Let $A$ be a symmetric real matrix. Let $m\in \mathbb{R}$ and consider $$M := A – m I$$ where $I$ denotes the identy matrix. We require that matrix $M$ be positive semidefinite. Why do we need that the minimum eigenvalue of $A$ be at least $m$?
Any help will be appreciated.
Best Answer
First, $\;A\;$ is diagonalizable, even orthogonally, as it is a symmetric matrix. Thus, there is an invertible matrix $\;P\;$ s.t. $\;P^{-1}AP=D\;$ is diagonal. Suppose
$$D=\begin{pmatrix}\lambda_1&0&\ldots&0\\0&\lambda_2&0\ldots&0\\ \ldots&\ldots&\ldots&0\\ 0&0&\ldots&\lambda_n\end{pmatrix}\;,\;\;\text{and we assume}\;\;\lambda_1\le\lambda_2\le\ldots\le\lambda_n$$
Then:
$$P^{-1}(A-mI)P=P^{-1}AP-mI=D-mI=\begin{pmatrix}\lambda_1-m&0&\ldots&0\\0&\lambda_2-m&0\ldots&0\\ \ldots&\ldots&\ldots&0\\ 0&0&\ldots&\lambda_n-m\end{pmatrix}$$
The above is similar to $\;A-mI\;$ , and it is positive semi-definite iff $\;\lambda_k-m\ge0\;,\;\;k=1,2,...,n$, thus...