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?
$$\det(A)=0 \implies A\mathbf{x}=\mathbf{0} \;\;\text{has a non-zero solution}$$
 A: 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.
A: $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 $Ax=0x$.
A: Here's a different proof of this statement that works over an arbitrary commutative unital integral domain $R$. One direction is rather uninspiring though, and the other direction reduces to the case over fields.
For our $n\times n$ matrix $A$ with entries in $R$, let $A^\alpha$ denote the classical adjoint of $A$. A nifty property of the classical adjoint is that $A^\alpha A = \det(A)I_n$. So if $A\mathbf{x}=0$ has a nonzero solution, we get that
$$
A\mathbf{x}=0 
\quad\implies\quad
A^\alpha A\mathbf{x}=0
\quad\implies\quad
\det(A)I_n \mathbf{x}=0\,
$$
and since we have no zero divisors in $R$, $\det(A)$ cannot be a zero divisor and we must have $\det(A)=0$. 
For the other direction, suppose that $\det(A)=0$. Let $\mathrm{Fr}(R)$ denote the field of fractions of $R$. Since $R$ is a domain, $\mathrm{Fr}(R)$ contains a copy of $R$ and we can think of the entries of $A$ as living in $\mathrm{Fr}(R)$. So since this result is true over fields, there $A\mathbf{x}=0$ has a nonzero solution 
$$
\mathbf{x} = \left\langle \frac{a_1}{b_1}, \dotsc, \frac{a_n}{b_n} \right\rangle
$$ 
with $a_i,b_i \in R$. Let $\beta$ be the product $b_1\dotsb b_n$. Then since $\mathbf{x}$ is a solution to $A\mathbf{x}=0$, $\beta\mathbf{x}$ will be a solution too, and the entries of $\beta \mathbf{x}$ live $R$, giving us a nontrivial solution in $R$.
This theorem is not true though, if we aren't working over an integral domain 
