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
Best Answer
The $(i,i)^{\text{th}}$ component of the square of an $n \times n$ symmetric matrix $A=(a_{ij})$ is given by $$\sum_{j=1}^n a_{ij}a_{ji} = \sum_{j=1}^n a_{ij}^2$$ If $A \ne 0$ then some $a_{ij} \ne 0$, and then $(A^2)_{ii} \ne 0$.