Eigenvalue perturbation theory of symplectic, orthogonal, and unitary matrices under generic structured rank one perturbations


We study the perturbation theory of structured matrices under structured rank one perturbations, with emphasis on matrices that are unitary, orthogonal, or symplectic with respect to an indefinite inner product. The rank one perturbations are not necessarily of arbitrary small size (in the sense of norm). In the case of sesquilinear forms, results on selfadjoint matrices can be applied to unitary matrices by using the Cayley transformation, but in the case of real or complex symmetric or skew-symmetric bilinear forms additional considerations are necessary. For complex symplectic matrices, it turns out that generically (with respect to the perturbations) the behavior of the Jordan form of the perturbed matrix follows the pattern established earlier for unstructured matrices and their unstructured perturbations, provided the specific properties of the Jordan form of complex symplectic matrices are accounted for. For instance, the number of Jordan blocks of fixed odd size corresponding to the eigenvalue 1 or −1 have to be even. For complex orthogonal matrices, it is shown that the behavior of the Jordan structures corresponding to the original eigenvalues that are not moved by perturbations follows again the pattern established earlier for unstructured matrices, taking into account the specifics of Jordan forms of complex orthogonal matrices. The proofs are based on general results developed in the paper concerning Jordan forms of structured matrices (which include in particular the classes of orthogonal and symplectic matrices) under structured rank one perturbations. These results are presented and proved in the framework of real as well as of complex matrices.

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Corresponding author

Correspondence to Christian Mehl.

Additional information

This research was supported by Deutsche Forschungsgemeinschaft, through the DFG Research Center Matheon Mathematics for key technologies in Berlin.

Communicated by Miloud Sadkane.

Appendix: Proof of Theorem 5.3

Appendix: Proof of Theorem 5.3

In this section we prove Theorem 5.3. The proof follows the same lines as the proof of Theorem 4.2 in [18], but is more general and extends the result that was obtained there. Before we prove Theorem 5.3, we quote two results from [18]. The first one follows from the Brunovsky canonical form, see [3], and also [4, 8], of general multi-input control systems \(\dot{x}=Ax+Bu\) under transformations

$$(A,B) \quad\mapsto\quad\bigl(C^{-1}(A+BR)C, C^{-1}BD\bigr), $$

with invertible matrices C,D and arbitrary matrix R of suitable sizes.

Theorem 10.1

Let \(A\in{\mathbb{C}}^{n\times n}\) be a matrix in Jordan canonical form

$$ A={\mathrm{J}}_{n_1}(\lambda_1)\oplus\cdots \oplus{\mathrm{J}}_{n_g}(\lambda_g) \oplus{\mathrm{J}}_{n_{g+1}}(\lambda_{g+1})\oplus\cdots\oplus{ \mathrm{J}}_{n_\nu}(\lambda_\nu), $$

where \(\lambda_{1}=\dots=\lambda_{g}=:\widehat{\lambda}\in\mathbb{C}\), \(\lambda_{g+1},\dots,\lambda_{\nu}\in\mathbb{C}\setminus\{\widehat{\lambda}\}\), n 1≥⋯≥n g . Moreover, let B=uv T, where

$$u=\left [ \begin{array}{c}u_1\\ \vdots\\ u_\nu \end{array} \right ],\quad v=\left [ \begin{array}{c}v_1\\ \vdots\\ v_\nu \end{array} \right ], \quad u_i,v_i \in\mathbb{C}^{n_i},\ i=1,\dots,\nu. $$

Assume that the first component of each vector v i , i=1,…,ν is nonzero. Then the matrix \(\operatorname{Toep} (v_{1})\oplus\cdots\oplus\operatorname{Toep} (v_{\nu})\) is invertible, and if we denote its inverse by S, then S −1 AS=A and

$$ S^{-1}BS= \bigl[we_{1,n_1}^T, \dots,we_{1,n_\nu}^T \bigr], $$

where w=S −1 u. Moreover, the matrix S −1(A+B)S has at least g−1 Jordan chains associated with \(\widehat{\lambda}\) of lengths at least n 2,…,n g given by

$$ \begin{array}{l@{\quad}l@{\quad}l} e_1-e_{n_1+1},&\dots,&e_{n_2}-e_{n_1+n_2};\\ e_1-e_{n_1+n_2+1},&\dots,&e_{n_3}-e_{n_1+n_2+n_3};\\ \vdots&\ddots&\vdots\\ e_1-e_{n_1+\cdots+n_{g-1}+1},&\dots,&e_{n_g}-e_{n_1+\cdots +n_{g-1}+n_g}.\\ \end{array} $$

