# prove that if a square matrix $A$ is invertible then $AA^T$ is invertible.

prove that if a square matrix $A$ is invertible then $AA^T$ is invertible.

and also prove the opposite, that if $AA^T$ is invertible, then $A$ is invertible.

i wrote that $det(A) = det(A^T)$

and that $det(A) \neq 0$ when $A$ is invertible

and $det(A) = det(A^T) \neq 0$

and since product of invertible matrices are also invertible, then $AA^T$ must be invertible.

but somehow I feel I will get told "I didn't do any work, I just said that the question was true" again like I did on my test. :| is there a perfect formal way I can prove this?

and also prove the opposite, that if $AA^T$ is invertible, then $A$ is invertible?

• you're correct, you can also show what's the inverse of $AA^t$ Commented Oct 12, 2014 at 4:55
• The argument is not well-organized, but if you can use properties of the determinant, then from $\operatorname{det}(AB)=\operatorname{det}(A)\operatorname{det}(B)$ you can quickly prove that if $AA^T$ is invertible, then so is $A$. Commented Oct 12, 2014 at 5:03

\begin{align*} A \ \text{is invertible} \ & \text{iff}\ det (A) \neq 0\\ & \text{iff} \ det (A)\ det(A) \neq 0 \\ & \text{iff} \ det(A)\ det(A^T) \neq 0 \\ & \text{iff} \ det (AA^T) \neq 0\\ & \text{iff} \ AA^T \text{is invertible} \end{align*}

• Well, for a field of order 4 we have 2=0. Commented Sep 8, 2019 at 6:31

If $A$ is invertible then so is $A^T$ (with inverse the transpose of $A^{-1}$). The product $BC$ of two invertible matrices $B,C$ is invertible, with inverse $C^{-1}B^{-1}$. Now combine these facts. No determinants required.

For the converse, if a product of square matrices in invertible, then both factors must be invertible (since $C(BC)^{-1}$ is a right-inverse of$~B$ and $(BC)^{-1}B$ a left-inverse of $C$; for square matrices a one-sided inverse is automatically a two-sided inverse). Here maybe determinants could simplify the argument a bit: from $0\neq\det(BC)=\det(B)\det(C)$ it follows that $\det(B)\neq0$ and $\det(C)\neq0$.