Accurate Eigenvalues of Some Generalized Sign Regular Matrices via Relatively Robust Representations

Abstract

In this paper, we consider how to accurately solve the nonsymmetric eigenvalue problem for a class of generalized sign regular matrices including extremely ill-conditioned quasi-Cauchy and quasi-Vandermonde matrices. The problem of performing accurate computations with structured matrices is very much a representation problem. We first develop a relatively robust representation (RRR) for this class of matrices by introducing a free parameter, which exceeds an essential threshold, into an indefinite factorization. We then design a new $O(n^{3})$ algorithm to compute all the eigenvalues of such matrices with high relative accuracy, as warranted by the RRR. Error analysis and numerical experiments are performed to illustrate the high relative accuracy.

Appendix: The Proof of Theorem 2

Denote by $Q_{k,n}$ the set of strictly increasing sequences of k positive integer numbers less than or equal to n. The following result is derived by using a similar argument as that of [27, Theorem 2.6].

Lemma 7

Let $T=:\mathbb {PM}(T)\in {{\mathbb {R}}}^{n \times n}$ ($n> 2$) be nonsingular tridiagonal with $\mathbb {PM}(T)\ge 0$, and let $\tilde{T}\in {{\mathbb {R}}}^{n \times n}$ be obtained from T only by replacing one entry x of $\mathbb {PM}(T)$ with $\tilde{x}=x(1+\epsilon _{x})$, where $|\epsilon _{x}|\le \epsilon$ and $7\epsilon <1$. Then

$$\begin{aligned} |\mathrm{det}\tilde{T}[\mu |\nu ]-\mathrm{det}T[\mu |\nu ]|\le \frac{7\epsilon }{1-7\epsilon }|\mathrm{det}T[\mu |\nu ]|,~\forall ~\mu ,\nu \in Q_{k,n},~1\le k\le n. \end{aligned}$$

Proof

Set $T\in {{\mathbb {R}}}^{n \times n}$ be of the form (3). Then $\mathrm{det}T=-d_{1}d_{2}\ldots d_{n}$, and trivially, $|\mathrm{det}\tilde{T}-\mathrm{det}T|\le \frac{\epsilon }{1-\epsilon }|\mathrm{det}T|$. Now consider any minor $\mathrm{det}T[\mu |\nu ]$ for $\mu =(\mu _{i}),\nu =(\nu _{i})\in Q_{k,n}$ with $1\le k \le n-1$. Assume that $1\le z_{1}<\ldots <z_{r}\le n$ are all indices such that $\mu _{z_{s}}\ne \nu _{z_{s}}$ for $s=1,2,\ldots ,r$, and let $\gamma =\mu {\setminus } \{\mu _{z_{1}},\ldots ,\mu _{z_{r}}\}$. Since $T=(t_{ij})\in {{\mathbb {R}}}^{n \times n}$ is tridiagonal, we have

$$\begin{aligned} \mathrm{det}T[\mu |\nu ]=t_{\mu _{z_{1}},\nu _{z_{1}}}\ldots t_{\mu _{z_{r}},\nu _{z_{r}}}\cdot \mathrm{det}T[\gamma ], \end{aligned}$$

(40)

where for all $1\le s\le r$,

$$\begin{aligned} \left\{ \begin{array}{lll}t_{\mu _{z_{s}},\nu _{z_{s}}}=0,&{}\quad \mathrm{if}~|\mu _{z_{s}}-\nu _{z_{s}}|>1,\\ t_{\mu _{z_{s}},\nu _{z_{s}}}=\beta _{\mu _{z_{s}},\nu _{z_{s}}}d_{\nu _{z_{s}}},&{}\quad \mathrm{if}~ \mu _{z_{s}}-\nu _{z_{s}}=1~\mathrm{and}~\mu _{z_{s}}\ne n,\\ t_{\mu _{z_{s}},\nu _{z_{s}}}= d_{n-1},&{}\quad \mathrm{if}~ \mu _{z_{s}}-\nu _{z_{s}}=1~\mathrm{and}~\mu _{z_{s}}= n,\\ t_{\mu _{z_{s}},\nu _{z_{s}}}=\alpha _{\mu _{z_{s}},\nu _{z_{s}}}d_{\mu _{z_{s}}},&{}\quad \mathrm{if}~ \nu _{z_{s}}-\mu _{z_{s}}=1~\mathrm{and}~\nu _{z_{s}}\ne n, \end{array}\right. \end{aligned}$$

