$A-mI$ as positive semidefinite matrix 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.
 A: 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...
A: The question makes sense because all eigenvalues of $A$ are real. Now apply the definitions.
Let $\lambda$ be an eigenvalue of $A$, with eigenvector $v$. Then
$$
v^T(A-mI)v=v^TAv-v^T(mv)=v^T(\lambda v)-v^T(mv)=(\lambda-m)v^Tv
$$
has to be $\ge0$. Therefore $\lambda\ge m$. In particular this holds for the minimum eigenvalue.

One could use the fact that the eigenvalues of $A-mI$ are of the form $\lambda-m$, where $\lambda$ is an eigenvalue of $A$, and that a symmetric matrix is positive definite if and only if its eigenvalues are $\ge0$, but it's overkill. The above proof only uses the definition and the important fact that the eigenvalues of $A$ are real.
A: We have the result,
A symmetric matrix $A$ is positive (semi)definite if and only if all eigenvalues $\lambda$ of $A$ satisfies $\lambda(\geq)>0$.
Another result,
For any polynomial $p(x)$, $p(\lambda)$ is an eigenvalue of a matrix $p(A)$ if and only if $\lambda$ is an eigenvalue of $A$.
From this result, $A-m\lambda$ has eigenvalues $\{\lambda-m:\lambda$ is an eigenvalue of $A$}. Rest you can do.
