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Let $A$ be a k-by-k matrix and $\sigma(A)$ its spectrum, or the collection of eigenvalues of $A$.

If we know $\lambda\notin\sigma(A)$, then $\lambda$ is at a positive distance to all points in the spectrum since the latter is compact.

I wonder whether there is a bound for the norm of the inverse of $\lambda I-A$, maybe in terms of the distance from $\lambda$ to the spectrum. You can use all kinds of norms on $\|(\lambda I-A)^{-1}\|$.

Thanks!

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What do you want such a bound for? Have you tried the spectral radius formula? –  Qiaochu Yuan Jun 15 '12 at 20:17
    
Is there any interest in cases where $\lambda $ is "near" some point in $\sigma (A)$? –  user17794 Jun 15 '12 at 20:35
    
@QiaochuYuan I am considering the spectrum of T, a direct sum of countably many matrices, A_n, of bounded sizes. If $\lambda$ is not in the spectra of any of the matrices, then each $\lambda-A_n$ would be invertible, but it remains to prove the norms of their inverses are uniformly bounded. –  Hui Yu Jun 15 '12 at 20:42
    
@Hui: there's no reason to believe that the norms of their inverses are uniformly bounded. Take a direct sum of $1 \times 1$ matrices with entries $1 - \frac{1}{n}, n \in \mathbb{N}$ and $\lambda = 1$. –  Qiaochu Yuan Jun 15 '12 at 20:45
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@Norbert Well, I checked that paper. The bad thing is that it gives only the estimates for self-adjoint operators and operators differ from self-adjoint ones 'not much', and thus is quite restrictive. I guess there would be some more general result when all operators are assumed to be k-by-k matrices. –  Hui Yu Jun 16 '12 at 13:28

1 Answer 1

up vote 2 down vote accepted

Here is a crude upper bound. It might not be good enough for what you are after, but without further constraints on your matrices, it is hard to see how one can do substantially better.

Throughout, I am using the operator norm. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Cplx}{{\bf C}}\newcommand{\lm}{\lambda}$

Fix $k$. Let $A$ be a $k\times k$ matrix with complex entries. Schur's theorem from linear algebra tells us that there is an upper triangular matrix $B$ and a unitary matrix $U$ such that $A=U^*BU$.

Let $d_1,\dots, d_n$ be the diagonal entries of $B$ (these form the spectrum of $B$, and hence of $A$, as we will shortly see). For $\lm\in\Cplx\setminus\{d_1,\dots,d_n\}$, let $D_\lambda$ be the diagonal matrix whose entries are $(\lm_1-d_1)^{-1}, \dots, (\lm_k-d_k)^{-1}$, and put $$ C_\lm = D_\lm^{-1} (\lm I - B) $$ Then $C_\lambda$ is an upper triangular matrix with each diagonal entry equal to $1$. Since $(C_\lm - I)^k=0$, $C_\lm$ is $I$ + a nilpotent matrix, and so is invertible. In fact, just by the usual formula for $(1+x)^{-1}$, we have $$ C_\lm^{-1} = \sum_{j=0}^{k-1} (-1)^j(C_\lm-I)^j $$ and thus $$ (\lm I- B)^{-1} =(D_\lm C_\lm)^{-1} = \sum_{j=0}^{k-1} (-1)^j(C_\lm-I)^j D_\lm^{-1} $$

Now we just use the triangle identity and submultiplicativity of the norm. Let $d(\lm) = {\rm dist}(\lm, \sigma(B)) = \min_j \vert d_j-\lm\vert$. Then $\norm{D_\lm^{-1}} = d(\lm)^{-1}$, and $$ \norm{ (\lm I- B)^{-1} } \leq \sum_{j=0}^{k-1} d(\lm)^{-j-1} \norm{\lm I - B}^j = d(\lm)^{-1} \frac{d(\lm)^{-k}\norm{\lm I - B}^k - 1}{d(\lm)^{-1}\norm{\lm I-B} -1 } . $$ Since $B$ is unitarily equivalent to $A$, the same inequality holds with $B$ replaced by $A$.

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Thanks, Yemon. Actually your answer suffices for my problem. As you said in your comment, we need to put a bound for the norm of the matrices in the direct sum, but there is already one bound for them. Since their direct sum of them is a bounded operator, the matrices has an uniform bound for their norms. Then we can use this norm in your last inequality to get a bound for the norm of their inverses. –  Hui Yu Jun 18 '12 at 15:45
    
On the other hand, I believe your bound is actually sharp. Maybe we can find a matrix for which the equality holds. –  Hui Yu Jun 18 '12 at 15:46

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