2 Preliminaries

Let $\mathbb {R}$ and $\mathbb {C}$ denote the fields of real and complex numbers, respectively, and let $\mathbb {K}$ represent either $\mathbb {R}$ or $\mathbb {C}$ . The corresponding sets of dual numbers and dual complex numbers are denoted by $\widehat {\mathbb {R}}$ and $\widehat {\mathbb {C}}$ , respectively. We write $\widehat {\mathbb {K}}$ to denote either of these two sets. An element of $\widehat {\mathbb {K}}$ will be called a dual element.

A dual element is an expression of the form $a = a_{s} + a_{d}\varepsilon $ , where $a_s,a_d \in \mathbb {K}$ , and $\varepsilon $ is an infinitesimal unit satisfying $\varepsilon ^2 = 0$ and commuting with all real or complex scalars. The quantities $a_s$ and $a_d$ are called the standard part and dual part of $a$ , respectively. The conjugate of $a$ is defined by $a^* = a_s^* + a_d^* \varepsilon $ , where $a_s^*$ and $a_d^*$ denote the ordinary conjugates of $a_s$ and $a_d$ . A dual element is said to be appreciable if $a_s \neq 0$ , and infinitesimal otherwise. Moreover, a dual element is said to be invertible if there exists a dual element $b$ such that $ab=ba=1$ . Equivalently, a dual element is invertible if and only if it is appreciable (see [Reference Qi and Cui21]).

A dual matrix $A = A_s + \varepsilon A_d \in \widehat {\mathbb {K}}^{n\times n}$ is said to be Hermitian if $A=A^{*}$ . A dual vector $x=x_{s}+x_{d}\varepsilon $ , where $x_{s},~x_d \in \mathbb {K}^{n}$ , is called a unit dual vector if $\|x_s\|_{2}=1$ and $x_{s}^{*}x_{d}+x_d^{*}x_s=0$ , where $\|\cdot \|_2$ denotes the standard Euclidean norm.

Let $A = A_s + \varepsilon A_d \in \widehat {\mathbb {K}}^{n \times n}$ be a dual Hermitian matrix. Suppose that $A v_i = \lambda _i v_i$ , where $\lambda _i = \lambda _{i,s} + \varepsilon \lambda _{i,d}$ is the $i$ -th eigenvalue of $A$ , and $v_i = u_i + \varepsilon w_i$ is a corresponding unit dual eigenvector, for $i=1,\dots ,n$ . Then, $\|u_i\|_2 = 1$ and $u_i^{*} w_i + w_i^{*} u_i = 0$ , for $i=1,\dots ,n$ . Throughout the manuscript, $\operatorname {Re}(z)$ denotes the real part of a complex number $z\in \mathbb {C}$ .

Let $p_A:\mathbb {C}\rightarrow \mathbb {C}$ denote the characteristic polynomial of $A$ . Then

$$ \begin{align*} p_{A}(\lambda) = \det(\lambda I_{n} - A) = \prod_{k=1}^{n} (\lambda - \lambda_k), \end{align*} $$

where $I_n$ (or $I$ ) represents the $n \times n$ identity matrix.

Evaluating the derivative of $p_A$ at $\lambda =\lambda _i$ gives

$$ \begin{align*}P_i:= p^{\prime}_{A}(\lambda_{i}) = \prod\limits_{k \ne i} (\lambda_{i} - \lambda_{k}). \end{align*} $$

For $j=1,\ldots ,n$ , let $M_j$ denote the $(n-1)\times (n-1)$ principal submatrix of $A$ obtained by deleting the $j$ -th row and column. Its characteristic polynomial is

$$ \begin{align*} p_{M_j}(\lambda) = \det(\lambda I_{n-1}-M_j) = \prod_{k=1}^{n-1} \big(\lambda-\lambda_k(M_j)\big). \end{align*} $$

The following result of Qi and Cui [Reference Qi and Cui21] provides the necessary framework for the subsequent results. Theorem 4.1 of [Reference Qi and Cui21] states that for a given dual Hermitian matrix $A = A_s + \varepsilon A_d \in \widehat {\mathbb {K}}^{n \times n}$ , let $\lambda _{i,s}$ be a simple eigenvalue of $A_s$ , and $u_i$ be the corresponding unit eigenvector of $\lambda _{i,s}$ for $i=1,\dots ,n$ . Then, $\lambda _i = \lambda _{i,s} + \varepsilon \lambda _{i,d}$ is a simple eigenvalue of $A$ , where $\lambda _{i,d} = u_i^{*} A_d u_i$ , with the corresponding eigenvector $v_i = u_i + \varepsilon w_i$ , where $w_i$ satisfies

$$ \begin{align*}(A_s - \lambda_{i,s} I) w_i = (\lambda_{i,d} I - A_d) u_i, \end{align*} $$

for $i=1,\dots ,n$ .