Theorem 10.2

(Partial Brunovsky form)


$$A= \bigl({\mathrm{J}}_{n_1}(\widehat{\lambda})^{\oplus \ell _1} \bigr) \oplus\cdots\oplus \bigl({\mathrm{J}}_{n_m}(\widehat{\lambda})^{\oplus \ell_m} \bigr)\oplus\widetilde{A} \in{\mathbb{C}}^{n\times n}, $$

where n 1>…>n m and \(\sigma(\widetilde{A})\subseteq\mathbb{C}\setminus\{\widehat{\lambda}\}\). Moreover, let a= 1 n 1+⋯+ m n m denote the algebraic multiplicity of \(\widehat{\lambda}\) and let B=uv T, where \(u,v\in\mathbb{C}^{n}\) and

$$v=\left [ \begin{array}{c} v^{(1)}\\ \vdots\\ v^{(m)}\\ \widetilde{v} \end{array} \right ],\quad v^{(i)}= \left [ \begin{array}{c} v^{(i,1)}\\ \vdots\\ v^{(i,\ell_i)} \end{array} \right ],\quad v^{(i,j)}\in \mathbb{C}^{n_i},\quad j=1,\dots,\ell_i,\ i=1,\dots,m. $$

Assume that the first component of each vector v (i,j), j=1,…, i , i=1,…,m is nonzero. Then the following statements hold:

  1. (1)

    The matrix \(S:= (\bigoplus_{j=1}^{\ell _{1}}\operatorname{Toep}(v^{(1,j)})\oplus\cdots\oplus \bigoplus_{j=1}^{\ell_{m}}\operatorname{Toep}(v^{(m,j)}) )^{-1} \oplus I_{n-a}\) exists and satisfies

    $$S^{-1}AS=A,\quad S^{-1}BS=w \bigl[\underbrace{e_{1,n_1}^T, \dots ,e_{1,n_1}^T}_{\ell_1~\mathit{times}},\dots, \underbrace{e_{1,n_m}^T,\dots,e_{1,n_m}^T}_{\ell_m~\mathit{times}},z^T \bigr], $$

    where w=S −1 u and for some appropriate vector \(z\in\mathbb{C}^{n-a}\).

  2. (2)

    The matrix S −1(A+B)S has at least 1+…+ m −1 Jordan chains associated with \(\widehat{\lambda}\) given as follows:

    1. (a)

      1−1 Jordan chains of length at least n 1:

      $$\begin{array}{l@{\quad}l@{\quad}l} e_1-e_{n_1+1},&\dots,&e_{n_1}-e_{2n_1};\\ \vdots&\ddots&\vdots\\ e_1-e_{(\ell_1-1)n_1+1},&\dots,&e_{n_1}-e_{\ell_1n_1};\\ \end{array} $$
    2. (b)

      i Jordan chains of length at least n i for i=2,…,m:

      $$\begin{array}{l@{\quad}l@{\quad}l} e_1-e_{\ell_1n_1+\dots+\ell_{i-1}n_{i-1}+1},&\dots,&e_{n_i}-e_{\ell _1n_1+\dots+\ell_{i-1}n_{i-1}+n_i};\\ e_1-e_{\ell_1n_1+\dots+\ell_{i-1}n_{i-1}+n_i+1},&\dots ,&e_{n_i}-e_{\ell_1n_1+\dots+\ell_{i-1}n_{i-1}+2n_i};\\ \vdots&\ddots&\vdots\\ e_1-e_{\ell_1n_1+\dots+\ell_{i-1}n_{i-1}+(\ell_i-1)n_i+1},&\dots,&e_{n_i}- e_{\ell_1n_1+\dots+\ell_{i-1}n_{i-1}+\ell_in_i}.\\ \end{array} $$
  3. (3)

