# How are these equivalent for normal matrices?

If $A+\epsilon$ (where $\epsilon$ is a perturbation on the order of machine precision,) has an eigenvalue $\lambda$, and $A$ is a normal matrix, then the eigendecomposition of $A$ is $A=Q \Lambda Q^{*}$, where $Q$ is unitary.

Under those conditions, I have seen it written in my textbook that $\|(\lambda I-A)^{-1}\|=\|Q(\lambda I-\Lambda )^{-1} Q^{*}\|$. But I cannot figure out why these two things are equal.

Can someone show me why/how these are equivalent?

Thank you.

The $\|\cdot\|$ denotes the $2$-norm.

• It's a fact that for both the spectral (2-) norm and the Frobenius norm, the norm is invariant under multiplication by an orthogonal matrix. This is not true for most other matrix norms. – Brian Borchers Dec 12 '17 at 6:02
• $A$ has eigenvalue $\lambda$, so $A-\lambda I$ is not injective, so why it is invertible? – GNUSupporter 8964民主女神 地下教會 Dec 12 '17 at 6:18
• Ah, really really good point GN Supporter. I think I copied the statement from the textbook wrong....let me check. – nundo Dec 12 '17 at 6:36
• @BrianBorchers yes I understand that the Q's do not affect the norm (2-norm, or Fro norm). But then why are we able to stick the $\Lambda$ in there? – nundo Dec 12 '17 at 6:36
• It would help to provide context if you cited the source that you got this from. – Brian Borchers Dec 12 '17 at 16:24

This clearly isn't an exact equation, but rather a relationship that holds in the limit as $\| \epsilon \|$ goes to 0.

There's a notational confusion here, in that $\lambda$ in the statement of the problem is an eigenvalue of the perturbed matrix, and $\Lambda$ is the diagonal matrix of eigenvalues of the original matrix. I'll change the notation to use $\mu$ as an eigenvalue of the perturbed matrix $A+\epsilon$, and restate the proposition as

Assume that $A$ and $A+\epsilon$ are normal matrices, $A$ has been diagonalized as $A=Q\Lambda Q^{*}$ and that $\| \epsilon \|$ is a small perturbation. In the limit as $\| \epsilon \|$ goes to 0, if $\mu$ is an eigenvalue of $A+\epsilon$, then to first order, $\| (\mu I - A)^{-1} \|_{2} \approx \| Q(\mu I - \Lambda)^{-1} Q^{*} \|_{2}$

Since $\mu$ is an eigenvalue of the perturbed matrix, it is close to an eigenvalue $\lambda_{i}$ of the original matrix $A$, with corresponding eigenvector $v_{i}$. The perturbation theory for eigenvalues tells us that to first order,

$\mu \approx \lambda_{i} + v_{i}^{T}\epsilon v_{i}$

Thus the smallest eigenvalue of $\mu I - A$ is to first order,

$\mu_{smallest}(\mu I - A) \approx v_{i}^{T} \epsilon v_{i}$

Then to first order,

$\| (\mu I - A)^{-1} \|_{2} \approx \frac{1}{| v_{i}^{T} \epsilon v_{i} |}$

On the other side, the smallest eigenvalue of $\mu I - \Lambda$ is $\mu - \lambda_{i} \approx v_{i}^{T} \epsilon v_{i}$. Thus to first order,

$\| (\mu I - \Lambda)^{-1} \|_{2} = \| Q(\mu I - \Lambda)^{-1} Q^{*} \|_{2} \approx \frac{1}{| v_{i}^{T} \epsilon v_{i} |}$

You can easily check this in MATLAB by creating random symmetric matrices $A$ and $\epsilon$ with $\| \epsilon \| \ll \| A \|$.

Note that this fails if for some reason $v_{i}^{T}\epsilon v_{i}=0$. I believe that repeated eigenvalues in $A$ can also cause problems.