3 Main results

The following proposition establishes the dual eigenvalue–eigenvector identity for dual Hermitian matrices with simple standard eigenvalues. Its proof adapts the adjugate-matrix argument of Denton et al. [Reference Denton, Parke, Tao and Zhang8] to the dual setting.

Proposition 3.1 Let $A=A_s+\varepsilon A_d\in \widehat {\mathbb K}^{\,n\times n}$ , where $\mathbb K=\mathbb R$ or $\mathbb C$ , be a dual Hermitian matrix. Assume that the eigenvalues $\lambda _{1,s},\ldots ,\lambda _{n,s}$ of $A_s$ are simple. Let

$$\begin{align*}Av_i=\lambda_i v_i,\quad \lambda_i=\lambda_{i,s}+\varepsilon \lambda_{i,d}, \end{align*}$$

where $v_i$ is a unit dual eigenvector corresponding to $\lambda _i$ . Let $M_j$ be the principal submatrix of $A$ obtained by deleting the $j$ -th row and column. Then, for each $i, j=1,\ldots ,n$ ,

(3.1) $$ \begin{align} p_{M_j}(\lambda_i)=|v_{i,j}|^2\,p_A'(\lambda_i), \end{align} $$

where $v_{i,j}$ is the $j$ -th component of the unit dual eigenvector $v_i$ associated with $\lambda _i$ .

Proof By Theorem 4.1 of Qi and Cui [Reference Qi and Cui21], for each simple eigenvalue $\lambda _{i,s}$ of $A_s$ , there exists a dual eigenvalue

$$\begin{align*}\lambda_i=\lambda_{i,s}+\varepsilon \lambda_{i,d} \end{align*}$$

and an associated appreciable dual eigenvector $v_i=u_i+\varepsilon w_i$ . The eigenvectors may be normalized so that $v_i^*v_i=1$ .

We first prove that $v_1,\ldots ,v_n$ are mutually orthogonal. Since $\lambda _k=\lambda _{k,s}+\varepsilon \lambda _{k,d}$ , we have

$$\begin{align*}\lambda_k^*=\lambda_{k,s}^*+\varepsilon \lambda_{k,d}^*. \end{align*}$$

Because $A_s$ is Hermitian, $\lambda _{k,s}\in \mathbb R$ . Moreover,

$$\begin{align*}\lambda_{k,d}=u_k^*A_d u_k. \end{align*}$$

Since $A_d$ is Hermitian, $\lambda _{k,d}\in \mathbb R$ . Hence,

$$\begin{align*}\lambda_k^*=\lambda_k. \end{align*}$$

Since $A$ is dual Hermitian, $A^*=A$ . For $i\ne k$ , we have

$$\begin{align*}Av_i=\lambda_i v_i,\qquad Av_k=\lambda_k v_k. \end{align*}$$

Thus,

$$\begin{align*}v_k^*Av_i=\lambda_i v_k^*v_i. \end{align*}$$

On the other hand,

$$\begin{align*}v_k^*Av_i=(Av_k)^*v_i=(\lambda_k v_k)^*v_i =\lambda_k^*v_k^*v_i=\lambda_k v_k^*v_i. \end{align*}$$

It follows that

$$\begin{align*}(\lambda_i-\lambda_k)v_k^*v_i=0. \end{align*}$$

Since $\lambda _{i,s}\ne \lambda _{k,s}$ , the dual number

$$\begin{align*}\lambda_i-\lambda_k = (\lambda_{i,s}-\lambda_{k,s}) +\varepsilon(\lambda_{i,d}-\lambda_{k,d}) \end{align*}$$

has nonzero standard part and is therefore invertible. Hence,

$$\begin{align*}v_k^*v_i=0 \qquad (i\ne k). \end{align*}$$

Together with $v_i^*v_i=1$ , this gives

$$\begin{align*}V^*V=I, \qquad V=[v_1,\ldots,v_n]. \end{align*}$$

Thus, $V$ is dual unitary. Indeed, $V^*V=I$ implies

$$\begin{align*}\det(V)^*\det(V)=1, \end{align*}$$

so $\det (V)$ is invertible in the dual-number algebra. Hence, $V$ is invertible and $V^{-1}=V^*$ .