    Partition w=S −1 u as

    $$w=\left [ \begin{array}{c} w^{(1)}\\ \vdots\\ w^{(m)}\\ \widetilde{w} \end{array} \right ],\qquad w^{(i)}= \left [ \begin{array}{c} w^{(i,1)}\\ \vdots\\ w^{(i,\ell_i)} \end{array} \right ],\qquad w^{(i,j)}= \left [ \begin{array}{c} w^{(i,j)}_1\\ \vdots\\ w^{(i,j)}_{n_i} \end{array} \right ]\in\mathbb{C}^{n_i}, $$

    and let λ 1,…,λ q be the pairwise distinct eigenvalues of A different from \(\widehat{\lambda}\) having the algebraic multiplicities r 1,…,r q , respectively. Set \(\mu_{i}=\lambda_{i}-\widehat{\lambda}\), i=1,2,…,q.

    Then the characteristic polynomial \(p_{\widehat{\lambda}}\) of \(A+B-\widehat{\lambda}I\) is given by

    $$\begin{aligned} p_{\widehat{\lambda}}(\lambda) =&(-\lambda)^aq(\lambda)+ \Biggl(\prod _{i=1}^q(\mu_{i}-\lambda )^{r_i} \Biggr) \\ &{}\cdot \Biggl((-\lambda)^a+(-1)^{a-1} \sum _{i=1}^m\sum_{j=1}^{\ell_i} \sum_{k=1}^{n_i}w^{(i,j)}_{k} \lambda ^{a-k} \Biggr), \end{aligned}$$

    where q(λ) is some polynomial;

  4. (4)

    Write \(p_{\widehat{\lambda}}(\lambda)=c_{n}\lambda^{n}+\cdots +c_{a-n_{1}+1}\lambda^{a-n_{1}+1}+c_{a-n_{1}}\lambda^{a-n_{1}}\). Then

    $$c_{a-n_1}=(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1} \Biggr); $$

    and in the case n 1>1 we have in addition that

    $$\begin{aligned} c_{a-n_1+1} =&(-1)^a \Biggl(\sum_{\nu=1}^qr_{\nu} \mu_\nu^{r_\nu-1} \prod_{\genfrac{}{}{0pt}{2}{i=1}{i\neq\nu}}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1} \Biggr) \\ &{}+(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1-1} \Biggr), \end{aligned}$$

    if n 1−1>n 2 or, if n 1−1=n 2, then

    $$\begin{aligned} c_{a-n_1+1} =&(-1)^a \Biggl(\sum_{\nu=1}^qr_{\nu} \mu_\nu^{r_\nu-1} \prod_{\genfrac{}{}{0pt}{2}{i=1}{i\neq\nu}}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1} \Biggr) \\ &{}+(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1-1}+\sum _{j=1}^{\ell _2}w^{(2,j)}_{n_2} \Biggr). \end{aligned}$$

The following notation of linear combinations of Jordan chains will be necessary.

Definition 10.1

Let \(A\in{\mathbb{C}}^{n\times n}\) and let X=(x 1,…,x p ) and Y=(y 1,…,y q ) be two Jordan chains of A associated with the same eigenvalue \(\widehat{\lambda}\) of (possibly different) lengths p and q. Then the sum X+Y of X and Y is defined to be the chain Z=(z 1,…,z max(p,q)), where

$$z_j=\left \{ \begin{array}{l@{\quad}l} x_j&\mbox{if $p\geq q$},\\ y_j&\mbox{if $p<q$}, \end{array} \right .\quad j=1,\dots,|p-q| $$


$$z_j=\left \{ \begin{array}{l@{\quad}l} x_j+y_{j-p+q}&\mbox{if $p\geq q$},\\ y_j+x_{j-q+p}&\mbox{if $p<q$}, \end{array} \right .\quad j=|p-q|+1,\dots,\max(p,q). $$

To illustrate this construction, consider e.g. X=(x 1,x 2,x 3,x 4) and Y=(y 1,y 2), then X+Y=(x 1,x 2,x 3+y 1,x 4+y 2).

