Prove that if $A$ is an invertible $n × n$ matrix, then the columns of $A$ span $\mathbb R^n$
How would I go about proving this?
So far my answer is like this but it's not sufficient enough apparently:
If $A$ is invertible that means its determinant does not equal $0$. This means the rows equals the number of columns. And so the row rank${}={}$column rank${}= n.$ So column spans all of $A$
How can I improve my explanation? Would it be better if I showed an example? Or would that not be considered a proof?
Best Answer
Your argument is basically good, though the half sentence 'This means the rows equals the number of columns' doesn't contain new information, as the matrix was assumed to be square.
The key connection you stated is that invertibility implies full rank.
For a direct approach, observe that for any $v\in\Bbb R^n$, the vector $Av$ is a linear combination of the columns of $A$.
Now if $A$ is invertible, then for any vector $w\in\Bbb R^n$ we have $Av=w$ where $v=A^{-1}w$.
So, every vector is in the column space of $A$.