Let \(A\) be an \(n \times n\) (square) matrix. Prove the following two statements:
If \(A\) is invertible and \(AB = 0\) for some \(n \times n\) matrix \(B\), then \(B = 0\) .
If \(A\) is not invertible, then there exists an \(n \times n\) matrix \(B\) such that \(AB = 0\) but \(B \neq 0\) .
Solution
1. If \(A\) is invertible and \(AB=0\), then \(B=0\).
Multiply on the left by \(A^{-1}\): \[ B=A^{-1}AB=A^{-1}0=0. \]
2. If \(A\) is not invertible, there is a nonzero matrix \(B\) with \(AB=0\).
Since \(A\) is singular, its nullspace contains some nonzero column vector \(v\) with \(Av=0\). Let \(B\) be the matrix whose first column is \(v\) and whose remaining columns are zero. Then \(B\ne0\), while every column of \(AB\) is zero; hence \(AB=0\).