It is straightforward to check that the sum Z=X+Y of two Jordan chains associated with an eigenvalue \(\widehat{\lambda}\) is again a Jordan chain associated with \(\widehat{\lambda}\) of the given matrix A, but it should be noted that this sum is not commutative.

Proof of Theorem 5.3

Let \(\tau\in\mathbb{F}\setminus\{0\}\) be arbitrary. We may assume without loss of generality that A and G are already in the forms (5.5) and (5.6). Furthermore, we may assume \(\widehat{\lambda}=0\), otherwise consider the matrix \(A-\widehat{\lambda}I\) instead of A. Then the algebraic and geometric multiplicity a and γ of the eigenvalue zero of A are given by

$$a=\sum_{s=1}^m\ell_sn_s, \qquad\gamma=\sum_{s=1}^m \ell_s, $$

respectively. Let us partition u conformably with the forms (5.5) and (5.6), i.e., we let

$$u=\left [ \begin{array}{c} u^{(1)}\\ \vdots\\ u^{(m)}\\ \widetilde{u} \end{array} \right ],\qquad u^{(i)}= \left [ \begin{array}{c} u^{(i,1)}\\ \vdots\\ u^{(i,\ell_i)} \end{array} \right ],\qquad u^{(i,j)}= \left [ \begin{array}{c} u^{(i,j)}_1\\ \vdots\\ u^{(i,j)}_{n_i} \end{array} \right ]\in\mathbb{C}^{n_i}, $$

for j=1,…, i ; i=1,…,m. Thus, \(\widetilde{u}\in \mathbb{C}^{n-a}\). Then the vector v T=u T G has the following structure:

$$v\,{=}\,\bigl(u^TG\bigr)^T\,{=}\,G^Tu\,{=}\,\left [ \begin{array}{c} v^{(1)}\\ \vdots\\ v^{(m)}\\ \widetilde{v} \end{array} \right ],\qquad v^{(i)}\,{=}\,\left [ \begin{array}{c} v^{(i,1)}\\ \vdots\\ v^{(i,\ell_i)} \end{array} \right ], \qquad v^{(i,j)}\,{=}\,\left [ \begin{array}{c} v^{(i,j)}_1\\ \vdots\\ v^{(i,j)}_{n_i} \end{array} \right ] \in\mathbb{C}^{n_i}, $$

for j=1,…, i and i=1,…,m, where

$$v^{(1,2s-1)}=\bigl(G^{(1,2s)}\bigr)^Tu^{(1,2s)}= \left [ \begin{array}{c} g_{n_1,1}^{(1,2s)}u^{(1,2s)}_{n_1}\\ g_{n_1-1,2}^{(1,2s)}u^{(1,2s)}_{n_1-1}+g_{n_1,2}^{(1,2s)}u^{(1,2s)}_{n_1}\\ \ast\\ \vdots\\ \ast \end{array} \right ] $$


$$v^{(1,2s)}=\bigl(G^{(1,2s-1)}\bigr)^Tu^{(1,2s-1)}= \left [ \begin{array}{c} g_{n_1,1}^{(1,2s-1)}u^{(1,2s-1)}_{n_1}\\ g_{n_1-1,2}^{(1,2s-1)}u^{(1,2s-1)}_{n_1-1}+g_{n_1,2}^{(1,2s-1)}u^{(1,2s-1)}_{n_1}\\ \ast\\ \vdots\\ \ast \end{array} \right ] $$

for s=1,…, 1/2. Generically, the hypothesis of Theorem 10.2 is satisfied, i.e., the first entries of the vectors v (i,j) are nonzero. Thus, generically the matrix S as in Theorem 10.2 exists so that S −1(A+τB)S is in partial Brunovsky form. In fact, S −1 takes the form

