If $A$ is an $m \times n$ matrix, prove that $Ax=0$ has an unique solution $iff$ $\text{rank}(A)=n$

Question

Prove that $Ax=0$ has an unique solution $\iff \text{rank}(A)=n$

Here is my proof of the "if" part:

If $\text{rank} (A)=n$, then $\mathcal{C}(A)=\mathbb{R^n}$ and therefore $Ax=0$ has a solution. If $Ax=0$ and $Ay=0$ then $A(x-y)=0$. But $\dim (\mathcal{N(A))}=0 $ therefore $x=y$. This proves uniqueness.

How do i prove the other way round?

Answer 1

$\text{Rank A}=n\iff \text{N(A)}=\{0\}\iff Ax=0\implies x=0$

where $\text{N(A)}$ denotes Nullity of $A$

Answer 2

Nullity and rank theorem: For $T: U \rightarrow V$ where $T(x) = Ax$, you have: $$\operatorname{dim}V + \operatorname{dim N}(A) = n$$ Where $\operatorname{N}(A)$ is the nullity of $A$. Note that $\operatorname{dim}V = \operatorname{Rank}(A)$, then you have: $$n + \operatorname{dim N}(A) = n$$ therefore $$ \operatorname{dim N}(A) = 0 $$ so $Ax=0$ only have 1 solution.

Answer 3

Suppose the rank of $A$ is less than $n$. Then the columns of $A$ are linearly dependent. Denote them by $a_1,a_2,\dots,a_n$. Then there are scalars, not all zero, such that $$ c_1a_1+c_2a_2+\dots+c_na_n=0 $$ Then $$ A\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}=0 $$ and the system $Ax=0$ has a non trivial solution.

If the rank of $A$ is $n$, then $\mathcal{N}(A)=\{0\}$, so the only vector $x$ such that $Ax=0$ is $x=0$, by definition of null space.