$n$ by $n$ invertible matrix $A$ has $\text{rank(A)}=n$

Question

I'm wondering why $\text{rank}(A)=n$ means $A$ is invertible.

Since invertible means one-to-one and onto, we have to prove that.
$\text{rank}(A)=n$ means $\dim(N(A))=0$ which means one-to-one.
Now, we have to show onto.
$\text{rank}(A)=\dim(R(A))=n$ and it means onto.
So we complete the proof.
Is that right?

[ADDITION]
I want to check this: $\text{rank}(A)=\text{rank}(L_A)=\dim(R(L_A))=n$ and it means onto?
I mean, range and codomain have same dimension means they are same?
I know even though they have same dimension but still their spaces could be different, like row space and column space.
But the above statement "range and codomain have the same dimension means they are same" is true?

I solve it. It is true since $R(L_A)$ is a subspace of $F^n$.
Refer to theorem $1.11$ in textbook "linear algebra" by friedberg.

For a linear map $f\colon V\to W$ we have $\dim \ker f+\dim\operatorname{im} f=\dim V$. — Hagen von Eitzen, Apr 04 '13 at 10:46
I know $rank(A)=n$ means $n$ linearly independent columns(rows). I'm a bit confused with the definition of onto and other things in here, so want to make it clear. — noname, Apr 04 '13 at 10:46
Proof - Square Matrix has maximal rank if and only if it is invertible — Martin Sleziak, Apr 04 '13 at 11:05

score 5 · Accepted Answer · answered Apr 04 '13 at 11:09

If $rank(A)=n$ then $A$ is an equivalent matrix to $I_{n\times n}$. This means that there are some elementary matrices $E_1,E_2,...,E_s$ such that $E_s...E_2E_1A=I$ and so $$A=E_1^{-1}...E_s^{-1}I$$ But $E_i^{-1}$ are also invertible as $I$ so $A$ is invertible as well.

Metin Y. · Answer 2 · 2013-04-04T11:01:32.400

If you know that "$rank(A)=n$ means $n$ linearly independent columns" then you shall consider that if you got at least one dependent column(or row), that column will be the $0$ column after some column operations. Now, calculate the determinant with respect to this $0$ column, you will clearly have $det(A)=0$, which is equivalent to $A$ is not invertible. Here I'm assuming you know the fact that $det(A)\neq0$ if and only if $A$ is invertible.

rschwieb · Answer 3 · 2013-04-04T10:58:08.463

Yes. Arguing as you have by the Rank-Nullity theorem, that is a perfectly valid way to show that the transformation is 1-1 and onto.

In fact, the rank nullity theorem helps you see that if a square matrix is 1-1 transformation, then it is also onto, and similarly if it is onto, it is also 1-1.

Maybe a key point to make sure is clear is the "$dim(A)=n$ means onto" point. This just says that the image of $A$ as a transformation from $F^n$ to $F^n$ is an $n$ dimensional subspace of $F^n$... but there is only one such subspace: the whole space! Thus the image of $A$ is all of $F^n$.

You can unroll a bit of the proof to see what is going on. If the rank of the matrix were less than $n$, then there would be a nonzero linear combination of the rows of $A$ which was zero. Lining up those coefficients into a row vector $x=[\alpha_1,\dots,\alpha_n]$ you can see that $xA=0$ but $x\neq 0$, showing the matrix does not produce a 1-1 transformation. (A similar thing must hold for the columns of $A$ if you like to work with column vectors instead.)

$n$ by $n$ invertible matrix $A$ has $\text{rank(A)}=n$

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