$$S^{-1}= \Biggl(\bigoplus_{j=1}^{\ell_1} \operatorname{Toep} \bigl(v^{(1,j)}\bigr) \Biggr)\oplus\cdots\oplus \Biggl( \bigoplus_{j=1}^{\ell_m}\operatorname{Toep} \bigl(v^{(m,j)}\bigr) \Biggr)\oplus I_{n-a}, $$

and it follows that

$$ S^{-1}\tau BS=w\bigl(\underbrace{e_{1,n_1}^T, \dots,e_{1,n_1}^T}_{{\ell_1~\mathrm{times}}},\dots, \underbrace{e_{1,n_m}^T,\dots,e_{1,n_m}^T}_{{\ell_m~\mathrm{times}}},z^T \bigr) $$

for some \(z\in\mathbb{C}^{n-a}\), where w=τS −1 u. Thus,

$$ w=\tau S^{-1}u=\left [ \begin{array}{c} w^{(1)}\\ \vdots\\ w^{(m)}\\ \widetilde{w} \end{array} \right ],\qquad w^{(i)}=\left [ \begin{array}{c}w^{(i,1)}\\ \vdots\\ w^{(i,\ell_i)} \end{array} \right ],\qquad w^{(i,s)}=\left [ \begin{array}{c}w^{(i,j)}_1\\ \vdots\\ w^{(i,j)}_{n_i} \end{array} \right ]\in\mathbb{C}^{n_i}, $$

for j=1,…, i and i=1,…,m, where

$$ \begin{aligned} w^{(1,2s-1)}_{n_1}&=\tau g_{n_1,1}^{(1,2s)}u^{(1,2s)}_{n_1}u^{(1,2s-1)}_{n_1}, \\ w^{(1,2s)}_{n_1}&=\tau g_{n_1,1}^{(1,2s-1)}u^{(1,2s-1)}_{n_1}u^{(1,2s)}_{n_1}. \end{aligned} $$

Thus, using hypothesis (2a) we obtain \(w^{(1,2s)}_{n_{1}}=-w^{(1,2s-1)}_{n_{1}}\). Furthermore,

$$\begin{aligned} w^{(1,2s-1)}_{n_1-1} =&\tau g_{n_1,1}^{(1,2s)}u^{(1,2s)}_{n_1}u^{(1,2s-1)}_{n_1-1} +\tau g_{n_1-1,2}^{(1,2s)}u^{(1,2s)}_{n_1-1}u^{(1,2s-1)}_{n_1} \\ &{}+\tau g_{n_1,2}^{(1,2s)}u^{(1,2s)}_{n_1}u^{(1,2s-1)}_{n_1}, \\ w^{(1,2s)}_{n_1-1} =&\tau g_{n_1,1}^{(1,2s-1)}u^{(1,2s-1)}_{n_1}u^{(1,2s)}_{n_1-1} +\tau g_{n_1-1,2}^{(1,2s-1)}u^{(1,2s-1)}_{n_1-1}u^{(1,2s)}_{n_1} \\ &{}+\tau g_{n_1,2}^{(1,2s-1)}u^{(1,2s-1)}_{n_1}u^{(1,2s)}_{n_1} \end{aligned}$$

for s=1,…, 1/2, provided that n 1>1. This implies that

$$\begin{aligned} w^{(1,2s-1)}_{n_1-1}+w^{(1,2s)}_{n_1-1} =&\tau \bigl(g_{n_1,1}^{(1,2s)}+g_{n_1-1,2}^{(1,2s-1)} \bigr)u^{(1,2s)}_{n_1}u^{(1,2s-1)}_{n_1-1} \\ &{}+\tau\bigl(g_{n_1-1,2}^{(1,2s)}+g_{n_1,1}^{(1,2s-1)} \bigr)u^{(1,2s)}_{n_1-1}u^{(1,2s-1)}_{n_1} \\ &{}+\tau\bigl(g_{n_1,2}^{(1,2s)}+g_{n_1,2}^{(1,2s-1)} \bigr)u^{(1,2s-1)}_{n_1}u^{(1,2s)}_{n_1} \end{aligned}$$

which, by the hypothesis (2b), is generically nonzero.

