Prove that $A\mathbf{x}=0$ has a non-zero solution $\mathbf{x}$ iff $\det(A)=0$.

I was reading the wiki page for eigenvalues and eigenvectors, and I found this statement as a fundamental linear algebra theorem.

$A\mathbf{x}=\mathbf{0}$ has a non-zero solution $\mathbf{x}$ iff $\det(A)=0$.

I know how to prove from left to right: Assuming $\det(A)\neq 0$, the only solution for $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=A^{-1}\mathbf{0}=\mathbf{0}$. This is a contradiction to the fact that $A\mathbf{x}=\mathbf{0}$ has a non-zero solution $\mathbf{x}$. Therefore, $A\mathbf{x}=\mathbf{0}$ has a non-zero solution $\implies$ $\det(A)=0$.

Can anybody show me how to prove the other direction? $$\det(A)=0 \implies A\mathbf{x}=\mathbf{0} \;\;\text{has a non-zero solution}$$

$0=\det(A)=\det(A-0\cdot I)\iff \text{$0$ is an eigenvalue of $A$}$, iff there's a vector $x\neq0$ such that $Ax=0x$.

The idea for the reverse is as follows: Since $\det A=0$, then it means any row of $A$ can be written as a linear combination of the other rows. If that is true then (suppose $A$ is $n\times n$) $\mathrm{rank} A<n$. In other words the linear transformation (Suppose the vector space is $V$, $n$-dimensional) $T:V\to V$ given by $x\mapsto Ax$, is not onto. So $\mathrm{Ker}T\neq 0$ meaning there is a vector other than zero being sent to $0$, i.e. $Ax=0$ has a nontrivial solution.

There is a more hands on way to prove this too. Basically using the fact that that row is a linear combination of other rows, you can do a series of basis transformations to obtain $B=UAU^{-1}$ ($U$ is the basis transformation), such that $B$ has a row equal to zero. Then $By=0$ has a nontrivial solution. Consequently define $x=U^{-1}y$, then $Ax = AU^{-1}y =0$. And $x$ is nontrivial.