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?

  • $\begingroup$ Do you mean $\|AQ\|_2$ instead of $\|QA\|_2$? Note that $\|A^T\|_2 = \|A\|_2$ and $(AQ)^T = Q^TA^T$. $\endgroup$ – amsmath Aug 27 at 13:29
  • $\begingroup$ Yes. I've edited it. $\endgroup$ – dorsh605 Aug 27 at 13:30

$$ \|AQ\|_2 = \|(AQ)^*\|_2 = \|Q^*A^*\|_2 = \|A^*\|_2 = \|A\|_2. $$ 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.

  • $\begingroup$ I thought it should be something that simple and like many other claims, make a statement about rows to be a statement about columns by using the transpose but I'm not sure why is it true that $\|A^T\|_2=\|A\|_2$? $\endgroup$ – dorsh605 Aug 27 at 13:33
  • $\begingroup$ I edited the answer, $\endgroup$ – amsmath Aug 27 at 13:37
  • $\begingroup$ Why is it true that $\sup_{\|x\|=1}\sup_{\|y\|=1}\langle Ax,y\rangle = \sup_{\|y\|=1}\sup_{\|x\|=1}\langle x,A^*y\rangle $? $\endgroup$ – dorsh605 Aug 27 at 14:06
  • $\begingroup$ I've edited the answer once more. the star means conjugate transpose, as usual. As to the middle equation: $\langle Ax,y\rangle = y^*Ax = (A^*y)^*x = \langle x,A^*y\rangle$. $\endgroup$ – amsmath Aug 27 at 14:07
  • $\begingroup$ Oh I see. In the last lines of your answer you were referring to the 2-norm as I asked. Because you didn't add the 2 subscript beneath the norms I thought that you prove that for a general vector norm (that is induced by some general inner product). So just to be sure, the middle equality is not true for some general norm (induced by an inner product)? $\endgroup$ – dorsh605 Aug 27 at 14:11

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