Is it correct:$\lambda_{\min}(A) \langle I,B\rangle \le \langle A,B \rangle \le \lambda_{\max}(A) \langle I,B\rangle$?

Question

Assume that $A$ is a semi-definite matrix and $B$ is not. Also $\langle,\rangle$ is the standard inner product on the semi-definite cone. $\lambda_{\min}$ and $\lambda_{\max}$ are minimum and maximum eigen value of the matrix repectively.

Is the following inequality is correct? if it's not what is the weakest condition that it is correct?

$$\lambda_{\min}(A) \langle I,B\rangle \le \langle A,B \rangle \le \lambda_{\max}(A) \langle I,B\rangle$$

It is well known that if $B$ is positive semidefinite it's true, but I want the weakest condition on $B$. Based on the comment and an answer, it is not true for the general $B$, also it's true for the semidefinte $B$. Is the weakest condition is semidefiniteness of $B$. If it is, isn't any other upper/lower bound for this quantity based on eigen values of $A, B$?

Answer 1

The inequality always holds in the scalar case. So, we may consider only matrices whose sizes are at least $2\times2$. Also, as both $A$ and $I$ are Hermitian, the skew-Hermitian part of $B$ is irrelevant in the inner products. Hence we may assume that $B$ is Hermitian. Now, we claim that:

Let $B$ be a Hermitian matrix of size at least $2\times2$. Then $\lambda_{\min}(A) \langle I,B\rangle \le \langle A,B \rangle \le \lambda_{\max}(A) \langle I,B\rangle$ for every semidefinite $A$ if and only if $B$ is positive semidefinite.

Proof. If $B$ is positive semidefinite, by a change of orthonormal basis, we may assume that $A$ is diagonal and the inequality follows immediately.

If $B$ is not positive semidefinite, let $u$ be a unit eigenvector corresponding to a negative eigenvalue $-\lambda$ of $B$ and let $A=uu^\ast$. Then $\lambda_\min(A)=0$ and hence $\lambda_\min(A)\langle I,B\rangle=0>-\lambda = \langle A,B \rangle$.

Answer 2

As the earlier comment states, this does not hold for arbitrary matrices $B$. However, the statement holds if $B$ is positive semidefinite.

In order to prove that the inequality holds for positive semidefinite $A,B$, it suffices to note that $A - \lambda_{min}I$ and $\lambda_{max}I-A$ are both positive semidefinite, and that $\langle M,N\rangle \geq 0$ for any $M,N$ in the semidefinite cone.