Proving equality of column space to field

Question

Question:

Let $A$ be a matrix of $m\times n$ and $B$ matrix of $n\times m$ over field $F$.

Given that $AB = I_m$ ($m\times m$ unit matrix), prove that the column space of $A$ is equal to $F^m$.

My Attempt:

So looking at the question I can understand that dim of $F^m$ is obviously $m$ so I can deduce that I need to show that $m=n$, and if I show it equals its pretty much the question.

Now from what I know $AB = I$ means that the number of rows in $A$ is less or equal to the number of columns which in formality means $m\le n$. I'm stuck at this point because the other way $n\le m$ is something I can't get from the question.

It's not necessarily true that $m=n$. You can come up with simple examples where $m=1$ and $n>1$. — angryavian, Jun 01 '21 at 19:47

score 2 · Accepted Answer · answered Jun 01 '21 at 19:53

2

Showing that the column space of $A$ is equal to $F^m$ means showing that for any $y \in F^m$, there exists a vector $x \in F^n$ for which $Ax = y$.

Show that for any $y \in F^m$, the vector $x = By \in F^n$ satisfies this requirement.

answered Jun 01 '21 at 19:53

Ben Grossmann

225,327

score 1 · Answer 2 · answered Jun 01 '21 at 19:50

1

If you are considering fields $\mathbb{R}$ or $\mathbb{C}$, you can use the fact that $$m = \text{rank}(AB) \le \text{rank}(A) \le \min\{m, n\} \le m.$$

answered Jun 01 '21 at 19:50

angryavian

89,882

1

This result holds over arbitrary fields as well – Ben Grossmann Jun 01 '21 at 19:51
@BenGrossmann Thanks, I should have just thought about linear combinations of columns of $A$. – angryavian Jun 01 '21 at 19:55

Proving equality of column space to field

Question:

My Attempt:

2 Answers2