Since $Av_i=\lambda _i v_i$ for all $i$ , we have

$$\begin{align*}AV=V\Lambda, \qquad \Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). \end{align*}$$

Multiplying on the right by $V^*$ , we obtain

$$\begin{align*}A=V\Lambda V^*. \end{align*}$$

Now let $\lambda $ be a dual number. Then

$$\begin{align*}\lambda I-A=V(\lambda I-\Lambda)V^*. \end{align*}$$

The determinant and adjugate are well defined over the commutative algebra of dual numbers. Since the entries belong to a commutative algebra, the standard polynomial identity

$$\begin{align*}\operatorname{adj}(PXP^{-1}) = P\operatorname{adj}(X)P^{-1} \end{align*}$$

holds for every invertible matrix $P$ . This follows, for instance, from

$$\begin{align*}\operatorname{adj}(AB)=\operatorname{adj}(B)\operatorname{adj}(A), \end{align*}$$

which is a polynomial identity valid over every commutative ring. Therefore,

$$\begin{align*}\operatorname{adj}(\lambda I-A) = V\operatorname{adj}(\lambda I-\Lambda)V^*. \end{align*}$$

Since $\lambda I-\Lambda $ is diagonal, we have

$$\begin{align*}\operatorname{adj}(\lambda I-\Lambda) = \operatorname{diag}\left( \prod_{m\ne 1}(\lambda-\lambda_m), \ldots, \prod_{m\ne n}(\lambda-\lambda_m) \right). \end{align*}$$

Putting $\lambda =\lambda _i$ , all diagonal entries of $\operatorname {adj}(\lambda _i I-\Lambda )$ vanish except the $i$ -th one. Hence,

$$\begin{align*}\operatorname{adj}(\lambda_i I-A) = \left(\prod_{m\ne i}(\lambda_i-\lambda_m)\right)v_i v_i^*. \end{align*}$$

Taking the $(j,j)$ -entry of both sides gives

$$\begin{align*}\det(\lambda_i I_{n-1}-M_j) = \left(\prod_{m\ne i}(\lambda_i-\lambda_m)\right)v_{i,j}v_{i,j}^*. \end{align*}$$

Since $v_{i,j}v_{i,j}^*=|v_{i,j}|^2$ , we get

$$\begin{align*}p_{M_j}(\lambda_i) = |v_{i,j}|^2\prod_{m\ne i}(\lambda_i-\lambda_m). \end{align*}$$

It remains to identify the product with $p_A'(\lambda _i)$ . Since $A=V\Lambda V^*$ and $V^*=V^{-1}$ , we have

$$\begin{align*}\begin{aligned} p_A(\lambda) &=\det(\lambda I-A)=\det\big(V(\lambda I-\Lambda)V^*\big) \\ &=\det(V)\det(\lambda I-\Lambda)\det(V^*)=\det(\lambda I-\Lambda)=\prod_{m=1}^n(\lambda-\lambda_m). \end{aligned} \end{align*}$$

Therefore,

$$\begin{align*}p_A'(\lambda_i)=\prod_{m\ne i}(\lambda_i-\lambda_m). \end{align*}$$

Hence,

$$\begin{align*}p_{M_j}(\lambda_i)=|v_{i,j}|^2p_A'(\lambda_i). \end{align*}$$

This completes the proof.

Assume that the standard eigenvalues $\lambda _{1,s},\ldots ,\lambda _{n,s}$ of $A_s$ are simple. The dual Hermitian principal submatrix $M_{j}$ of order $(n-1)\times (n-1)$ , obtained by removing the $j$ -th row and column of $A$ , can be written in standard and dual parts as $M_{j} = M_{j,s} + \varepsilon M_{j,d}$ for $j=1,\dots ,n$ . For $i=1,\dots ,n$ , denote

$$ \begin{align*}p_{M_{j,s}}(\lambda_{i,s}) &= \det(\lambda_{i,s} I - M_{j,s}),\\p^{\prime}_{M_{j,s}}(\lambda_{i,s}) &= \frac{d}{d\lambda}(\det(\lambda I - M_{j,s}))\big|_{\lambda = \lambda_{i,s}},\\p^{\prime}_{A_{s}}(\lambda_{i,s}) 7&= \prod\limits_{k \ne i}(\lambda_{i,s} - \lambda_{k,s}) := P_{s,i}. \end{align*} $$

