a) Proof: If \(A B=0\), then it follows that every entry of the product matrix is 0 . Hence, either all entries of the first matrix \(A\) are 0 , or all entries of the second matrix \(B\) are 0 , or both.

b) Proof: If \(A B=0\), then we have

\[B A = (A B)^T = 0^T = 0\]

c) Counterexample: Consider the matrices

\[A = \begin{bmatrix}-3 & 0 \\ 0 & -3\end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\]

It follows that \(A B=\left[\begin{array}{cc}-3 & 0 \\ 0 & -6\end{array}\right]=0\), but \(B \equiv 0\). Hence, the statement is false.

d) Counterexample: Consider the matrices

\[A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\]

It follows that \(A B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0\), and \(B\) is invertible, but \(A \equiv 0\). Hence, the statement is false.

e) Proof: If \(A B=0\), then it follows that for every vector \(V\), the product \(B A V=0\). Hence, there exists a non-zero vector \(V\) such that \(B A V=0\).