If the product of two matrices is invertible, are each matrices in the product invertible?

Question

True or flase? If the product of two matrices is invertible then each matrices in the product is invertible.

It was said to be true but I didn't understood the argument: a matrix A \begin{pmatrix} 1 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 1 \end{pmatrix}

is non-invertible, its determinant equals $0$, furthermore it is equal to the product of diagonal coefficients of $A$, $A$ not being triangular.

Okay but $A*A$ isn't invertible neither, it gives:

\begin{pmatrix} 3 & 2 & 3\\ 2 & 2 & 2\\ 3 & 2 & 3 \end{pmatrix}

So I don't undertand the point, it should have demonstrate that $A*A$ was invertible and that $A$ wasn't.

Answer 1

We can say that also without using determinants. Let $f_A$ and $f_B$ be the linear applications associated with our matrices and let $g=f_{AB}=f_A\circ f_B$. If $AB$ is invertible, $g$ is a isomorphism, hence it is both injective and surjective, hence $f_A$ is surjective and $f_B$ is injective.

But that is the same as stating that both $A$ and $B$ are invertible matrices.

Answer 2

There is an additional property of matrices which is how I think of this. $$\det(AB)=\det(A)\det(B)$$ This means that if the product $AB$ has det $0$, then necessarily at least one of $A$ and $B$ does.