By the Jacobi formula [Reference Horn and Johnson15], we obtain

(3.2) $$ \begin{align} \operatorname{tr}(\operatorname{adj}(\lambda_{i,s} I - M_{j,s})) = \frac{d}{d\lambda}(\det(\lambda I - M_{j,s}))\big|_{\lambda = \lambda_{i,s}} = p^{\prime}_{M_{j,s}}(\lambda_{i,s}). \end{align} $$

The following theorem decomposes the dual eigenvalue–eigenvector identity into its standard and dual components. The standard component coincides with the classical identity, whereas the dual component provides an explicit first-order correction formula describing the interaction between the perturbations of the eigenvalues, eigenvectors, and principal submatrices.

Proof Let $X_{s}:=\lambda _{i,s} I-M_{j,s}$ and $Y:=\lambda _{i,d} I-M_{j,d}$ . Then

$$ \begin{align*} \begin{aligned} p_{M_j}(\lambda_i)~&=~ \det(\lambda_i I-M_j) \\&=~ \det\big((\lambda_{i,s}+\varepsilon \lambda_{i,d})I-(M_{j,s}+\varepsilon M_{j,d})\big)\\ &=~ \det(X_s+\varepsilon Y). \end{aligned} \end{align*} $$

From Jacobi’s formula [Reference Horn and Johnson15], which states that for any differentiable matrix function $X(t)$ ,

$$\begin{align*}\frac{d}{dt}\det(X(t)) = \operatorname{tr}\!\left( \operatorname{adj}(X(t)) \frac{dX(t)}{dt} \right). \end{align*}$$

Consider

$$\begin{align*}X(t)=X_s+tY. \end{align*}$$

Then the scalar function $f(t):=\det (X(t))$ has the first-order Taylor expansion about $t=0$ ,

$$\begin{align*}f(t)=f(0)+tf'(0)+O(t^2). \end{align*}$$

Since $f(0)=\det (X_s)$ and, by Jacobi’s formula,

$$\begin{align*}f'(0)= \operatorname{tr}\!\left( \operatorname{adj}(X_s)Y \right), \end{align*}$$

we obtain

$$\begin{align*}\det(X_s+tY) = \det(X_s) + t\operatorname{tr}\!\left( \operatorname{adj}(X_s)Y \right) + O(t^2). \end{align*}$$

Substituting $t=\varepsilon $ and using $\varepsilon ^2=0$ yields

$$\begin{align*}\det(X_s+\varepsilon Y) = \det(X_s) + \varepsilon\operatorname{tr}\!\left( \operatorname{adj}(X_s)Y \right). \end{align*}$$

Now

$$ \begin{align*}\det(X_s)=\det(\lambda_{i,s} I-M_{j,s})=p_{M_{j,s}}(\lambda_{i,s}), \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \operatorname{tr}\big(\hspace{-1pt}\operatorname{adj}(X_s)Y\big) ~&=~\operatorname{tr}\big(\operatorname{adj}(\lambda_{i,s} I-M_{j,s})(\lambda_{i,d}I-M_{j,d})\big)\\&=~\lambda_{i,d}\operatorname{tr}\big(\operatorname{adj}(\lambda_{i,s}I-M_{j,s})\big) -\operatorname{tr}\big(\operatorname{adj}(\lambda_{i,s}I-M_{j,s})M_{j,d}\big). \end{aligned} \end{align*} $$

By (3.2),

$$ \begin{align*}\operatorname{tr}\big(\hspace{-1pt}\operatorname{adj}(\lambda_{i,s} I-M_{j,s})\big) = \left.\frac{d}{d\lambda}\det(\lambda I-M_{j,s})\bigg|_{\lambda=\lambda_{i,s}} = p^{\prime}_{M_{j,s}}(\lambda_{i,s})\right.\!. \end{align*} $$

Hence,

(3.5) $$ \begin{align} p_{M_j}(\lambda_i) ~=~ p_{M_{j,s}}(\lambda_{i,s}) + \varepsilon\left( \lambda_{i,d} p^{\prime}_{M_{j,s}}(\lambda_{i,s}) - \operatorname{tr}\big(\operatorname{adj}(\lambda_{i,s} I-M_{j,s})M_{j,d}\big) \right). \end{align} $$

