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Question:

let $$V=\{A|A\in F^{n\times n}\}$$,define linear transformation $$\sigma:A\mapsto 2A-3A^T$$ show that

the matrix $\sigma$ Characteristic polynomial $$f(x)=(x+1)^n(x-5)^{\frac{n(n-1)}{2}}(x+1)^{\frac{n(n-1)}{2}}?$$

show that $\sigma$ can diagonalization

My try:this Characteristic polynomial reslut is my frend tell me,

so I can't ensure this is true.

Thank you very much!

2 Answers2

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Let $\cal S$ ($\cal A$) 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 $(E_{ij})$ is the canonical basis of $V$.

For any $i < j$, we have ${\sf span}(E_{ij},E_{ji})={\sf span}(S_{ij},A_{ij})$ where $S_{ij}=\frac{E_{ij}+E_{ji}}{2}$ and $A_{ij}=\frac{E_{ij}-E_{ji}}{2}$.

So the family $\lbrace E_{ii} | 1 \leq i\leq n \rbrace \cup \lbrace S_{ij} ; A_{ij} | 1 \leq i < j \leq n \rbrace $ is another basis of $A$.

The transpose map $\tau : A \mapsto A^{T}$ 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 $\cal S$ is the eigenspace of $\tau$ corresponding to the eigenvalue $1$. So $\lbrace E_{ii} | 1 \leq i\leq n \rbrace \cup \lbrace S_{ij} | 1 \leq i < j \leq n \rbrace $ is a basis of $\cal S$, and we deduce ${\sf dim}(\cal S)=\frac{n(n+1)}{2}$.

Similarly $\lbrace S_{ij} | 1 \leq i < j \leq n \rbrace $ is a basis of $\cal A$, and we deduce ${\sf dim}(\cal A)=\frac{n(n-1)}{2}$.

For $A\in \cal S$ we have $A^{T}=A$ so $\sigma(A)=-A$.

For $A\in \cal A$ we have $A^{T}=-A$ so $\sigma(A)=5A$.

So the characteristic polynomial $\chi_A$ of $A$ is

$$ \chi_A=(X+1)^{{\sf dim}({\cal S})} (X-5)^{{\sf dim}({\cal A})}= (X+1)^{\frac{n(n+1)}{2}} (X-5)^{\frac{n(n+1)}{2}} $$

Ewan Delanoy
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  • Thank you,I can't understand you anwser,why $V=S\bigoplus A$? and why $\sigma{(A)}=-A,A\in S$ and $\sigma{(A)}=5A,A\in A$ –  Dec 17 '13 at 16:22
  • We know that $(X+1)$ annihilates $\mathcal{S}$ and $(X-5)$ annihilates $\mathcal{A}$, so the minimal polynomial of $\sigma$ is $(X+1)(X-5)$. So the characteristic polynomial of $\sigma$ is $(X+1)^p(X-5)^q$, some $p,q$ such that $p+q=n^2$. Can you say a little bit about how we know $p=\text{dim}(\mathcal{S})$ and $q=\text{dim}(\mathcal{A})$? – Eric Auld Dec 17 '13 at 16:23
  • Oh, perhaps by the following argument: $\sigma$ is self-adjoint so diagonalizable, so it has diagonal form with $-1$ on the diagonal for the block corresponding to $\mathcal{S}$ and $5$ on on the block corresponding to $\mathcal{A}$. – Eric Auld Dec 17 '13 at 16:27
  • @EricAuld,why the minimal polynomial is $(X+1)(X+5)$ –  Dec 17 '13 at 16:45
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We have $$\sigma(\sigma(A))=\sigma(2A-3A^T)=2(2A-3A^T)-3(2A^T-3A)\\=13A-12A^T=13A+4(\sigma(A)-2A)=5A+4\sigma(A)$$ hence the polynomial $x^2-4x-5=(x+1)(x-5)$ with simple roots $-1$ and $5$ annihilates $\sigma$ so $\sigma$ is diagonalized and since $\sigma\ne k\mathrm {id}$ hence $-1$ and $5$ are two eigenvalues of $\sigma$. Now to find the multiplicity of the both eigenvalues we solve the equality $$\sigma(A)=kA\quad k=-1,5$$ for example for $k=-1$ we find $$2A-3A^T=-A\iff A=A^T\iff A\in S_n$$ and we know that $$\dim S_n=\frac{n(n+1)}{2}$$ and we find also that the multiplicity of $5$ is $\dim A_n=\frac{n(n-1)}{2}$ and we conclude.