A problem about positive-define

Question

Suppose $A$ is a real symmetric matrix, and let $h$ be the maximal eigenvalue of $A$.

Claim: $hE-A$ is positive-define, where E is the identity matrix.

I know that if $x$ is a eigenvector of $A$, then $x^T(hE-A)x \ge 0.$

But I have no idea that how to prove the case when $x$ is not a eigenvector.

Can someone give me a hint or a solution to this problem? Thanks a lot.

Answer 1

If you look at the characteristic polynomial of $hE-A$, you have $$\det(hE-A-\lambda E)=\det((h-\lambda )E-A).$$ So the eigenvalues of $hE-A$ are of the form $h-\lambda$, where $\lambda$ is an eigenvalue of $A$.

Thus $hE-A$ is symmetric and all its eigenvalues are nonnegative.

Answer 2

Since $A\in\mathbb{R}^{n\times n}$ is symmetric, there exists a orthogonal basis of eigenvectors of $A$ so every vector $x\in\mathbb{R}^n$ can be written as the linear combination of eigenvectors of $A$. This follows from the spectral theorem for symmetric matrices.