Difference between matrices with altered eigenvalues

Question

Given two p.s.d. matrices $X_1$ and $X_2$ with eigen decomposition $X_1 = U_1V_1U_1^T$ and $X_2 = U_2V_2U_2^T$ and a constant $\lambda > 0$

Now consider an altered version of the eigenvalue $\hat{V}_i = diag(\frac{\{ V_i \}_{1,1}}{\{ V_i \}_{1,1} + \lambda}, \cdots, \frac{\{ V_i \}_{N,N}}{\{ V_i \}_{N,N} + \lambda})$ for $i\in \{1,2\}$.

Is $|| U_1\hat{V}_1U_1^T - U_2 \hat{V}_2U_2^T ||_F \leq || X_1 - X_2 ||_F$?

If yes, how to show this?

p.s. $||\cdot||_F$ is the Frobenius norm defined as $||X||_F=\sqrt{trace(X^TX)}$

Answer 1

The answer is no just looking at scalars.

Take $X_2 = 0$, $X_1 = {1 \over 2}$, and $\lambda >0$.

Then $\hat{V}_1 = {1 \over 1+ 2 \lambda}$, and $\| \hat{V}_1 \|_F = {1 \over 1+ 2 \lambda}$, whereas $\|X_1\|_F = {1 \over 2}$. By choosing $\lambda>0$ sufficiently small we see that $\|\hat{V}_1\|_F \not\le \|X_1\|_F$.