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Given a diagonal matrix $\bf D$ with diagonal entries $d_{ii} \in [0,1]$ and a matrix $\bf W$ with singular values $\sigma_i ({\bf W}) \in [0,1]$, can it be proven that the eigenvalues of $\bf W D W^\top$ lie in the interval $[0,1]$?

It's worth noting that $\bf W D W^\top$ is positive semidefinite, ensuring real non-negative eigenvalues.

Any insights or proofs would be greatly appreciated.

1 Answers1

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The answer is YES. To prove it, you can prove that $$\mathbf I - \mathbf W \mathbf D\mathbf W^\intercal \succeq 0$$

Indeed,

  • $\mathbf W\mathbf W^{\intercal} - \mathbf W \mathbf D\mathbf W^\intercal = \mathbf W \underbrace{\left(\mathbf I - \mathbf D\right)}_{\succeq 0}\mathbf W^\intercal \succeq 0$

  • The eigenvalues of $\mathbf W\mathbf W^\intercal$ are the singular values of $\mathbf W$ squared. Since these values lie in $[0, 1]$ then the eigenvalues of $\mathbf W\mathbf W^\intercal$ will also lie in $[0, 1]$ and that proves that, $$\mathbf I - \mathbf W\mathbf W^\intercal \succeq 0.$$

Kroki
  • 13,135
  • Thank you for your response. Could you explain why the inequality $\mathbf{I} - \mathbf{W} \mathbf{D} \mathbf{W}^\intercal \succeq 0$ leads to the conclusion that the eigenvalues of $\mathbf{W} \mathbf{D} \mathbf{W}^\intercal$ are all less than 1? – koren abitbul Jul 19 '23 at 20:47
  • You are using $\mathbf I \succeq \mathbf W\mathbf W^\intercal$ and $\mathbf W\mathbf W^{\intercal} \succeq \mathbf W \mathbf D\mathbf W^\intercal$ to conclude that $\mathbf I \succeq \mathbf W \mathbf D\mathbf W^\intercal$, right? – Rodrigo de Azevedo Jul 19 '23 at 22:01
  • Yes or you can see the sum of the two matrices and you will find the first matrix – Kroki Jul 20 '23 at 01:03
  • @korenabitbul since the eigenvalues of $\mathbf I$ are all equal to $1$ and the matrix that I show is a semidefinite positive matrix then all eigenvalues of your matrix are smaller than $1$ – Kroki Jul 20 '23 at 01:05
  • Thank you very much! – koren abitbul Jul 20 '23 at 08:21