(41)

and

$$\begin{aligned} t_{n-1,n}=d_{n}+d_{n-1}\beta _{n,n-1}\alpha _{n-1,n},~\mathrm{if}~ \nu _{z_{s}}-\mu _{z_{s}}=1~\mathrm{and}~\nu _{z_{s}}= n. \end{aligned}$$

Remind that only one parameter in $\mathbb {PM}(T)$ is perturbed. Thus, there is at most one entry $t_{\mu _{z_{s}},\nu _{z_{s}}}$ of (41) to be perturbed as

$$\begin{aligned} |\tilde{t}_{\mu _{z_{s}},\nu _{z_{s}}}-t_{\mu _{z_{s}},\nu _{z_{s}}}|\le \frac{\epsilon }{1-\epsilon }|t_{\mu _{z_{s}},\nu _{z_{s}}}|. \end{aligned}$$

(42)

For the entry $t_{n-1,n}$, by considering that $\alpha _{n-1,n}=\theta +\mathbb {PM}(T)_{n-1,n}$, where $\theta >0$ is computed in a subtraction-free manner by applying (8) and (9) to $\mathbb {PM}(T)$, we get that $|\tilde{\theta }-\theta |\le \frac{2\epsilon }{1-2\epsilon }|\theta |$, thus,

$$\begin{aligned} |\tilde{\alpha }_{n-1,n}-\alpha _{n-1,n}|\le \frac{2\epsilon }{1-2\epsilon }|\alpha _{n-1,n}|, \end{aligned}$$

consequently,

$$\begin{aligned} |\tilde{t}_{n-1,n}-t_{n-1,n}|\le \frac{3\epsilon }{1-3\epsilon }|t_{n-1,n}|. \end{aligned}$$

(43)

In addition, for the minor $\mathrm{det}T[\gamma ]$ with $\gamma =(\gamma _{i})\in \mathbb {R}^{k-r}$, the following statements hold.

• The case $\gamma _{k-r}\ne n$. By [27, the equalities (2.12) and (2.13)], considering that only one parameter in $\mathbb {PM}(T)$ is perturbed, we have

  $$\begin{aligned} |\mathrm{det}\tilde{T}[\gamma ]-\mathrm{det}T[\gamma ]|\le \frac{\epsilon }{1-\epsilon }|\mathrm{det}T[\gamma ]|. \end{aligned}$$

  (44)

• The case $\gamma _{k-r}=n$. By using (10) and (7) in a subtraction-free manner,

  $$\begin{aligned} {\left\{ \begin{array}{ll}\mathrm{det}T[2:n]=\delta d_{n-1} \prod \nolimits _{t=3}^{n}d_{t-2}\alpha _{t-2,t-1}\beta _{t-1,t-2},\\ \mathrm{det}T[t:n]=\frac{\mathrm{det}T[t-1:n]+\prod \limits _{i=t-1}^{n}d_{i}}{d_{t-2}\beta _{t-1,t-2}\alpha _{t-2,t-1}},~t=3,\ldots ,n,\end{array}\right. } \end{aligned}$$

  we have

  $$\begin{aligned} |\mathrm{det}\tilde{T}[t:n]-\mathrm{det}T[t:n]|\le \frac{2\epsilon }{1-2\epsilon }|\mathrm{det}T[t:n]|,\quad t=2,\ldots ,n. \end{aligned}$$

  (45)

  Thus, since

  $$\begin{aligned} \mathrm{det}T[\gamma ]=\mathrm{det}T[l:n]\cdot \mathrm{det}T[\gamma '], \quad \gamma '=\gamma \backslash \{l,\ldots ,n\} \end{aligned}$$

  for some $n-k+r+1\le l\le n$, we have by (44) and (45) that

  $$\begin{aligned} |\mathrm{det}\tilde{T}[\gamma ]-\mathrm{det}T[\gamma ]|\le \frac{3\epsilon }{1-3\epsilon }|\mathrm{det}T[\gamma ]|. \end{aligned}$$

  (46)

Therefore, combining (40) with (42), (43), (44) and (46), we get that

$$\begin{aligned} |\mathrm{det}\tilde{T}[\mu |\nu ]-\mathrm{det}T[\mu |\nu ]|\le \frac{7\epsilon }{1-7\epsilon }|\mathrm{det}T[\mu |\nu ]|. \end{aligned}$$

The result is proved. $\square$

Now we are ready to prove Theorem 2.

