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

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.

• True${}{}{}{}{}{}{}$ , and both matrices in your question are non-invertible, so what is odd here? – DonAntonio Mar 30 '16 at 22:11
• $\det(AB) = \det(A)\det(B)$. If either matrix is singular, then it has a zero determinant. Hence, the determinant of the product is 0. – ÍgjøgnumMeg Mar 30 '16 at 22:14
• You do need that both matrices in the product are square-- you can have a 2x3 $A$ and a 3x2 $B$, so neither is invertible, with $AB$ an invertible 2x2 (e.g. start with 2x2 identity matrices and add a column (resp. row) of $0$'s to get $A$ (reps.$B$) ). – Ned Mar 30 '16 at 22:36

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.
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.