# Inequality for trace of product of matrices

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 $$B$$.

• What does the subscript “>” mean? – Avi Steiner Feb 7 at 15:56
• Sorry, I edited the question to state it more clearly. – sleepingrabbit Feb 7 at 16:01
• No. Take $A=-I_n$... If $A$ (Instead of $B$) is positive semidefinite, I am confident that the inequality will hold though. – Mindlack Feb 7 at 16:25

## 1 Answer

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