## Invertible Matrices

December 2020

**Theorem.** If \(A\) is an \(m \times n\) matrix, \(B\) is an \(n \times m\) matrix, and \(n < m\), then \(AB\) is not invertible.

*Proof.* If \(AB\) is invertible, then the homogenous system \(ABX = 0\) has only the trivial solution \(X=0\).
By Theorem 6 [1], \(B\) must have a non-trivial solution since it has less rows than columns. Since \(AB\) is a product of \(B\) with \(B\) on the right side,
the rows of \(AB\) are linear combinations of the rows of \(B\), and so it must have the same solutions as \(B\) (and therefore a non-trivial one).
Thus, \(AB\) does not have \(X=0\) as its only solution, and so it is not invertible.

[1] Theorem 6 is the following: If \(A\) is an \(m \times n\) matrix with \(m < n\), then the homogenous system \(AX = 0\) has a non-trivial solution.