Proof of Theorem 2

By Lemma 1, A or $-A$ is similar to $\bar{A}=:|\mathbb {PM}(A)|$. According to the form (3) of A, denote

$$\begin{aligned} K=|B_{1}| \ldots |B_{n-2}|,~ T=|B_{n-1}| P |B_{n}| |D| |C_{n-1}|,~ M=|C_{n-2}|\ldots |C_{1}|. \end{aligned}$$

Then $\bar{A}=KTM$, where K and M are TN, and T is SR with signature $(1,\ldots ,1,-1)$. So, by [2, Theorem 3.1], $\bar{A}$ is SR with signature $(1,\ldots ,1,-1)$. Thus, by [2, Corollary 6.6], all the eigenvalues of $\bar{A}$, and so A, are real. Moreover, by the Cauchy-Binet identity,

$$\begin{aligned} \mathrm{det}\bar{A}[\alpha |\beta ]= & {} \sum \limits _{\omega \in Q_{k,n}}\sum \limits _{\nu \in Q_{k,n}}\mathrm{det}K[\alpha |\omega ]\mathrm{det}T[\omega |\nu ]\mathrm{det}M[\nu |\beta ]\\&\ge 0,~\forall ~\alpha ,\beta \in Q_{k,n},~1\le k\le n-1. \end{aligned}$$

Denote by $\tilde{\bar{A}}=\tilde{K}\tilde{T}\tilde{M}$ the matrix obtained from $\bar{A}$ by perturbing one entry of $|\mathbb {PM}(A)|$. Observe that if the perturbed parameter is from the factor of K or M, then

$$\begin{aligned} |\mathrm{det}\tilde{K}[\alpha |\beta ]-\mathrm{det}K[\alpha |\beta ]|\le & {} \frac{\epsilon }{1-\epsilon }|\mathrm{det}K[\alpha |\beta ]|,~\mathrm{or}~\\ |\mathrm{det}\tilde{M}[\alpha |\beta ]-\mathrm{det}M[\alpha |\beta ]|\le & {} \frac{\epsilon }{1-\epsilon }|\mathrm{det}M[\alpha |\beta ]|; \end{aligned}$$

otherwise, Lemma 7 implies that

$$\begin{aligned} |\mathrm{det}\tilde{T}[\alpha |\beta ]-\mathrm{det}T[\alpha |\beta ]|\le \frac{7\epsilon }{1-7\epsilon }|\mathrm{det}T[\alpha |\beta ]|, \end{aligned}$$

where $\alpha ,\beta \in Q_{k,n}$ with any $1\le k\le n$. So,

$$\begin{aligned} |\mathrm{det}\tilde{\bar{A}}[\alpha |\beta ]-\mathrm{det}\bar{A}[\alpha |\beta ]|\le \frac{7\epsilon }{1-7\epsilon }|\mathrm{det}\bar{A}[\alpha |\beta ]|,~\forall ~\alpha ,\beta \in Q_{k,n},~1\le k\le n. \end{aligned}$$

This also means that all the minors of orders less than n of both $\bar{A}$ and $\tilde{\bar{A}}$ are nonnegative. Thus, the kth ($1\le k\le n-1$) compound matrices $\mathcal {B}^{(k)}=(b^{(k)}_{ij})$ and $\tilde{\mathcal {B}}^{(k)}=(\tilde{b}^{(k)}_{ij})$ of $\bar{A}$ and $\tilde{\bar{A}}$ are nonnegative satisfying that

$$\begin{aligned} |\tilde{b}^{(k)}_{ij}-b^{(k)}_{ij}|\le \frac{7\epsilon }{1-7\epsilon }b^{(k)}_{ij},\quad \forall ~ i,j. \end{aligned}$$

Therefore, using a similar argument as that of [27, Theorem 3.4], we conclude that the result is true by considering Lemma 1. $\square$