Now

$$ \begin{align*}P_i= \prod\limits_{k\ne i}(\lambda_i-\lambda_k) = \prod\limits_{k\ne i}\big((\lambda_{i,s}-\lambda_{k,s})+\varepsilon(\lambda_{i,d}-\lambda_{k,d})\big). \end{align*} $$

Since the standard eigenvalues are simple, we have

$$\begin{align*}\lambda_{i,s}-\lambda_{k,s}\neq 0,\quad k\neq i. \end{align*}$$

Hence, each dual number

$$\begin{align*}(\lambda_{i,s}-\lambda_{k,s}) +\varepsilon(\lambda_{i,d}-\lambda_{k,d}) \end{align*}$$

has a nonzero standard part and is therefore invertible in the dual-number algebra. Factoring out the standard part from each factor, we obtain

$$ \begin{align*}P_i =\prod_{k\neq i}(\lambda_{i,s}-\lambda_{k,s})\prod_{k\neq i}\left(1+\varepsilon\frac{\lambda_{i,d}-\lambda_{k,d}}{\lambda_{i,s}-\lambda_{k,s}}\right). \end{align*} $$

Setting

$$ \begin{align*}P_{s,i}=\prod_{k\neq i}(\lambda_{i,s}-\lambda_{k,s}). \end{align*} $$

Since $\varepsilon ^2=0$ , every term in the product containing two or more factors of $\varepsilon $ vanishes. Consequently, only the constant term and the terms containing exactly one factor of $\varepsilon $ contribute to the expansion. Therefore,

$$\begin{align*}\prod_{k\neq i} \left( 1+\varepsilon \frac{\lambda_{i,d}-\lambda_{k,d}} {\lambda_{i,s}-\lambda_{k,s}} \right) = 1+ \varepsilon \sum_{k\neq i} \frac{\lambda_{i,d}-\lambda_{k,d}} {\lambda_{i,s}-\lambda_{k,s}}. \end{align*}$$

Hence,

$$\begin{align*}P_i = P_{s,i} + \varepsilon P_{s,i} \sum_{k\neq i} \frac{\lambda_{i,d}-\lambda_{k,d}} {\lambda_{i,s}-\lambda_{k,s}}. \end{align*}$$

Setting

$$ \begin{align*}P_{d,i}:=P_{s,i}\sum_{k\neq i}\frac{\lambda_{i,d}-\lambda_{k,d}}{\lambda_{i,s}-\lambda_{k,s}}, \end{align*} $$

we obtain

(3.6) $$ \begin{align} P_i= P_{s,i}+\varepsilon P_{d,i}. \end{align} $$

Now, we expand $|v_{i,j}|^{2}$ as

$$ \begin{align*} \begin{aligned} |v_{i,j}|^{2}&=(u_{i,j}+ \varepsilon w_{i,j})(\overline{u_{i,j}}+ \varepsilon \overline{w_{i,j}}) \\ & = u_{i,j}\overline{u_{i,j}} +\varepsilon(u_{i,j}\overline{w_{i,j}}+w_{i,j}\overline{u_{i,j}}) + \varepsilon^{2}(w_{i,j}\overline{w_{i,j}}). \end{aligned} \end{align*} $$

Since $\varepsilon ^{2}=0$ , the last term vanishes. Therefore,

(3.7) $$ \begin{align} |v_{i,j}|^{2}= u_{i,j}\overline{u_{i,j}} +\varepsilon(u_{i,j}\overline{w_{i,j}}+w_{i,j}\overline{u_{i,j}}) =|u_{i,j}|^{2}+2\varepsilon\operatorname{Re}(u_{i,j}\overline{w_{i,j}}). \end{align} $$

Using (3.6) and (3.7), we obtain

(3.8) $$ \begin{align} |v_{i,j}|^2\,p_A'(\lambda_i) &= (P_{s,i}+\varepsilon P_{d,i})(|u_{i,j}|^{2}+2\varepsilon\operatorname{Re}(u_{i,j}\overline{w_{i,j}})) \notag \\ &= P_{s,i}|u_{i,j}|^{2}+ \varepsilon\big(2\operatorname{Re}(u_{i,j}\overline{w_{i,j}})P_{s,i}+|u_{i,j}|^{2}P_{d,i}\big). \end{align} $$

