How can I show $\|XMY\|=\|M\|$?

Question

Let $M\in \operatorname{Mat}_{d\times d}(\Bbb{C})$ and define $\|M\|:=\sqrt{\sum_{i,j} |M_{i,j}|^2}$. Let $X,Y\in \operatorname{GL_{d\times d}}(\Bbb{C})$ s.t. $X^HX=I, Y^HY=I$. I want to show that $\|XMY\|=\|M\|$

My Idea was the following. First we know that $\langle AX,Y\rangle=\langle X, A^HY\rangle$ and $I^H=I$. Therefore $$\begin{align}\|XMY\|^2&=\langle XMY, XMY\rangle \\&=\langle X^HXM,MYY^H\rangle \\&=\langle M, M(Y^HY)^H\rangle\\&=\langle M,M\rangle =\|M\|^2\end{align}$$

Which proves the statement. Does this wok?

Answer 1

I'm afraid your second step $\langle XMY, XMY\rangle=\langle X^HXM,MYY^H\rangle $ is not valid, since we can only do the conversion $X\to X^H$ from the left side, not the right side. The correct proof should be \begin{align}\|XMY\|^2&=\langle XMY, XMY\rangle \\ &=\langle X^HXMY,MY\rangle \\ &=\langle MY,MY\rangle\\ &=\langle Y^HM^H,Y^HM^H\rangle\\&=\langle YY^HM^H,M^H\rangle \\&=\langle M^H,M^H\rangle \\&=\|M\|^2\end{align}

Answer 2

The only way I can prove it is using tensor notation.

Start with the definition, $$ \|M\|^2 = {M^j}_i {{M^*}_j}^i = {M_j}^i {{M^*}^j}_i $$ since if $M = {M^j}_i$ then $M^H := {{M^*}_j}^i$ (in words, a Hermitian matrix is the conjugate transpose of a matrix).

From the definition of $X$ and $Y$, $$ {X^j}_i {{X^*}_j}^{i'} = \delta^{i'}_i $$ and, $$ {Y^j}_i {{Y^*}_j}^{i'} = \delta^{i'}_i $$

Since, $$ {(XMY)^j}_i = {X^k}_i {M^l}_k {Y^j}_l $$ then, $$ {(XMY)_j}^i = {(XMY)^{j'}}_{i'} g_{j'j} g^{i'i} $$ where $g_{j'j}$ (resp. $g^{i'i}$) is the metric tensor of the vector space (resp. dual space). If we use the formula for $XMY$ the last equation can be rewritten as, $$ {X^{k'}}_{i'} {M^{l'}}_{k'} {Y^{j'}}_{l'} g_{j'j} g^{i'i} $$ Using these last two equations we can write, $$ \|XMY\|^2 = {X^k}_i {M^l}_k {Y^j}_l {{X^*}^{k'}}_{i'} {{M^*}^{l'}}_{k'} {{Y^*}^{j'}}_{l'} g_{j'j} g^{i'i} $$

First apply $g^{i'i}$ to the product ${X^k}_i{{X^*}^{k'}}_{i'}$: $$ {X^k}_i{{X^*}^{k'}}_{i'}g^{i'i} = {X^k}_i {X^*}^{k'i} $$ Next we repeat the same for the product of $Y$s: $$ {Y^j}_l {{Y^*}^{j'}}_{l'} g_{j'j} = {Y^j}_l {Y^*}_{jl'} $$ If we rewrite ${X^*}^{k'i}$ as ${{X^*}_{k''}}^ig^{k''k'}$ then, $$ {X^k}_i {X^*}^{k'i} = {X^k}_i {{X^*}_{k''}}^i g^{k''k'} = \delta^k_{k''} g^{k''k'} $$ In the same way we can show that, $$ {Y^j}_l {Y^*}_{jl'} = \delta^{l''}_l g_{l''l'} $$ We are now left with the following expression: $$ \|XMY\|^2 = {M^{l}}_k {{M^*}^{l'}}_{k'} \delta^k_{k''} g^{k''k'} \delta^{l''}_l g_{l''l'} $$ Since, $$ {M^{l}}_k \delta^k_{k''} \delta^{l''}_l= {M^{l''}}_{k''} $$ we have, $$ \|XMY\|^2 = {M^{l''}}_{k''}g^{k''k'}g_{l''l'}{{M^*}^{l'}}_{k'} = {M_{l'}}^{k'}{{M^*}^{l'}}_{k'} = \| M \|^2 $$