I have proved that if a square $n$ x $n$ matrix $A$ has a right and left inverse, then these are equal and form an inverse matrix of $A$.

However I'm interested in the following implication:

Suppose a left inverse $B$ of a square $n$ x $n$ matrix $A$ exist. Does this imply that a right inverse $C$ of $A$ exist ?

Also, if this is true - is the implication also true in the case of a right inverse $B$ ?


  • $\begingroup$ Yes, and yes. The easiest way to see it is probably to consider the linear maps induced by the matrices. $\endgroup$ – Daniel Fischer Nov 18 '13 at 9:26
  • $\begingroup$ At the risk of repeating other answers already given, a function has a left inverse iff it's injective; a function has a right inverse iff it's surjective; and a linear transformation between finite dimensional vector spaces of the same dimension is injective iff it's surjective. $\endgroup$ – littleO Nov 18 '13 at 10:09

Yes, If $A$ has a left-inverse $B$, then $BA = I$, and so $A$ is injective (as a linear map). But it is a linear operator on a finite dimensional space, so it is also surjective, and so it has a right-inverse.

| cite | improve this answer | |
  • $\begingroup$ Hi Prahlad. I've never been thinking of matrices like this. However if $A$ was a traditional function, yes if it is surjective indeed it has a right inverse. However how do I construct the right inverse of $A$ in the matrix case ? Doing it for a traditional function I choose some element in the domain that has image equal to the input. $\endgroup$ – Shuzheng Nov 18 '13 at 16:32
  • $\begingroup$ Once you know it has a right inverse, then you can check that the right-inverse must equal the left-inverse, and so $B$ is the right-inverse! $\endgroup$ – Prahlad Vaidyanathan Nov 18 '13 at 16:34
  • $\begingroup$ But you have just proved that it is a matrix. In fact, you have proved that it is $B$! $\endgroup$ – Prahlad Vaidyanathan Nov 18 '13 at 16:43
  • $\begingroup$ So I can say $f: R^{n \cdot n} \rightarrow R^{n \cdot n}$ and then $BX = B(Af(X)) = (BA)f(X) = (I)f(X) = f(X)$, where $R^{n \cdot n}$ denote the set of $n$ x $n$ matrices and $X \in R^{n \cdot n}$. The value of the theoretical right inverse $f(X)$ is the same as the the matrix product $BX$ for every $X \in R^{n \cdot n}$. Hence the linear map $B$ and $f$ are equal and can be substituted for one another. What I proved before needed some work to prove that $f$ is in fact $B$ - This is what I've done now ? Also are there other ways of proving this ? If my proof is right, please tell me. $\endgroup$ – Shuzheng Nov 18 '13 at 17:17

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.