Joshua Foley

2022-07-16

representative matrix of a linear transformation
given a linear transformation: $T:{M}_{n}\left(\mathbb{C}\right)\to {M}_{n}\left(\mathbb{C}\right)$, $T\left(A\right)=A-2{A}^{T}$, what is the

Nicolas Calhoun

Let $\mathcal{A}$ denote the subspace of symmetric (antisymmetric matrices). Denote by ${E}_{ij}$ the matrix all of whose coefficients are zero except in the entry at the intersection of the i-th line and the $j$-th column, equal to 1. Thus ($\left({E}_{ij}\right)$) is the canonical basis of $V$.
For any $i we have $\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}\left({E}_{ij},{E}_{ji}\right)=\mathsf{s}\mathsf{p}\mathsf{a}\mathsf{n}\left({S}_{ij},{A}_{ij}\right)$ where ${S}_{ij}=\frac{{E}_{ij}+{E}_{ji}}{2}$ and ${A}_{ij}=\frac{{E}_{ij}-{E}_{ji}}{2}$
The transpose map $\tau :A↦{A}^{T}$ is easily seen to be diagonal in that basis : we have
$\tau {E}_{ii}={E}_{ii},\tau {S}_{ij}={S}_{ij},\tau {A}_{ij}=-{A}_{ij}$
Note that $\mathcal{S}$ is the eigenspace of $\tau$ corresponding to the eigenvalue $1$. So
$\left\{{E}_{ii}|1\le i\le n\right\}\cup \left\{{S}_{ij}|1\le i is a basis of $\mathcal{S}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{S}\mathcal{\right)}\mathcal{=}\frac{\mathcal{n}\mathcal{\left(}\mathcal{n}\mathcal{+}\mathcal{1}\mathcal{\right)}}{\mathcal{2}}$<brSimilarly $\left\{{S}_{ij}|1\le i is a basis of $\mathcal{A}$, and we deduce $\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{A}\mathcal{\right)}\mathcal{=}\frac{\mathcal{n}\mathcal{\left(}\mathcal{n}\mathcal{-}\mathcal{1}\mathcal{\right)}}{\mathcal{2}}$
For $A\in \mathcal{S}$ we have ${A}^{T}=A$ so $\sigma \left(A\right)=-A$.
For $A\in \mathcal{A}$ we have ${A}^{T}=-A$ so $\sigma \left(A\right)=5A$
So the characteristic polynomial ${\chi }_{A}$ of $A$ is
${\chi }_{A}=\left(X+1{\right)}^{\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{S}\right)}\left(X-5{\right)}^{\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{A}\right)}=\left(X+1{\right)}^{\frac{n\left(n+1\right)}{2}}\left(X-5{\right)}^{\frac{n\left(n+1\right)}{2}}$${\chi }_{A}=\left(X+1{\right)}^{\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{S}\right)}\left(X-5{\right)}^{\mathsf{d}\mathsf{i}\mathsf{m}\left(\mathcal{A}\right)}=\left(X+1{\right)}^{\frac{n\left(n+1\right)}{2}}\left(X-5{\right)}^{\frac{n\left(n+1\right)}{2}}$

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