Let $A, B \in \mathbb{R}^{n \times n}$ be symmetric positive semidefinite matrices and:
\begin{equation}
\lambda_{max} (A) \leq \lambda_{max} (B)
\end{equation}
Can we prove that the matrix $C = A – B$ is negative semi-definite ?
Proving this statement would imply proving the following:
\begin{equation}
\sup_{x \neq 0} x^* (A – B)x = \sup_{x \neq 0} \left[ x^* A x – x^*Bx \right] \leq 0
\end{equation}
While we can only show that:
\begin{equation}
\sup_{x \neq 0} x^* A x – \sup_{y \neq 0} y^*B y \leq 0
\end{equation}
Am I missing something ? Thank you !
Best Answer
This is not true.
$$A=\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}, B=\begin{bmatrix}2 & 0 \\ 0 & 1\end{bmatrix}$$
We have $\lambda_{max}(A)=\lambda_{max}(B)$.
We have $A-B=\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$ which is positive semidefinite.
$$A=\begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix}, B= \begin{bmatrix}3 & 0 \\ 0 & 1 \end{bmatrix}$$
We have $\lambda_{max}(A) < \lambda_{max}(B)$ but
$$A-B=\begin{bmatrix} -1 & 0 \\ 0 & 1\end{bmatrix}$$
is not negative semidefinite.