We will now show in two steps that generically A+τB has the Jordan canonical form (5.7). By Theorem 10.2 we know that generically A+τB has 1−1 Jordan chains of length n 1 and j Jordan chains of length n j , j=2,…,m associated with the eigenvalue zero. (These chains are linearly independent but need not form a basis of the corresponding root subspace of A+τB yet, as it may be possible to extend some of the chains.) In the first step, we will show that generically there exists a Jordan chain of length n 1+1. In the second step, we will show that the algebraic multiplicity of the eigenvalue zero of A+τB generically is \(\widetilde{a}=(\sum_{s=1}^{m}\ell _{s}n_{s})-n_{1}+1=a-n_{1}+1\). Both steps together obviously imply that (5.7) represents the only possible Jordan canonical form for A+τB.

Step 1: Existence of a Jordan chain of length n 1+1.

Consider the following Jordan chains of S −1(A+τB)S associated with the eigenvalue zero and denoted by C 1,s and C i,j , respectively:

$$\begin{aligned} &\mbox{length } n_1:\quad C_{1,s}:\quad e_{2(s-1)n_1+1}-e_{(2s-1)n_1+1},\dots,e_{(2s-1)n_1}-e_{2sn_1}, \\ &\quad{} s=1,\dots,\frac{\ell_1}{2} \\ &\mbox{length } n_i:\quad C_{i,j}:\quad -e_{1}+e_{\varSigma_{k=1}^{i-1}\ell_kn_k+(j-1)n_i+1},\dots,-e_{n_i}+e_{\varSigma_{k=1}^{i-1} \ell_kn_k+jn_i}, \\ &\quad{} j=1,\dots,\ell_i, \end{aligned}$$

where i=2,…,m. Observe that C i,j , i≠1, are just the Jordan chains from Theorem 10.2 multiplied by −1 while the chains C 1,s are linear combinations of the Jordan chains from Theorem 10.2. Namely, in the notation of (10.3), and numbering the chains in (10.3) first, second, etc., from the top to the bottom, we see that the chains \(C_{1,1}, \ldots, C_{1, \ell_{1}/2}\) are the negative of the second chain plus the first chain, the negative of the fourth chain plus the third chain, …, the negative of the ( 1−1)-th chain plus the ( 1)-th chain, respectively. Now consider the Jordan chain

$$C:= \Biggl(\sum_{s=1}^{\ell_1/2} \alpha_{1,s}C_{1,s} \Biggr)+\sum_{i=2}^m \sum_{j=1}^{\ell_i}\alpha_{i,j}C_{i,j} $$

of length n 1 (see Definition 10.1), and let y denote the n 1-th (and thus last) vector of this chain. We next show that the Jordan chain C can be extended by a certain vector to a Jordan chain of length n 1+1 associated with the eigenvalue zero, for some particular choice of the parameters α i,s (depending on u) such that generically at least one of \(\alpha_{1,1}, \ldots, \alpha_{1,\ell_{1}/2}\) is nonzero. To see this, we have to show that y is in the range of S −1(A+τB)S. First, partition

$$y=\left [ \begin{array}{c} y^{(1)}\\ \vdots\\ y^{(m)}\\ \widetilde{y} \end{array} \right ],\qquad y^{(i)}= \left [ \begin{array}{c} y^{(i,1)}\\ \vdots\\ y^{(i,\ell_i)} \end{array} \right ],\qquad y^{(i,j)}= \left [ \begin{array}{c} y^{(i,j)}_1\\ \vdots\\ y^{(i,j)}_{n_i} \end{array} \right ]\in\mathbb{C}^{n_i}, $$

for j=1,…, i ; i=1,…,m. Then by the definition of y, we have \(\widetilde{y}=0\in\mathbb{C}^{n-a}\),

$$\begin{aligned} &y^{(1,2s-1)}_{n_1}=\alpha_{1,s},\quad y^{(1,2s)}_{n_1}= -\alpha_{1,s},\quad s=1,\dots, \ell_1/2, \\ &y^{(i,j)}_{n_i}=\alpha_{i,j},\quad j=1,\dots, \ell_i; \ i=2,\dots,m. \end{aligned}$$

