inversible matrix and it is rank proof

Question

can anyone please help me with that proof? thanks in advance

let $A$ be a square matrix ($n\times n$), I want to prove that $A$ is invertible if and only if $\text{rank}(A)=n$

I did this, is that right? I'm not sure about my prove and the way I got from one direction to the other. let $\text{rank}(A)=n \Rightarrow$ means we have no zero rows $\iff$ means A has one solution (no free values) $\iff$ means A has an inverse.

Answer 1

Hint:

Look at $A:\mathbb{R}^n\rightarrow \mathbb{R}^n$ and use Rank-nullity.

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Your idea is right, and this is just nitpicking:

• If $\text{rank}(A)=n$, then the equation $Ax=b$ has atleast one solution for every $b\in\mathbb{R}^n$. This means that the columns of $A$ span $\mathbb{R}^n$, so $A$ is row equivalent to $I_n$ and so $A$ is invertible.
• If $A$ is invertible then $Ax=b\Rightarrow A^{-1}Ax=x=A^{-1}b$, so the equation $Ax=b$ has atleast one solution for every $b\in\mathbb{R}^n$. This means that the columns of $A$ spans $\mathbb{R}^n$, so $\text{rank}(A)=n$.

Another way to prove it is to note that $\text{rank}(A)=n\iff A:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is a bijection $\iff$ $A$ is invertible.