# If $AB=A$, does B have to be the identity matrix?

Suppose $$A$$ and $$B$$ are square matrices and that $$AB=A$$ with $$B \neq I$$. What does this say about the invertibility of $$A$$?

This question showed up on an exam I took this past spring. I got stuck on it, but I thought about it for a while and think I figured it out. Here's something similar to what I got:

Suppose $$A$$ is invertible. Then:

\begin{align} AB &= A \\ A^{-1}AB &= A^{-1}A \\ IB &= I \\ B &= I \end{align}

This shows that if $$AB=A$$, then $$B$$ must be an identity matrix if $$A$$ is invertible.

Conclusion: If $$AB=A$$ and $$B \neq I$$, then $$A$$ must be singular.

An obvious example would be making $$A$$ a zero matrix.

Is what I've got correct?

• You're correct. – littleO May 23 at 0:52
• You may also need to prove that every non-invertible square matrix is singular. – user1952500 May 23 at 1:04