We have to solve the linear system

$$ S^{-1}(A+\tau B)Sx=y. $$


$$x=\left [ \begin{array}{c} x^{(1)}\\ \vdots\\ x^{(m)}\\ \widetilde{x} \end{array} \right ],\qquad x^{(i)}= \left [ \begin{array}{c} x^{(i,1)}\\ \vdots\\ x^{(i,\ell_i)} \end{array} \right ],\qquad x^{(i,j)}= \left [ \begin{array}{c}x^{(i,j)}_1\\ \vdots\\ x^{(i,j)}_{n_i} \end{array} \right ]\in\mathbb{C}^{n_i}, $$

and making the ansatz \(\widetilde{x}=0\), then (10.8) becomes (here we use (10.4) and (10.5)):

$$\begin{aligned} &w^{(i,j)}_{k} \Biggl(\sum_{\nu=1}^m \sum_{\mu=1}^{\ell_\nu}x^{(\nu ,\mu)}_{1} \Biggr)+ x^{(i,j)}_{k+1}=y^{(i,j)}_{k}, \\ &\quad{}k=1,\ldots,n_i-1; j=1,\ldots,\ell_i; i=1, \ldots,m, \end{aligned}$$
$$\begin{aligned} &w^{(i,j)}_{n_i} \Biggl(\sum_{\nu=1}^m \sum_{\mu=1}^{\ell_\nu }x^{(\nu,\mu)}_{1} \Biggr)=\alpha_{i,j},\quad j=1,\ldots,\ell_i; i=2, \ldots,m, \end{aligned}$$
$$\begin{aligned} &w^{(1,2s-1)}_{n_1} \Biggl(\sum_{\nu=1}^m \sum_{\mu=1}^{\ell_\nu}x^{(\nu,\mu)}_{1} \Biggr)=\alpha_{1,s},\quad s=1,\ldots,\ell_1/2, \end{aligned}$$
$$\begin{aligned} &w^{(1,2s)}_{n_1} \Biggl(\sum_{\nu=1}^m \sum_{\mu=1}^{\ell_\nu}x^{(\nu,\mu)}_{1} \Biggr)=-\alpha_{1,s},\quad s=1,\ldots,\ell_1/2. \end{aligned}$$

Set \(x^{(1,1)}_{1}=1\) and \(x^{(\nu,\mu)}_{1}=0\), for μ=1,…, ν ; ν=1,…,m; (ν,μ)≠(1,1), as well as \(\alpha_{i,j}=w^{(i,j)}_{n_{i}}\) for j=1,…, i ; i=2,…,m and \(\alpha_{1,s}=w^{(1,2s-1)}_{n_{1}}\) for s=1,…, 1/2. Then (10.10) and (10.11) are satisfied and so is (10.12), because \(w^{(1,2s)}_{n_{1}}=-w^{(1,2s-1)}_{n_{1}}\) by (10.6). Finally, (10.9) can be solved by choosing \(x^{(i,j)}_{k+1}=y^{(i,j)}_{k}-w^{(i,j)}_{k}\) for k=1,…,n i −1; j=1,…, i ; i=1,…,m.

Step 2: We show that the algebraic multiplicity of the eigenvalue zero of A+τB generically is \(\widetilde{a}=(\sum_{s=1}^{m}\ell _{s}n_{s})-n_{1}+1=a-n_{1}+1\).

