# finding the shortest distance of a hermitian matrix to a set of hermitian matricies with specific eigenvalues 2-norm

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!

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