A symmetric matrix whose square is zero

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 $(A^2)_{ii} \ne 0$.

As Pete said I used that a symmetric square matrix is orthogonal diagonalizable

A symmetric square matrix is diagonalizable hence $$A=Q^{-1} D Q$$ Hence $$A^2 =Q^{-1} D\cdot D Q $$ We multiply with $Q^T$ from right and with $Q$ from left $$Q A^2 Q^{-1} = D^2 $$ As $A^2=0$ we have $$0=D^2$$ So the square of every eigenvalue is $0$ hence all eigenvalues are $0$ hence $A$ must be $0$.

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 $A=0$.