# Thread: 2-norm is invariant with respect to orthogonal transformations

1. ## 2-norm is invariant with respect to orthogonal transformations

Hi,

my book says that it is easy to show that for all orthogonal Q and Z of appropriate dimensions we have

$||QAZ||_2 = ||A||_2$.

I'm thinking that $Q\in \mathbb{C}^{m\times m}$, $A\in \mathbb{C}^{m\times n}$ and $Z\in \mathbb{C}^{n\times n}$.

The 2-norm of a matrix is defined as,

$||A||_2 = \sup_{||x||_2=1}||Ax||_2$

and so

$||QAZ||_2 = \sup_{||x||_2=1}||(QAZ)x||_2$.

I do not know what to do here.. Any tips are greatly appreciated. Thanks.

2. Where have you used the fact that Q and Z are orthogonal?

3. Hint: Orthogonal matrices preserve the (usual) inner product (or equivalently the norm).

4. Here's an attempt..

\begin{aligned}
||QAZ||^2_2 =& \sup_{||x||_2=1}||(QAZ)x||^2_2 \\
=& \sup_{||x||_2=1}(QAZx)^*(QAZx) \\
=& \sup_{||x||_2=1}(AZx)^*(AZx) \\
=& \sup_{||x||_2=1}||(AZx)||^2_2
\end{aligned}

Since $||Zx||^2_2=(Zx)^*(Zx)=x^*x=||x||^2_2$, we get that

$\sup_{||x||_2=1}||(AZx)||^2_2 = \sup_{||x||_2=1}||(Ax)||^2_2 = ||A||^2_2.$

That last part looks dodgy...

5. I don't see clear your third equality. Denote $\left\|{\;\;}\right\|_2= \left\|{\;\;}\right\|$

I'd better use:

$\left\|{M}\right\|=\sqrt{\rho(M^*M)}$

Then,

$(QAZ)^*(QAZ)=Z^*A^*Q^*QAZ=Z^*A^*AZ=Z^{-1}(A^*A)Z$

That is,

$(QAZ)^*(QAZ),\;A^*A$

are similar matrices. As a consequence:

$\rho[(QAZ)^*(QAZ)]=\rho(A^*A)$

which implies:

$\left\|{QAZ}\right\|=\left\|{A}\right\|$

Fernando Revilla