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?

  • 9
    $\begingroup$ You're correct. $\endgroup$ – littleO May 23 at 0:52
  • $\begingroup$ You may also need to prove that every non-invertible square matrix is singular. $\endgroup$ – user1952500 May 23 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.