Using (3.5) and (3.8) in (3.1), we obtain

(3.9) $$ \begin{align} \begin{aligned} &p_{M_{j,s}}(\lambda_{i,s}) +\varepsilon\Big( \lambda_{i,d}\,p^{\prime}_{M_{j,s}}(\lambda_{i,s}) -\operatorname{tr}\!\big( \operatorname{adj}(\lambda_{i,s}I-M_{j,s})M_{j,d} \big) \Big) \qquad \\ &\quad = P_{s,i}|u_{i,j}|^{2} +\varepsilon\Big( 2\operatorname{Re}\left(u_{i,j}\overline{w_{i,j}}\right)P_{s,i} +|u_{i,j}|^{2}P_{d,i} \Big). \end{aligned} \end{align} $$

Equating the standard (non $\varepsilon $ ) parts of (3.9) gives the standard identity (3.3). Substituting the value $P_{s,i}$ and $P_{d,i}$ , and equating the $\varepsilon $ parts of (3.9) gives the dual correction identity (3.4).

Proof From the proof of Theorem 3.2, equation (3.9) is precisely the decomposition of

$$\begin{align*}p_{M_j}(\lambda_i) = |v_{i,j}|^2\,p_A'(\lambda_i) \end{align*}$$

into its standard and dual parts. By equating the standard and dual parts of both sides, we obtain (3.3) and (3.4). Conversely, if (3.3) and (3.4) hold, then the standard and dual parts of both sides coincide, and therefore identity (3.1) holds.

4 Illustrative examples

The following examples illustrate the main ideas of Theorem 3.2 in both symbolic and numerical settings. We first consider a symbolic $2\times 2$ example to highlight the structure of the standard and dual correction identities in a simple setting. We then present a numerical $3\times 3$ example to demonstrate how the identities can be verified concretely for nontrivial dual Hermitian matrices.

Example 4.1 Consider the dual Hermitian matrix

$$ \begin{align*}A=\begin{pmatrix}a & c\\\overline{c} & d\end{pmatrix}+\varepsilon \begin{pmatrix}\alpha & \gamma\\\overline{\gamma} & \delta\end{pmatrix}, \end{align*} $$

where $a,d,\alpha ,\delta \in \mathbb {R}$ and $c,\gamma \in \mathbb {C}$ .

Assume that the standard part $A_s$ has distinct eigenvalues

$$ \begin{align*}\lambda_{1,s},~\lambda_{2,s}, \end{align*} $$

with corresponding orthonormal eigenvectors

$$ \begin{align*}u_1=\begin{pmatrix}u_{1,1}\\u_{1,2}\end{pmatrix}, ~\text{and}~ u_2=\begin{pmatrix}u_{2,1}\\u_{2,2}\end{pmatrix}. \end{align*} $$

Let

$$ \begin{align*}\lambda_i=\lambda_{i,s}+\varepsilon\lambda_{i,d},~\text{and}~v_i=u_i+\varepsilon w_i, \end{align*} $$

for $i=1,2$ . For $i=j=1$ , the principal submatrix is

$$ \begin{align*}M_1=d+\varepsilon\delta, \end{align*} $$

so that

$$ \begin{align*}p_{M_1}(\lambda_1)=(\lambda_{1,s}-d)+\varepsilon(\lambda_{1,d}-\delta). \end{align*} $$

Since

$$ \begin{align*}p_A'(\lambda_1)= \lambda_1-\lambda_2, \end{align*} $$

we have

$$ \begin{align*}p_A'(\lambda_1)=(\lambda_{1,s}-\lambda_{2,s})+\varepsilon(\lambda_{1,d}-\lambda_{2,d}). \end{align*} $$

Moreover,

$$ \begin{align*}|v_{1,1}|^2=|u_{1,1}|^2+2\varepsilon\operatorname{Re}(u_{1,1}\overline{w_{1,1}}). \end{align*} $$

Using

$$ \begin{align*}p_{M_1}(\lambda_1)=|v_{1,1}|^2p_A'(\lambda_1), \end{align*} $$

and comparing standard and dual parts, we obtain

$$ \begin{align*}\lambda_{1,s}-d=|u_{1,1}|^2(\lambda_{1,s}-\lambda_{2,s}), \end{align*} $$

which is the classical eigenvector–eigenvalue identity, together with the dual correction identity

$$ \begin{align*}\lambda_{1,d}-\delta=2\operatorname{Re}(u_{1,1}\overline{w_{1,1}})(\lambda_{1,s}-\lambda_{2,s})+|u_{1,1}|^2(\lambda_{1,d}-\lambda_{2,d}). \end{align*} $$

This example illustrates how the dual identity naturally extends the classical relation through an additional correction term.

