Suppose $\|.\|_2$ is induced by vector 2 norm s.t. $\|A\|_2 = \max_{\|x\|_2=1}\|Ax\|_2$.
Suppose $P and Q$ are matrices with orthonormal columns(not necessarily orthogonal though). Say $P\in R^{p\times m}$, $Q\in R^{q\times n}$. edit: $A\in R^{m\times n}$.
How do we prove that $\|PAQ^T\|_2 = \|A\|_2$. I think I can prove the case where $P, Q$ are orthogonal but it seems that the same proof does not hold for the more general matrices.
So I start with $\|A\|_2 = \max_{\|x\|_2=1}\|Ax\|_2 = \max_{\|x\|_2=1}((Ax)^TAx)^{1/2} = \max_{\|x\|_2=1}(x^TA^TAx)^{1/2} $.
And similarly, $\|PAQ^T\|_2 = \max_{\|x\|_2=1}(x^TQA^TAQ^Tx)^{1/2} $. I will not be able to replace $Q^Tx$ with another unit value vector since it is not with norm 1.
Can anyone help me with this?
Thanks!
(1) If $Q^T : \mathbb{R}^{k}\rightarrow\mathbb{R}^{k+l}$, then it is not orthonormal.
Hence $Q^T : \mathbb{R}^{k+l}\rightarrow\mathbb{R}^{k}$ and $ P^T:\mathbb{R}^{k'+l'}\rightarrow\mathbb{R}^{k'}$ Here through $Q^T$, image of a unit sphere contains unit sphere.
Proof : If $Q=[v_1\cdots v_k]$, then for $a=\sum_{i=1}^k\ a_iv_i, \ \|(a_1,\cdots, a_k)\|_2=1$, $$Q^Ta=(a_1,\cdots, a_k)$$
(2) Hence $\|PAQ^T\|_2\geq \|A\|_2$.
Proof : $\| A\|_2$ is maximum of the map $$f:S'\times S\rightarrow \mathbb{R},\ f(x,y)=\langle x,Ay\rangle $$ where $S',\ S$ are unit sphere in $\mathbb{R}^{k'},\ \mathbb{R}^k$.
Here if $\| A\|_2=\langle y,Ax\rangle$, then $$\| PAQ^T\|_2\geq \langle P^Ty',AQ^T x'\rangle =\| A\|_2$$ where $P^Ty'=y,\ Q^Tx'=x$.
(3) In further note that image of a unit sphere is contained in unit ball so that we prove the converse.