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Let $x \in \mathbb{R}^d$, and suppose that $A$ is symmetric positive definite with $A = \mathbb{E}[xx^T]+ \delta I$, $\delta$ is a small positive constant. The diagonal of $A$ is $D = \text{diag}(A)$.

Suppose $\lambda_{\min}(A)$, $\lambda_{\max}(A)$ are the smallest and the largest eigenvalue of $A$ respectively, and $\lambda_{\min}(D)$, $\lambda_{\max}(D)$ are the smallest and the largest eigenvalue of $D$ respectively.

It is obviously that $\text{tr}(A) = \text{tr}(D)$. By Hadamard inequality, we have $\text{det}(A) \leq \text{det}(D)$. Do the following inequalities hold: \begin{gather} \lambda_{\min}(A) \leq \lambda_{\min}(D) \\ \lambda_{\max}(A) \geq \lambda_{\max}(D) \end{gather}

I have tested it by programming and found these inequalities hold. If true, how to prove it?

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2 Answers 2

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It is true for any matrix $A$ hermitian. It is called the maximum principle.

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$$\lambda_{\max}(A) = \max_{u^Tu=1}u^TAu \geq \max_{i \in \{1, \ldots,d \}}e_i^TAe_i=\lambda_{\max}(D)$$

$$\lambda_{\min}(A) = \min_{u^Tu=1}u^TAu \leq \min_{i \in \{1, \ldots,d \}}e_i^TAe_i=\lambda_{\min}(D)$$

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