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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!

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1 Answer 1

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Guide:

Let $v$ be an eigenvector,

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

Hence

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

Can you solve for $\lambda$?

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  • $\begingroup$ I would probably emphasize that a symmetric matrix has a full collection of genuine real eigenvalues, and a basis of eigenvectors $\endgroup$
    – Will Jagy
    Jan 29, 2018 at 2:35
  • $\begingroup$ 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 $\endgroup$
    – Anthony
    Jan 29, 2018 at 2:39
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    $\begingroup$ @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. $\endgroup$
    – imranfat
    Jan 29, 2018 at 3:14

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