Zero diagonal of product of skew-symmetric and symmetric matrix with strictly positive identical diagonal elements

Question

I encountered the following problem in a proof which I couldn't solve so far:

Let $\mathbf{A}$ be a skew-symmetric matrix, and $\mathbf{B}$ a symmetric matrix. There is at least one off-diagonal element in $\mathbf{B}$ which is non-zero, i.e. $\exists i,j,i \neq j: B_{ij} \neq 0$.

I now look at the diagonal elements of the product of the two matrices, $(\mathbf{A}\mathbf{B})_{ii}$. I need to show: If all diagonal elements are zero, i.e. $\forall i: (\mathbf{A}\mathbf{B})_{ii} = 0$, then $\mathbf{A} = \mathbf{0}$.

Alternatively one could also show the negation: If $\mathbf{A} \neq \mathbf{0}$ then $\exists i: (\mathbf{A}\mathbf{B})_{ii} \neq 0$.

Any ideas?

Edit: $\mathbf{B}$ also has strictly positive and identical diagonal entries. I didn't think this condition was required, but apparently it is (as the counter-example in the answer shows).

2nd edit: That additional prerequisite doesn't change the counter-example below. I think what I wanted to show doesn't hold.

3rd edit: I probably asked the wrong question for my problem. I tried it again in a different question: product of skew-symmetric and symmetic matrix: diagonal elements.

Answer 1

What you want to show isn't true. Counterexample: $$ A=\pmatrix{0&-1&-1\\ 1&0&-1\\ 1&1&0},\ B=\pmatrix{0&1&-1\\ 1&0&1\\ -1&1&0},\ AB=A. $$