Example 4.2 Consider a dual Hermitian matrix $A=A_s+\varepsilon A_d \in \widehat {\mathbb {R}}^{3 \times 3}$ with $A_s$ and $A_d$ given as

(4.1) $$ \begin{align} A_{s}=\begin{pmatrix} 2 & ~1 &~ 0.5\\ 1 &~ 3 &~ 0.2 \\ 0.5 & ~0.2 &~ 1 \end{pmatrix} \qquad \text{and} \qquad A_d=\begin{pmatrix} 0.10 &~ 0.50 &~ -0.20 \\ 0.50 &~ 0 &~ 0.3 \\ -0.20 &~ 0.3 &~ -0.05 \end{pmatrix}. \end{align} $$

For each simple standard eigenvalue $\lambda _{i,s}$ , $i=1,2,3$ , a unit eigenvector $u_i$ of $A_s$ is computed. The corresponding dual eigenvalue correction is given by

$$\begin{align*}\lambda_{i,d}=u_i^*A_du_i. \end{align*}$$

The eigenvector correction $w_i$ is then obtained by solving

$$\begin{align*}(A_s-\lambda_{i,s}I)w_i = (\lambda_{i,d}I-A_d)u_i, \end{align*}$$

as given by Theorem 4.1 of Qi and Cui [Reference Qi and Cui21]. The corresponding dual eigenvectors are

$$\begin{align*}v_i=u_i+\varepsilon w_i,\quad i=1,2,3, \end{align*}$$

where $v_i$ is chosen so that $v_i^*v_i=1$ . We now verify Theorem 3.2 for the dominant eigenvalue

$$\begin{align*}\lambda_1=3.68893048+\varepsilon\,0.51805479. \end{align*}$$

The purpose of this example is to illustrate explicitly how the standard identity and the dual correction identity interact with the principal submatrices of $A$ (Table 1).

Table 1

Computation of eigenvalues and eigenvectors of $A = A_s + \varepsilon A_d$ , where $A_s$ and $A_d$ are given in (4.1).

Table 1 Long description

The table consists of four columns and three data rows.

Column headers are:

1. i

2. Eigenvalue lambda sub i = lambda sub i,s + varepsilon lambda sub i,d

3. u sub i

4. w sub i

Row 1 (i = 1):

- Eigenvalue: 3.68893048 + varepsilon 0.51805479

- u sub 1: A column vector with elements -0.53776387, -0.82747448, and -0.16154260.

- w sub 1: A column vector with elements -0.07054694, 0.05121676, and -0.02750337.

Row 2 (i = 2):

- Eigenvalue: 1.54365251 minus varepsilon 0.62078441

- u sub 2: A column vector with elements -0.70673578, 0.54691183, and -0.44878947.

- w sub 2: A column vector with elements -0.00574458, 0.08965822, and 0.11830724.

Row 3 (i = 3):

- Eigenvalue: 0.76741701 + varepsilon 0.15272962

- u sub 3: A column vector with elements -0.45971139, 0.12717482, and 0.87891524.

- w sub 3: A column vector with elements 0.09135623, -0.05232624, and 0.05535469.

We first consider the case $j=1$ , corresponding to the removal of the first row and column of $A$ . In this case, the quantities appearing in equation (3.4) are computed directly from the associated principal submatrices $M_{1,s}$ and $M_{1,d}$ .

Corresponding to $i=j=1$ , the matrices $M_{1,s}$ and $M_{1,d}$ are obtained as

$$ \begin{align*}M_{1,s}= \begin{pmatrix} 3 & 0.2 \\ 0.2 & 1 \end{pmatrix}~ \text{and}~ M_{1,d}= \begin{pmatrix} 0 & 0.3 \\ 0.3 & -0.05 \end{pmatrix}, \end{align*} $$

respectively. For the dual correction identity (3.4),

$$ \begin{align*}\operatorname{adj}(\lambda_{1,s}I-M_{1,s})=\begin{pmatrix}\lambda_{1,s}-1 & 0.2\\0.2 & \lambda_{1,s}-3\end{pmatrix}. \end{align*} $$

