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 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.

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 $A$}\}$.
