1
$\begingroup$

I was recently asked to find the right inverse of some matrixes. I found that all three of them were invertible, so it was just a matter of finding their inverses, which would be exactly the same as the right inverses.

What if a matrix is not invertible?

Basically, what I want to know is, if a matrix is not invertible, does it mean that there are no left and/or right inverses at all? That is,

$$|A| = 0 \iff \not \exists B(AB = I \ \lor \ BA = I)$$

$\endgroup$
  • $\begingroup$ For square matrices, if $AB = I$, then also $BA=I$. Therefore, if a singular square matrix had a right inverse, that would contradict its singularity. $\endgroup$ – Rustyn Apr 6 '14 at 7:45
4
$\begingroup$

If your matrix $A$ is square, then $A$ has a left inverse if and only if $A$ is invertible. Also, if $A$ is square, then $A$ has a right inverse if and only if $A$ is invertible.

It is possible to construct noninvertible nonsquare matrices with a right inverse. For example, consider the projection $\Bbb R^2\to \Bbb R$. This is given by the matrix $$\begin{bmatrix}1 &0\end{bmatrix}$$ Its right inverse is $$\begin{bmatrix}1\\ 0\end{bmatrix}$$ Of course, the second matrix in this example is an example of a noninvertible matrix with a left inverse.

$\endgroup$
  • 1
    $\begingroup$ So, for a square matrix, if it isn't invertible, then there is no such thing as left/right inverses? $\endgroup$ – Zol Tun Kul Apr 6 '14 at 7:51
  • 2
    $\begingroup$ @Omega That is correct. $\endgroup$ – Brian Fitzpatrick Apr 6 '14 at 7:52
3
$\begingroup$

If $$AB=I_n$$ and since $I_n$ is bijective then it's a well known result that $A$ is surjective and $B$ is injective but this is equivalent to say that $A$ and $B$ are bijective in finite dimenesional space due to the rank-nullity theorem. Hence right or left inverse is equivalent to the inverse matrix.

$\endgroup$
  • $\begingroup$ You are assuming that the matrices are square. While OP might be assuming that as well, it is not said in the question, so I think you should qualify your answer by mentioning this additional hypothesis. $\endgroup$ – Marc van Leeuwen Apr 6 '14 at 10:52

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.