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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?

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  • $\begingroup$ Do you know what "span" means ? $\endgroup$ Oct 10, 2020 at 19:04
  • $\begingroup$ "The rows equal the number of columns"? If you meant the number of linearly independent rows equals the number of linearly independent columns, then that is true regardless of whether the determinant is $0.$ $\endgroup$ Oct 10, 2020 at 19:23
  • $\begingroup$ @TheSilverDoe yea it means the combos of the vectors all the possibilities and operations and scale things $\endgroup$
    – Si Random
    Oct 10, 2020 at 19:41

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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 $A$.

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