert } K - (B \oplus 0) {\Vert } < {\varepsilon }$ . This implies that , as required.

   The spectrum of any compact operator in $\mathcal B ( \mathcal H)$ is either finite or a sequence converging to $0$ . Suppose that . It easily follows that we must have $\sigma (K) = \{ \beta _i, -\beta _i\}_{i=1}^\eta \cup \{ 0\}$ , where $\eta \in \mathbb N \cup \{ 0, \infty \}$ , $\beta _i \ne 0$ if $1 \le i < \eta +1$ , and $\{ \beta _i, -\beta _i\} \cap \{ \beta _j, -\beta _j\} = \varnothing $ if $1 \le i \ne j < \eta +1$ . Note that if $\eta = 0$ , then this should be interpreted as saying that $\sigma (K) = \{ 0\}$ . Moreover, if $\eta = \infty $ , then $\lim _i \beta _i = 0$ , and in all cases, $\mu (\beta _i) = \mu (-\beta _i)$ for all $1 \le i < \eta + 1$ .

   Let $ {\varepsilon }> 0$ . Fix $0 \le p < \eta + 1$ such that $| \beta _i| \ge \frac { {\varepsilon }}{4}$ if $1 \le i \le p$ and $| \beta _i | < \frac { {\varepsilon }}{4}$ if $p < i < \eta + 1$ .

   By considering the Riesz subspaces $\mathcal H_i$ corresponding to the clopen sets $\{ \beta _i, -\beta _i\}$ , $1 \le i \le p$ , we may decompose $\mathcal H = \mathcal H_1 \oplus \mathcal H_2 \oplus \cdots \oplus \mathcal H_p \oplus \mathcal H_0$ such that with respect to this decomposition,

   $$\begin{align*}K = \begin{bmatrix} K_{11} & K_{12} & K_{13} & \cdots & K_{1p} & K_{1 0} \\ & K_{22} & K_{23} & \cdots & K_{2p} & K_{2 0} \\ & & \ddots & & & \vdots \\ & & & \ddots & \vdots & \vdots \\ & & & & K_{p p} & K_{p 0} \\ & & & & & K_{0 0} \end{bmatrix}, \end{align*}$$

   where $\sigma (K_{ii}) = \{ \beta _i, -\beta _i\}$ , $1 \le i \le p$ and $\sigma (K_{00}) = \sigma (K) \setminus \{ \beta _j, -\beta _j\}_{j=1}^p \subseteq \mathbb D_{ {\varepsilon }/4} := \{ z \in \mathbb C: |z| < \frac { {\varepsilon }}{4} \}$ .

   Let P denote the orthogonal projection of $\mathcal H$ onto $\oplus _{j=1}^p \mathcal H_j$ , and observe that $\mathrm {rank}\, \, P < \infty $ . Since K and $K_{00}$ are compact, we can find a finite-rank orthogonal projection $Q = (I-P)Q(I-P)$ such that

   • • $ {\Vert } (P+Q) K (P+Q) - K {\Vert } < \frac { {\varepsilon }}{4}$ , and

   • • by furthermore appealing to the upper-semicontinuity of the spectrum,

     $$\begin{align*}\sigma(Q K_{00} Q) \subseteq \mathbb D_{ {\varepsilon}/4}. \end{align*}$$

   (To achieve the second step, choose $\delta> 0$ such that $ {\Vert } X - K_{00} {\Vert } < \delta $ implies that $\sigma (X) \subseteq \mathbb D_{ {\varepsilon }/4}$ , and then choose Q sufficiently “large” such that $ {\Vert } Q K_{00} Q - K_{00} {\Vert } < \delta $ .)

   Write $\mathcal J_0 = {\mathrm {ran}}\, Q$ and let $\mathcal J_1 := \mathcal H_0 \ominus \mathcal J_0$ . Relative to the decomposition $\mathcal H = \oplus _{j=1}^p \mathcal H_j \oplus \mathcal J_0 \oplus \mathcal J_1$ , we have

