Exercise 33 of Section 8.3, from Linear Algebra with Applications (5e) by Otto Bretscher:
If the singular values of an $n{\times}n$ matrix $A$ are all $1$, is $A$ necessarily orthogonal?
I understand the converse is true, and I thought the above statement was true as well, but the solution in the back of the book says
No; consider $A = \left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right)$.
However, by my calculation, the singular values of this matrix are 2 and 1, and I fail to see why this matrix is a counter example.
Also, here is my false proof. Would anyone mind pointing out the flaw in my reasoning?
Thank you.
If the singular values are all $1$ (where our convention includes the possibility of zeros as singular values), then $\Sigma=I_n$ so $A=UV^T$ for some orthogonal $U,V$ so $A^T A=V U^T U V^T=V V^T=I_n$ and you can do the same for $A A^T$. So yes, $A$ is orthogonal in this situation. I don't see what $\begin{bmatrix} 0 & 1 \\ 2 & 0 \end{bmatrix}$ has to do with the problem.
The only way I can understand the statement being interpreted as false is if your convention for the SVD is the reduced SVD and thus zero is not an admissible singular value. In this case $A$ will be orthogonal if and only if it has full rank (but the above example is still irrelevant because it violates the hypothesis).