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What is the best way to show if $A$ is symmetric then $A^2$ is as well using eigenvalues?

Turbo
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  • Hint: It might actually be easier to see what to do if I ask you to show that if $A$ and $B$ are matrices which are symmetric and commute then also $AB$ is symmetric. – Tobias Kildetoft Oct 31 '16 at 08:52

1 Answers1

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Just recall that for two matrices $B,C$ it holds that: $(BC)^T = C^TB^T$. Applying this with $B=A$, $C=A$, we get:

$$ (A^2)^T = (AA)^T = A^TA^T = AA = A^2$$

You can also do it with eigenvalues but it seems a lot more tedious to me (and requires much stronger results):

Let $A=VDV^T$ be the eigenvalue decomposition of $A$ (with $D$ diagonal, $V$ orthonormal, the decomposition exists by spectral theorem), then:

$$ A^2 = VD^2V^T $$

And it is easy to see that this is symmetric as well. In particular, apply the identity I cited at the beginning and note that $D^2$ is symmetric since it is diagonal.

air
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