Prove that if $AA^{*}$ is Orthogonal $\Rightarrow$ A is Orthogonal. ($A\in\mathbb{R}_{nxn}$)

Question

Prove that if $AA^{*}$ is Orthogonal $\Rightarrow$ A is Orthogonal. ($A\in\mathbb{R}_{nxn}$)

I should probably state that the first thing I did was to prove $AA^*$ is self-adjoint and all eigenvalues are non-negative.

Here's my approach:

$AA^{*}$ is Orthogonal, so $(AA^{*})\cdot(AA^{*})^{*}=(AA^{*})^2=I$.

so $(AA^{*})^2-I=0$ $\Rightarrow (AA^{*}+I)\cdot(AA^{*}-I)=0$

if $(AA^{*}+I)=0$ then $AA^*=-I$ but $AA^*$ is a positive matrix.

so $AA^*-I=0 \Rightarrow AA^*=I \Rightarrow A^*=A^{-1} \Rightarrow A$ is Orthogonal.

does it make sense?

thanks!

Answer 1

Your reasoning is not correct because $(B - I)(B + I) = 0$ does not imply that $B = -I$ or $B = I$.

Instead, simply note that $(A^*A)$ is a self-adjoint matrix with positive eigenvalues for which $(A^*A)^2 = I$. Show that if $B^2 = I$, then any eigenvalue $\lambda$ of $B$ must satisfy $\lambda^2 = 1$. From the positivity of the eigenvalues of $A^*A$, we know that all eigenvalues are equal to $1$. Because self-adjoint operators are diagonalizable (by the spectral theorem), we can conclude from this that $A^*A = I$.

Answer 2

Your proof can be corrected by noting that $AA^\ast+I$ is positive definite and hence nonsingular. Therefore $(AA^\ast+I)(AA^\ast-I)=0$ implies that $AA^\ast-I=0$.