I'm reading the book "Numerical Linear Algebra" by Trefethen.
On the bottom of page 23 under the section "Invariance under unitary multiplication" he states the following theorem:
for any $A\in \mathbb{C}^{m\times n}$ and unitary $Q\in \mathbb{C}^{m\times m}$ we have
$$\|QA\|_2=\|A\|_2 \quad,\quad \|QA\|_F=\|A\|_F$$
where $\|\cdot\|_2$ represents the matrix norm induced by the 2-norm of the vector spaces $\mathbb{C}^m$ and $\mathbb{C}^n$ and where $\|\cdot\|_F$ represents the Frobenius norm on $\mathbb{C}^{m\times n}$
After a short proof of the above he states a generalization of the above theorem which goes as follows:
The above remains valid if $Q$ is generalized to a rectangular matrix with orthonormal columns, that is, $Q\in \mathbb{C}^{p\times m}$ with $p>m$. Analogous identities also hold for multiplication by unitary matrices on the right, or more generally, by rectangular matrices with orthonormal rows
I tried to prove the last generalization and I managed to do so for the first 2 cases (for the rectangular matrix with orthonormal columns when we multiply from the left and for the unitary matrix when we multiply by the right). I also managed to prove the equality for the Frobenius norm when we multiply from the right by a rectangular matrix with orthonormal rows.
The only thing I didn't manage to prove is that if we multiply $A$ from the right by a rectangular matrix $Q\in \mathbb{C}^{n\times p}$ with orthonormal rows then $\|AQ\|_2=\|A\|_2$. I tried to prove it by mixing the methods used to prove it for the previous cases but to no avail.
basically I'm stuck because I don't think it's true that $\|Qx\|_2=\|x\|_2$ in this case. I used that in the case of multiplication from the right by unitary matrix so I'm not sure how to continue.
Anyone have any suggestions?
Best Answer
$$ \|AQ\|_2 = \|(AQ)^*\|_2 = \|Q^*A^*\|_2 = \|A^*\|_2 = \|A\|_2. $$ Here is why $\|A^*\| = \|A\|$: First, \begin{align*} \|A\| = \sup_{\|x\|=1}\|Ax\| = \sup_{\|x\|=1}\sup_{\|y\|=1}|\langle Ax,y\rangle| = \sup_{\|y\|=1}\sup_{\|x\|=1}|\langle x,A^*y\rangle| = \sup_{\|y\|=1}\|A^*y\| = \|A^*\|. \end{align*} Here, I used that $\|x\| = \sup_{\|y\|=1}|\langle x,y\rangle|$, which follows easily from Cauchy's inequality.