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?
Best Answer
\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*}