Prove that if a real matrix $A$ is symmetric and $A^2=0$, then $A=0$

Question

Prove that if a real matrix $A$ is symmetric and $A^2=0$, then $A=0$.

Didn't really understand how a squared matrix could be equal to $0$.

Answer 1

The entry $(i, j)$ of $A^2$ can be written out as $$ [A^2]_{ij} = \sum_{k=1}^n A_{ik} A_{kj} $$ Along the diagonal $(i=j)$ we can use that $A$ is symmetric, $A_{ik} = A_{ki}$, and simplify the formula above $$ [A^2]_{ii} = \sum_{k=1}^n A_{ik}^2. $$ For the real-valued $A$ this is a sum of non-negative numbers meaning that if the sum is zero then all elements of the sum must also be zero. $$ [A^2]_{ii} = 0 \Longrightarrow A_{ik} = A_{ki} = 0 \quad \forall k $$ Using this and if the diagonal of $A^2$ is zero then we can prove that all elements of $A$ must also be zero (assuming, of course, that $A$ is symmetric).