Why is a matrix $A$ that fulfils $AA^t = I$ invertible?

Question

Given a square matrix $A$ that fulfils

$$AA^t = I$$

Justify why must $A$ be invertible.

The answer, according to my book, is simply

$$AA^t = I$$

$$A^t = A^{-1}$$

I don't understand why does $A^t=A^{-1}$ must imply that $A$ is necessarily invertible. Why is that?

Answer 1

A matrix $A$ is invertible if there exists some matrix $B$ such that $AB=I=BA$ (I'm ignoring here the sizes of the matrices, assuming it is clear). It is in fact a theorem that for (finite) matrices the existence of $B$ with $AB=I$ automatically implies that $BA=I$ and thus that $A$ is invertible. We then denote $A^{-1}=B$.

Now, you are given that $AA^t=I$, so call $B=A^t$ and apply the above.

Answer 2

The key thing to remember is that for any matrices A,B, $\det(AB)=\det(A)\det(B)$ and $\det(A^t)=\det(A)$.

Then when we take determinants in the equation $AA^t=I$, we see that $\det(AA^t)=\det(A)\det(A^t)=(\det(A))^2=\det(I)=1$ so that $\det(A)=\pm 1$. A matrix is invertible if and only if it has a non-zero determinant, so this shows that A is invertible.

Answer 3

$AA^t=I$ means that $A$ is right invertible, hence surjective, and the rank-nullity theorem says that surjectivity and invertibility of finite dimensional matrices are equivalent, hence $A$ is invertible.