# Problem on symmetric matrices

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)?

• Your proof is correct. Why do you think b is wrong? – marwalix May 14 '15 at 8:29
• I have answers and they says (2) is false, but I falied to come up with a counterexample. – Mirak May 14 '15 at 8:30

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.