Find a Orthogonal Matrix Q

Question

Assume we are in the real case instead of complex. Suppose $A$ and $B$ have $AA^T = BB^T=C$ where $C$ is non-singular.Show that there is an orthogonal matrix $Q$ such that $A=BQ$.

My attempt is: $AA^T=C$ is symmetric, so by Spectral theorem, there exists orthogonal $Q$ such that $AA^T = QDQ^{-1}$ where $D$ is diagonal. Then since I have got a $Q$ here, I can define $B=AQ^T$ so that $A = BQ$. And it's easy to verify $AA^T = BB^T=C$ with this set-up. However, the matrix $B$ was actually given in the first place. So what I did is probably incorrect. But without assuming $A$ or $B$ is invertible, I really don't see any ways we can come up with such a $Q$.

Thanks so much in advance for any help.

Answer 1

Hint. $C$ is positive definite. Put $X=C^{-1/2}A$ and $Y=C^{-1/2}B$, we get $XX^T=YY^T=I_n$, meaning that each of $X$ and $Y$ has orthonormal rows. If you can show that $X=YQ$ for some real orthogonal matrix $Q$, then $A=BQ$.

Answer 2

Write SVDs, $A=U_1\Sigma_1 V_1^T$, $B=U_2\Sigma_2V_2^T$ where the $U$s are $n\times n$, $\Sigma$s are $n\times m$, and $V$s are $m\times m$. Note that $m\geq n$ is a necessary condition for $C$ to be full rank.

Then, $$ U_1 \Sigma_1\Sigma_1^T U_1^T = (U_1\Sigma_1V_1^T)(V_1\Sigma_1^TU_1^T) = AA^T = BB^T = (U_2\Sigma_2V_2^T)(V_2\Sigma_2^TU_2^T) = U_2 \Sigma_2\Sigma_2^T U_2^T $$

What kind of decomposition is on the left and right of the above inequality. What can you say about the uniqueness of this decompostiion, and therefore about the relationship between $U_1$ and $U_2$ and $\Sigma_1$ and $\Sigma_2$?

Now can you relate $A$ and $B$?