We define :
- $n \in \mathbb{N}$ is fixed
- all the matrix are squared of size $n$, with real coefficients.
- if $A$ is a matrix $(n,n)$ we define $A'$ its transposed, instead of $^t{A}$
- $\sigma(A): = tr(AA') $ where $tr$ is the trace.
- for $q\leq n$, for $p \leq n$, we define $E_{p,q}$ is the matrix with all zeros except $1$ at intersection of row $p$ and column $q$.
- $\mid X \mid$ is the euclidian norm of the vector $X$ : $ (X'X)^{\frac{1}{2}}$
- We suppose that there is a particular matrix $S$ of size $(n,n)$ such that : $$\forall B, \sigma(BS)= \sigma(SB) \tag{1}$$
Our goal is to prove that $\forall(p,q), \mid C_q \mid = \mid L_p \mid$ where $C_q$ is the row $q$ of $S$ and the $L_p$ is the row $p$ of $S$. To do that, we apply the hypothesis $(1)$ with $B= E_{p,q}$.
My attempt :
We choose $p\leq n$ and $q \leq n$. We note $(e_1,...e_n)$ the basis. $f$ is the endorphism represented by $S$ and $h$ the endomorphism represented by $E_{p,q}$
$
h(e_i)=
\begin{cases}
e_p \quad \text{if} \quad i=q \\
0 \quad \text{else}
\end{cases}
$
$\forall i ,\quad s(e_i)= \sum_{k=1}^{n} s_{k,i}e_k$
$h \circ f(e_i)= s_{q,i} e_p$
so $E_{p,q} S$ is null except of its row $p$ that is equal to the row $q$ of $S$ written $L_q$
so $(E_{p,q} S)'$ is null except for its column $p$ that is equal to $L_q'$
so $ E_{p,q} S (E_{p,q} S)' = E_{p,p} \mid L_q \mid ^2$ so $\sigma( E_{p,q} S )= \mid L_q \mid ^2 $. We do a product per block. $ E_{p,q} S = ( \underbrace{0,0,0,..0}_{p-1},L_q,0,0)$
$ f \circ h(e_j) = \begin{cases} f(e_p) \quad \text{if} \quad i=q \\ 0 \quad \text{sinon} \end{cases} $
so $f(e_p)= \sum_{k=1}^{n} s_{k,p} e_p$ so $ S E_{p,q}$ is the null matrix, except that its column $q$ is equal to $C_p$ ( the column $p$ of $S$).
I would like to do a product per block to conclude that
$\sigma(S E_{p,q})= \mid C_p \mid ^ 2$
but the product per block is not available here.
Thanks,