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Assume that $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix and $B \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix. Is the following statement true $$ \lambda_{\mathrm{min}} \operatorname{tr} A \le \operatorname{tr} (AB) \le \lambda_{\mathrm{max}} \operatorname{tr} A \, ? $$ Here, $\lambda_{\mathrm{min}}$ denotes the smallest eigenvalue of $B$ and $\lambda_{\mathrm{max}}$ denotes the largest eigenvalue of $B$.

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  • $\begingroup$ What does the subscript “>” mean? $\endgroup$ – Avi Steiner Feb 7 at 15:56
  • $\begingroup$ Sorry, I edited the question to state it more clearly. $\endgroup$ – sleepingrabbit Feb 7 at 16:01
  • $\begingroup$ No. Take $A=-I_n$... If $A$ (Instead of $B$) is positive semidefinite, I am confident that the inequality will hold though. $\endgroup$ – Mindlack Feb 7 at 16:25
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The statement is not true, since $A=-I_2$ and $B=\text{diag}(1,2)$ is an obvious counterexample. But if we assume that $A$ is positive definite and $B$ is symmetric, then the statement is true. (In fact, $A\ge O$ is also necessary for the statement to be true.) We begin with the following observation.

If $A,B$ are symmetric, positive definite matrices, then $\text{tr}(AB)\ge 0$.

Let $\sqrt{A}$ denote the square root of $A$. Then, $$\begin{align*} \text{tr}(AB)&=\text{tr}(\sqrt{A}\cdot\sqrt{A}^TB)\\&=\text{tr}(\sqrt{A}^TB\cdot\sqrt{A})\\ &=\sum_{i=1}^n e_i^T \sqrt{A}^TB\sqrt{A}e_i\ge 0 \end{align*}$$ since $B$ is positive definite. $\blacksquare$

Knowing this, note that if $A\ge O$ and $B$ is symmetric, then$$ B-\lambda_{\text{min}}I\ge O,\quad \ \lambda_{\text{max}}I-B\ge 0. $$ Thus,$$ \text{tr}(A(B-\lambda_{\text{min}}I))=\text{tr}(AB)-\lambda_{\text{min}}\text{tr}(A)\ge 0, $$ and $$ \text{tr}(A(\lambda_{\text{max}}I-B))=\lambda_{\text{max}}\text{tr}(A)-\text{tr}(AB)\ge 0. $$

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