The title is more general, and all that I require is to show an inequality that I already have verified using random matrices in matlab.

Let $\lambda_1 \leq ... \leq \lambda$ and $\mu_1 \leq ... \leq \mu_n$ all in $\mathbb{R}$

let $X=(x_{ij}) \in \mathbb{C}^{n \times n}$ a unitary matrix ($\in U(n)$)

show: $$\max_{|v|=1}|\langle ( diag(\lambda_1,...,\lambda_n) -X^*diag(\mu_1,...,\mu_n)X)v , v \rangle| \geq \max_{i=1,...,n}|\lambda_i - \mu_i| $$

or equivalently show: $$ \text{absolute maximum eigenvalue of } \\ diag(\lambda_1,...,\lambda_n) -X^*diag(\mu_1,...,\mu_n)X \text{ is } \geq \max_{i=1,...,n}|\lambda_i - \mu_i|$$

Equality happens if for example $X=Id$. I would be happy enough to show this for $\mathbb{R}$ and orthogonal matrices $O(n)$ instead of $\mathbb{C}$ and $U(n)$.

Background: this is the final step for showing that if a hermitian matrix $M$ has eigenvalues $\mu_i$ to eigenvectors $w_i, \ W=(w_1,...,w_n)$, then the closest hermitian matrix to $M$ with given eigenvalues $\lambda_i$ is $W^*diag(\lambda_1,...,\lambda_n)W$.

For n = 3 to 6 I have generated random eigenvalues and then 100.000 random orthogonal matrices in matlab. The computations confirm this inequality. Please help proving it. I gladly cite the solution where I need it and share the link to the finished work here.

thank you!

  • $\begingroup$ If you are interested. I have also verfied by experiments that: $\endgroup$ – RandomMathDude Aug 21 '15 at 11:08
  • $\begingroup$ $$ \max_{i,j=1,...,n}|\lambda_i - \mu_j| \geq \max_{|v|=1}|\langle ( diag(\lambda_1,...,\lambda_n) -X^*diag(\mu_1,...,\mu_n)X)x , x \rangle| $$ $\endgroup$ – RandomMathDude Aug 21 '15 at 11:10
  • $\begingroup$ $diag()$ is the diagonal matrix with the mentioned entries $\endgroup$ – RandomMathDude Aug 21 '15 at 11:15
  • $\begingroup$ Is $v$ a typo and should be $x$? or vice versa $\endgroup$ – user251257 Aug 21 '15 at 12:37
  • $\begingroup$ ah yes its a typo. thanks $\endgroup$ – RandomMathDude Aug 21 '15 at 13:27

Here is the answer:


Check the eigenvalue inequality with it's proof. It is based on Weyl's inequality.

  • $\begingroup$ Though the link does point to a paper which sheds light on the question, it would be best to include essential parts in the answer itself. $\endgroup$ – Shailesh Mar 13 '17 at 15:37

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