TITLE: $A$ is invertible matrix iff $Ax=0$ has the trivial solution only.
QUESTION [7 upvotes]: Why does the following statemnet true?

$A$ is invertible matrix iff $Ax=0$ has only the trivial solution.

My try:
Let $x$ a solution of $Ax = 0$.
Then, because $A$ is invertible there is $A^{-1}$. Hence,
$$AA^{-1}x = 0$$ $$I_nx = 0$$ $$x=0$$
I used the associative property of matrix multiplication.

REPLY [7 votes]: If $A$ is invertible just multiply by $A^{-1}$ we get $x=0$. Now assume the other way round that $x=0$ is the only solution. For the sake of contradiction assume that $A$ is not invertible. Then columns are not independent. Therfore $\Sigma \lambda_n C_n=0:all $ $\lambda_n$ are not identically $0$. So now $(\lambda_1,...,\lambda_n)$ is a solution. Contradicting the hypothesis.