if $A\in M_{n×n}^{\mathbb{C}}$ and self-adjoint then $\exists t\in \mathbb{R}$ such that $A-tI$ is a negative-definite matrix

Question

I know that if $A$ is self-adjoint then all the eigenvalues of $A$ are real.And also $A$ is unitary diagonalization over the complex numbers. Therefore $A$ has a bases $B=\{v_1, v_2,..., v_n\}$ of eigenvectors of $A$.

I would like to know please how to use this information and proceed with the proof.

Answer 1

Notice that $B$ is negative-definite if and only if

$$\langle Bv, v\rangle < 0$$

for all $v\neq 0$. With $B=A-tI$ we get

$$\langle Av, v\rangle -t\langle v, v\rangle < 0.$$

Now, using the operator norm (where the ambient space has the usual Euclidean norm) we have

\begin{align} \langle Av, v\rangle -t\langle v, v\rangle &\leqslant \lVert A\rVert_{op}\,\lVert v\rVert^2 - t\lVert v\rVert^2 \\&= \lVert v\rVert^2\left(\lVert A\rVert_{op} - t\right) \end{align}

It is well known, and a good exercise, to show that $\lVert A\rVert_{op}^2 = \max\{|\lambda|\,;\, \lambda \text{ is an eigenvalue of $A$}\}$.

Answer 2

Hint: If $Av=\lambda v$, then $(A-tI)v=(\lambda-t)v$.

Now, since $A$ has finitely many real eigenvalues, it has a maximum eigenvalue.