Consider the linear transormation . By definition of the rank, is an injection, hence is left-invertible; hence there exists a transformation such that . Writing we have ; but so .
Hi, just reviewing some linear algebra this summer. Came across this in one of my exercise books, and was looking for a proof.
Let A be an m x n matrix with rank m. Prove that there exists an n x m matrix B such that AB = I(m).
[That last part reads "AB equals I sub m."]