   $$\begin{align*}(P+Q) K (P + Q) = \begin{bmatrix} K_{11} & K_{12} & K_{13} & \cdots & K_{1p} & L_{1 0} & 0 \\ & K_{22} & K_{23} & \cdots & K_{2p} & L_{2 0} & 0 \\ & & \ddots & & & \vdots & \vdots \\ & & & \ddots & \vdots & \vdots & \vdots \\ & & & & K_{p p} & L_{p 0} & 0 \\ & & & & & L_{0 0} & 0 \\ & & & & & 0 & 0 \end{bmatrix}, \end{align*}$$
   where $\sigma (L_{00} \oplus 0) = \sigma (Q K_{00} Q) \subseteq \mathbb D_{ {\varepsilon }/4}$ . Since $\dim \, \mathcal J_0 = \mathrm {rank}\, \, Q < \infty $ , we can upper-triangularise $L_{00}$ with respect to some orthonormal basis for $\mathcal J_0$ . Let D denote the diagonal of $L_{00}$ , and observe that $ {\Vert } D {\Vert } < \frac { {\varepsilon }}{4}$ , since the norm of a diagonal matrix is equal to its spectral radius. Set $L := L_{00} - D$ , so that $\sigma (L) = \{ 0 \}$ , and set
   $$\begin{align*}F := \begin{bmatrix} K_{11} & K_{12} & K_{13} & \cdots & K_{1p} & L_{1 0} \\ & K_{22} & K_{23} & \cdots & K_{2p} & L_{2 0} \\ & & \ddots & & & \vdots \\ & & & \ddots & \vdots & \vdots \\ & & & & K_{p p} & L_{p 0} \\ & & & & & L \end{bmatrix} \in \mathcal B(\mathcal M), \end{align*}$$
   where $\mathcal M = \oplus _{j=1}^p \mathcal H_j \oplus \mathcal J_0$ . Then $\sigma (F) = \cup _{j=1}^p \sigma (K_{jj}) \cup \sigma (L)$ , noting that the eigenvalues of $K_{jj}$ appear with the same algebraic multiplicity in $\sigma (F)$ . In other words, F is balanced. Finally,
   $$ \begin{align*} \ \ \ \ \ {\Vert} K - (F\oplus 0) {\Vert} &\le {\Vert} K - (P+Q) K (P+Q) {\Vert} + {\Vert} (P+Q) K (P+Q) - (F \oplus 0) {\Vert} \\ &< \frac{ {\varepsilon}}{4} + {\Vert} L - L_{00} {\Vert} \\ &< \frac{ {\varepsilon}}{4} + \frac{ {\varepsilon}}{4} \\ &= \frac{ {\varepsilon}}{2}. \end{align*} $$
   As seen above, this completes the proof.

5. (e) implies (a). This is trivial.

Proof Since the set of Fredholm operators of index is open in $\mathcal B ( \mathcal H)$ , we can choose $\delta> 0$ such that if $C \in \mathcal B ( \mathcal H)$ and $ {\Vert } C - T {\Vert } < \delta $ , then .

Suppose that there exist and $B \in \mathcal B ( \mathcal H)$ such that $ {\Vert } T - (M B - BM) {\Vert } < \delta $ . Set $T_1 : = M B - B M$ , implying that .

Writing $M = \begin {bmatrix} 0 & A \\ 0 & 0 \end {bmatrix}$ and $B =\begin {bmatrix} W & X \\ Y & Z \end {bmatrix}$ relative to $\ker \, M \oplus (\ker \, M)^\perp $ yields $T_1 =\begin {bmatrix} AY & AZ-WA\\ 0 & -YA \end {bmatrix}$ . Since $T_1$ is Fredholm, $\dim \ker \, M =\dim (\ker \, M)^\perp =\infty $ , and $\pi (AY)$ is left invertible in the Calkin algebra and $\pi (-YA)$ is right invertible. Hence, $\pi (A)$ and $\pi (Y)$ are left and right invertible in the Calkin algebra; that is, $\pi (A)$ and $\pi (Y)$ are invertible in the Calkin algebra. Therefore, A and Y are both Fredholm operators. By [Reference Han, Lee and Lee18, Corollary 5],

$$\begin{align*}\textrm{ind}(T_1)=\textrm{ind}(AY)+\textrm{ind}(-YA)=2[\textrm{ind}(A)+\textrm{ind}(Y)]\in 2\mathbb{Z},\end{align*}$$

a contradiction.

The second statement is an immediate consequence of the first.

Proof Without loss of generality, we may assume that $w_n>0$ , $n\geq 0$ . Moreover, $W= S D$ , where S is the unilateral shift, and D is the diagonal operator $\text {diag} \{ w_n\}_n$ . Since $ \inf _n |w_n|>0$ , D is invertible. As S is a Fredholm operator with index $-1$ , so is W. By Proposition 4.17,

Proof Let $K \in \mathcal K ( \mathcal H)$ and $ {\varepsilon }> 0$ . Choose a finite-rank operator F such that $ {\Vert } K - F {\Vert } < \frac { {\varepsilon }}{2}$ , and write

$$\begin{align*}F= \begin{bmatrix} F_0 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}$$

relative to $\mathcal H = \mathcal H_1 \oplus \mathcal H_2$ , where $\dim \, \mathcal H_1 < \infty $ .

Let , and choose $m> 2$ such that $\frac {|\alpha |}{m} < \frac { {\varepsilon }}{2}$ . Write $\mathcal H_2 := \mathcal H_{3} \oplus \mathcal H_4$ , where $\dim \, \mathcal H_3 = m$ , and – relative to the decomposition $\mathcal H = \mathcal H_1 \oplus \mathcal H_3 \oplus \mathcal H_4$ – define

$$\begin{align*}G := \begin{bmatrix} F_0 & 0 & 0 \\ 0 & \frac{-\alpha}{m} I_m & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*}$$

Then $ {\Vert } G- F {\Vert } < \frac { {\varepsilon }}{2}$ , and therefore,

$$\begin{align*}{\Vert} K - G {\Vert} \le {\Vert} K - F {\Vert} + {\Vert} F - G {\Vert} < {\varepsilon}. \end{align*}$$

Since $F_0 \oplus \frac {-\alpha }{m} I_m \in \mathcal B(\mathcal H_1 \oplus \mathcal H_3)$ has trace zero, it follows from Smith’s Theorem 3.18 that $F_0 \oplus \frac {-\alpha }{m} I_m$ is a commutator of nilpotent matrices in $\mathcal B(\mathcal H_1 \oplus \mathcal H_3)$ , and thus, G is a commutator of finite-rank nilpotent operators. But then K is a limit of such operators.