Proof that the 2-norm of orthogonal transformation of a matrix is invariant

Question

For any matrix A and an orthogonal matrix Q, I can prove in the standard way that $$\|QA\|_2 = \|A\|_2 $$ using $$\|QA\|_2^2 = (QAx)^T(QAx) = (Ax)^T(Ax) = \|A\|_2^2 $$

However, I am unable to cancel Q, when the transformation is AQ, i.e. $$\|AQ\|_2^2 = (AQx)^T(AQx) = x^TQ^TA^TAQx$$ After this point I am unable to prove the same that $$\|AQ\|_2 = \|A\|_2 $$

Where does the $x$ come from? – Rodrigo de Azevedo Jun 17 '19 at 06:09 — Rodrigo de Azevedo, Jun 17 '19 at 06:09

score 6 · Accepted Answer · edited Jun 17 '19 at 06:07

6

Recall that the 2-norm for matrices is defined as $$\|B\|_2 \;\; =\;\; \sup_{\|x\|=1} \|Bx\|.$$

But for any orthogonal matrix $Q$ we have that $\|Qx\| = \|x\|$. Thus in your computation we can write $$\|AQ\|_2 \;\; =\;\; \sup_{\|x\|=1} \|AQx\| \;\; =\;\; \sup_{\|Qx\|=1} \|AQx\| \;\; =\;\; \sup_{\|y\|=1} \|Ay\| \;\; =\;\; \|A\|_2.$$

edited Jun 17 '19 at 06:07

Hanno

6,951

answered Jan 20 '17 at 03:54

Mnifldz

13,306

2

How do you know that the range of $Q$ is the entire vector space? – user 6663629 Jan 26 '20 at 21:25
3

@user42493 $Q$ is orthogonal, hence invertible. – Mnifldz Jan 26 '20 at 21:34

Proof that the 2-norm of orthogonal transformation of a matrix is invariant

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