Matrix $A \in \mathbb{R}^{17, 17}$ satisfies condition $A^2 = 0$. Prove that matrix $A + A^T$ is uninvertible (singular) matrix.

Question

I know that matrix is uninvertible (singular) when it's determinant equals to zero. So I used this and we know that det$(A \cdot A) = 0$ and then det$A = 0$. Then we know that det$(A) =$ det$(A^T) = 0$.

But I doubt if from everything I described above it follows that det$(A + A^T) = 0$.

Answer 1

We let 17=2n+1 . If $A^2=0$ $im(A)\subset \ker(A)$. As the dimension of the space is $2n+1$, $\dim(\ker A) \geq n+1$. But $\dim \ker(A)= \dim (\ker (A^T)$, we see that both $\ker A$ and $\ker (A^T)$ have dimension $>n$, so the intersection $\ker(A) \cap \ker (A^T)$ has dimension at least 1, in particular $A+A^T$ has a non trivial kernel.