1

I am trying to show that

$| \lambda_n(A) - \lambda_n(B)| \leq \lVert{\mathbf{A} - \mathbf{B}} \rVert_F $

where $A, B \in \mathbb{R}_{sym}^{nxn}$ and $\lambda_n(A)$ denotes the largest eigenvalue of $A$.

The largest eigenvalue for $A$ is defined as $\max_{v \in \mathbb{R}^n, \lVert v \rVert_2 = 1} v^TAv$, and for B, is similar.

I know this is related to Weyl's Inequality in some sense but cannot construct proof to show this.

nondefinite
  • 93
  • If you use the Rayleigh quotient you'll see $ v^TAv =\lambda_{n}(A)$ and is maximized by the corresponding eigenvector. Under the norm $| A |{F} = \lambda{n}(A)$ – Ryan Howe Dec 09 '20 at 14:17
  • @RyanHowe I am not familiar with Rayleigh quotient, can you elaborate that equality? – nondefinite Dec 09 '20 at 14:23
  • I was just pointing out that bit about the largest eigenvalue https://en.wikipedia.org/wiki/Rayleigh_quotient is this. It's explained here (https://math.stackexchange.com/questions/9302/norm-of-a-symmetric-matrix). I'll try to answer more later. This is more extensive (https://terrytao.wordpress.com/2010/01/12/254a-notes-3a-eigenvalues-and-sums-of-hermitian-matrices/) – Ryan Howe Dec 09 '20 at 14:41
  • @RyanHowe I checked all three links but still could not understand how the Frobenius norm of a matrix is equal to the largest eigenvalue. – nondefinite Dec 09 '20 at 17:00
  • It's not in general. It is for a symmetric matrix. When it's symmetric then it's equal to the 2-norm.

    https://math.stackexchange.com/questions/9302/norm-of-a-symmetric-matrix https://math.stackexchange.com/questions/252819/why-is-frobenius-norm-of-a-matrix-greater-than-or-equal-to-the-2-norm

     – Ryan Howe Dec 09 '20 at 17:42

2 Answers2

2

If we use the property you have it says

"The largest eigenavalue for A is defined as $ \max_{v \in \mathbb{R} , \|v\|=1} v^{T}Av$ "

Now if $A,B$ are symmetric then $A-B$ is also symmetric and using that we have

$$ \lambda_{\text{max}}(A-B) = \max_{v \in \mathbb{R} , \|v\|=1} v^{T}(A-B)v $$

Now you have to show the rest using properties of $\max$ note that

$$ \max\{ f(x) + g(x) \} \leq \max \{ f(x)\} + \max \{ g(x) \}$$

and if $\theta < 0$ $$ \max \theta f(x) = \theta \min f(x) $$

I think this may help.

Ryan Howe
  • 402
2

With an additional assumption, you don't even need to argue with $\operatorname{max}$.

Suppose that A and B are positively semidefinite. We have [1, 2], $A=M^2, \ B=N^2$ for some $M,N$ symmetric. $\Vert{A-B}\Vert_F \\ = \Vert M^2-N^2\Vert_F \\ \geq \Vert M^2-N^2\Vert_2 \\ \geq \left| \Vert M^2 \Vert_2 - \Vert N^2\Vert_2 \right| \\ = \left| \Vert M \Vert_2^2-\Vert N \Vert_2^2\right| \\ = \left| \sqrt{\lambda_{max}(M^2)}^2 - \sqrt{\lambda_{max}(N^2)}^2\right| \\ = \left |\lambda_{max}(A) - \lambda_{max}(B)\right|$.

テレビ スクリーン
  • 641