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?
$$\det(A)=0 \implies A\mathbf{x}=\mathbf{0} \;\;\text{has a non-zero solution}$$
Best Answer
$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 $Ax=0x$.