TITLE: A symmetric matrix whose square is zero
QUESTION [20 upvotes]: I was once asked in an oral exam whether there can be a symmetric non zero matrix whose square is zero. After some thought I replied that there couldn't be because the minimal polynomial of such a matrix is guaranteed to be $x^2$ which shows that it isn't diagonalizable. I had to further clarify that a matrix is diagonalizable iff its minimal polynomial is a product of distinct linear factors, and that every symmetric matrix is diagonalizable.
While all this is correct, the examiner mentioned that there is a simpler argument possible but he didn't elaborate on it. I have since been wondering what that simpler argument could be. Can someone give a simpler proof?
Thanks

REPLY [10 votes]: This really depends on the underlying field.

As Erick Wong has pointed out in a comment to another answer here, there exist complex symmetric matrices whose squares are zero. The example he gave is $\pmatrix{1&i\\ i&-1}$.
If the underlying field is $GF(2)$, we have $\pmatrix{1&1\\ 1&1}^2=0$.
If you are talking about real symmetric matrices, then $A^2=0\Rightarrow A=0$. Many answers here have explained why this is the case. Here I will add another one: if $A^2=0$, then $0=x^TA^2x=x^TA^TAx=\|Ax\|^2$ for all vector $x$, i.e. $Ax=0$ for all $x$. Hence $A=0$.