Using

$$ \begin{align*}p^{\prime}_{M_{1,s}}(\lambda_{1,s})=\operatorname{tr}(\operatorname{adj}(\lambda_{1,s}I-M_{1,s})), \end{align*} $$

we obtain

$$ \begin{align*}p^{\prime}_{M_{1,s}}(\lambda_{1,s})=(\lambda_{1,s}-1)+(\lambda_{1,s}-3)=3.377861. \end{align*} $$

Moreover,

$$ \begin{align*}\operatorname{tr}\left(\operatorname{adj}(\lambda_{1,s}I-M_{1,s})M_{1,d}\right)=\frac{3}{25} - \frac{\lambda_{1,s}-3}{20}=0.085553. \end{align*} $$

• • The quantity

  $$ \begin{align*}P_{s,1}=\prod_{k\neq1}(\lambda_{1,s}-\lambda_{k,s})=p^{\prime}_{A_s}(\lambda_{1,s}) \end{align*} $$
  measures the spectral separation of the eigenvalue $\lambda _{1,s}$ from the remaining eigenvalues of $A_s$ . Numerically,
  $$ \begin{align*}P_{s,1}=6.267458. \end{align*} $$

• • The corresponding dual correction term is

  $$ \begin{align*}P_{d,1}=P_{s,1}\sum_{k\neq1}\frac{\lambda_{1,d}-\lambda_{k,d}}{\lambda_{1,s}-\lambda_{k,s}}, \end{align*} $$
  which captures the perturbative contribution of the dual eigenvalue corrections. In this example,
  $$ \begin{align*}P_{d,1}=4.110858. \end{align*} $$

• • The left-hand side of the dual correction identity (3.4) is therefore

  $$ \begin{align*}\text{LHS}_{\varepsilon}(1,1):=\lambda_{1,d}p^{\prime}_{M_{1,s}}(\lambda_{1,s})-\operatorname{tr}(\operatorname{adj}(\lambda_{1,s}I-M_{1,s})M_{1,d})=1.664364. \end{align*} $$

• • The right-hand side becomes

  $$ \begin{align*}\text{RHS}_{\varepsilon}(1,1):=2\operatorname{Re}(u_{1,1}\overline{w_{1,1}})P_{s,1}+|u_{1,1}|^{2}P_{d,1}=1.664364.\end{align*} $$

Since $\text {LHS}_{\varepsilon }(1,1) = \text {RHS}_{\varepsilon }(1,1)=1.664364$ , the dual identity is verified for $i=j=1$ .

For the standard identity (3.3), we compute

$$ \begin{align*}p_{M_{1,s}}(\lambda_{1,s})=\det(\lambda_{1,s}I-M_{1,s}) =1.812486, \end{align*} $$

while

$$ \begin{align*}P_{s,1}|u_{1,1}|^2=1.812486. \end{align*} $$

Thus, the standard identity is also verified. Since $p_{M_{1,s}}(\lambda _{1,s})=P_{s,1}|u_{1,1}|^{2}=1.812486$ , the standard identity is verified for $i=j=1$ .

The same procedure may be repeated for $j=2$ . In this case,

$$ \begin{align*}M_{2,s}=\begin{pmatrix}2 & 0.5\\0.5 & 1\end{pmatrix},~ \text{and}~ M_{2,d}=\begin{pmatrix}0.10 & -0.20\\-0.20 & -0.05\end{pmatrix}. \end{align*} $$

Substituting these matrices into (3.4) gives

$$ \begin{align*}\text{LHS}_{\varepsilon}(1,2)=2.283519,\end{align*} $$

and

$$ \begin{align*}\text{RHS}_{\varepsilon}(1,2)=2.283525, \end{align*} $$

which agree up to rounding error, thereby confirming the dual correction identity.

For standard identity (3.3),

• • $p_{M_{2,s}}(\lambda _{1,s})=\det (\lambda _{1,s}I-M_{2,s})=(\lambda _{1,s}-3)(\lambda _{1,s}-1)-0.04=1.812486$ ;

• • $P_{s,1}|u_{1,2}|^{2}=1.812486$ .

Since $p_{M_{2,s}}(\lambda _{1,s})=P_{s,1}|u_{1,2}|^{2}=1.812486$ , the standard identity is verified for $i=1,~j=2$ . Similarly, the results can be verified for other values of $i,j=1,2,3$ .

This example shows how the classical eigenvector–eigenvalue identity for the standard part $A_s$ extends to the dual setting through an additional correction term involving both the dual perturbation matrix and the dual eigenvector corrections.