# Proving symmetric matrix has positive eigenvalues

I'm trying to solve the following problem

Let $$A$$ be $$n \times n$$ symmetric matrix such that $$A^2-5A+6Id_n = 0$$

where $$Id_n$$ stands for the identity matrix.

Show that $$A$$ has only positive eigenvalues.

I know what eigenvalues are, what a symmetric matrix is, and everything else in the problem. I'm just pretty confused on how to start this proof/what it would look like. Any help would be great!

Guide:

Let $v$ be an eigenvector,

$$A^2v-5Av + 6v = 0$$

Hence

$$(\lambda^2-5\lambda+6)v=0$$

Can you solve for $\lambda$?

• I would probably emphasize that a symmetric matrix has a full collection of genuine real eigenvalues, and a basis of eigenvectors Jan 29, 2018 at 2:35
• wow that's such a clever and quick trick to do that. So I'd factor it to (λ-3)(λ-2), so both eigenvalues are positive. I love that Jan 29, 2018 at 2:39
• @Anthony That trick is explained through the Cayley Hamilton theorem. In short that theorem states that a matrix satisfies its own Characteristic Equation. You can look up on Wikipedia. Jan 29, 2018 at 3:14