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Consider an m×n matrix A (with m rows, n columns) of rank r. What relations between m, n and r are necessary and sufficient for the existence of:

1) a right inverse B such that AB=I

2) a left inverse C such that CA=I

3) prove that B=C when both inverses exist.

EDIT

Some of my work:

First suppose that AB = I. Then B must be nxm.

If n$\ne m$, then

$$A_{mxn} *B_{nxm}(x) = I_{mxm}(x)$$ $$=x$$ => $AB_{mxm}$ has full rank = m, and trivial null-space, by the Rank-Nullity Theorem.

Then (AB) has m linearly-independent rows (and columns, since row-rank = column-rank).

Similar argument for 2).

What more can I say?

Any hints or solutions are welcome.

Thanks,

User001
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    what you meaning by det(A) where A in an $m\times n$ matrix – David Sep 29 '15 at 05:25
  • Good point...the generalization of invertibility is a bit confusing. In our linear algebra classes, we've only ever dealt with square matrices when talking about invertibility. I will edit that part out. Thanks, @gloom. Please feel free to comment further :-) – User001 Sep 29 '15 at 05:27

1 Answers1

1

Suppose that $A$ is $m \times n$ with rank $r$. Note that $r \leq \min\{m,n\}$.

  1. There is a matrix $B$ such that $AB = I_{m \times m}$ if and only if $r = m$. This can be proven nicely as a consequence of (2), using the transpose. By the above inequality, this can only occur when $m \leq n$
  2. There is a matrix $B$ such that $BA = I_{n \times n}$ if and only if $r = n$. This can be proven using row-reduction and reduced row-echelon form. By the above inequality, this can only occur when $m \geq n$.
  3. Several such proofs are given here. Of particular importance to us is the fact that the underlying vector spaces are finite-dimensional.
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Ben Grossmann
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