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Let $A$ be square non-singular matrix of order $n \geq 2$.

  1. If $A$ is symmetric, then $A^2$ is symmetric positive definite.

  2. If $A^2$ is symmetric positive definite, then $A$ is symmetric.

I think I have an idea how to prove (1). Consider $x'A^2x$. If A is symmetric, then $x'A^2x = x'AAx = x'A'Ax = (Ax)'Ax = $ sum of squares that cannot be negative. Hence, $A^2$ is positive semidefinite for sure. Now, given $\det A^2 = ( \det A )^2 \neq 0$, we may conclude that $A^2$ is positive definite (I used the result that states “positive semidefiniteness + non-singularity = positive definiteness”). QED

So, my question is: is my proof correct and how to find a counterexample to (2)?

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  • $\begingroup$ Your proof is correct. Why do you think b is wrong? $\endgroup$ – marwalix May 14 '15 at 8:29
  • $\begingroup$ I have answers and they says (2) is false, but I falied to come up with a counterexample. $\endgroup$ – Mirak May 14 '15 at 8:30
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Your proof for 1. is correct. For 2., you can consider an involution matrix like $$A:=\pmatrix{1&2\\ 1&-1}.$$ We have $A^2=3I_2$ but $A$ is not symmetric.

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    $\begingroup$ How did you come up with such a matrix? $\endgroup$ – Mirak May 14 '15 at 8:41
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    $\begingroup$ @Mirak this is called a involution matrix. See here for yet some other examples:math.stackexchange.com/q/1222359/191404 $\endgroup$ – Vim May 14 '15 at 8:44

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