Let μ 1,…,μ q be the pairwise distinct nonzero eigenvalues of A and let r 1,…,r q be their algebraic multiplicities. Denote by p 0(λ) the characteristic polynomial of A+τB. By Theorem 10.2, the lowest possible power of λ associated with a nonzero coefficient in p 0(λ) is an 1 and the corresponding coefficient \(c_{a-n_{1}}\) is

$$c_{a-n_1}=(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1} \Biggr)=0, $$

because of (10.6). If n 1=1 then \(\widetilde{a}=a\) and there is nothing to show as the algebraic multiplicity of the eigenvalue zero cannot increase when a generic perturbation is applied. Otherwise, we distinguish the two cases n 2<n 1−1 and n 2=n 1−1. If n 2<n 1−1, then by Theorem 10.2 the coefficient \(c_{a-n_{1}+1}\) of \(\lambda^{a-n_{1}+1}\) in p 0(λ) is

$$\begin{aligned} c_{a-n_1+1} =&(-1)^a \Biggl(\sum _{\nu=1}^qr_{\nu}\mu_\nu^{r_\nu-1} \prod_{\genfrac{}{}{0pt}{1}{i=1}{i\neq\nu}}^q\mu_{i}^{r_i} \Biggr) \Biggl(\sum_{j=1}^{\ell_1}w^{(1,j)}_{n_1} \Biggr) \\ &{}+(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1-1} \Biggr) \\ =&(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{j=1}^{\ell_1}w^{(1,j)}_{n_1-1} \Biggr) \\ =&(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{s=1}^{\ell_1/2} \bigl(\tau\bigl(g_{n_1,1}^{(1,2s)}+g_{n_1-1,2}^{(1,2s-1)} \bigr)u^{(1,2s)}_{n_1}u^{(1,2s-1)}_{n_1-1} \\ &{}+\tau\bigl(g_{n_1-1,2}^{(1,2s)}+g_{n_1,1}^{(1,2s-1)} \bigr)u^{(1,2s)}_{n_1-1}u^{(1,2s-1)}_{n_1} \\ &{}+\tau\bigl(g_{n_1,2}^{(1,2s)}+g_{n_1,2}^{(1,2s-1)} \bigr)u^{(1,2s-1)}_{n_1}u^{(1,2s)}_{n_1} \bigr) \Biggr) \end{aligned}$$

by (10.7), where we have used (10.6) to conclude that \(\sum_{j=1}^{\ell_{1}}w^{(1,j)}_{n_{1}}=0\) in the second equation. By the hypothesis (2b), it follows that \(c_{a-n_{1}+1}\) generically is nonzero. If, on the other hand, n 2=n 1−1, then again by Theorem 10.2 (and using (10.6)) the coefficient \(c_{a-n_{1}+1}\) of \(\lambda^{a-n_{1}+1}\) in p 0(λ) is

$$c_{a-n_1+1}=(-1)^{a-1} \Biggl(\prod_{i=1}^q \mu_{i}^{r_i} \Biggr) \Biggl(\sum _{s=1}^{\ell_1}w^{(1,s)}_{n_1-1} +\sum _{j=1}^{\ell_2}w^{(2,j)}_{n_2} \Biggr), $$

so in comparison to the case n 2<n 1−1, there is an extra term in \(c_{a-n_{1}+1}\) depending on \(w^{(2,j)}_{n_{2}}\), j=1,…, 2. However, each entry \(w^{(2,j)}_{n_{2}}\) only depends on the entries of the vectors u (2,s), s=1,…, 2, so still \(c_{a-n_{1}+1}\) is nonzero generically. In all cases, we have shown that zero is a root of p 0(λ) with multiplicity an 1+1. Thus, the algebraic multiplicity of the eigenvalue zero of A+τB is an 1+1. Together with Step 1, we obtain that (5.7) generically is the only possible Jordan canonical form of A+τB. □

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Mehl, C., Mehrmann, V., Ran, A.C.M. et al. Eigenvalue perturbation theory of symplectic, orthogonal, and unitary matrices under generic structured rank one perturbations. Bit Numer Math 54, 219–255 (2014). https://doi.org/10.1007/s10543-013-0451-3

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  • Symplectic matrix
  • Orthogonal matrix
  • Unitary matrix
  • Indefinite inner product
  • Cayley transformation
  • Perturbation analysis
  • Generic perturbation
  • Rank one perturbation

Mathematics Subject Classification (2010)

  • 15A63
  • 15A21
  • 15A57
  • 47A55
  • 93B10
  • 93B35
  • 93C73