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 $B$.

  • $\begingroup$ 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

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 $$ \text{tr}(A(\lambda_{\text{max}}I-B))=\lambda_{\text{max}}\text{tr}(A)-\text{tr}(AB)\ge 0. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.