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Let $V$ be the matrix space $4 \times 4$ over $\Bbb R$. $T: V \to V$ is a linear transformation defined by: $$T(M)=-2M^t+M$$ for all $M \in V$.

  1. Find the minimal polynomial of T.
  2. For every eigenvalue $\lambda$ of $T$, find the eigenspace $V_\lambda$ and calculate its dimension. Find $T$'s characteristic polynomial.

I have solved section 1 this way:

Let $S(M)=M^t$. Therefore $S^2(M)=M \Rightarrow S^2=I$.
$$T=-2S(M)+S^2$$ Since $$ S^2=M$$

$$\Rightarrow T=-2S+I$$

$$\Rightarrow 2S=I-T$$

$$\Rightarrow 4S^2=(I-T)^2=I^2-2T+T^2=I-2T+T^2$$ Since $$4S^2=4I$$ $$\Rightarrow T^2-2T-3I=0$$ $$\Rightarrow(T+1)(T-3)=0$$

When further explanations we get the the eigenvalues are $$\lambda=-1$$ $$\lambda =3$$

However, to find eigenspace I need the original matrix, to calculate $$(A-\lambda I)$$ How do I find such a matrix for calculation?

Thanks,

Alan

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Let me point out something useful:

If $T \colon V \rightarrow V$ is diagonalizable with eigenvalues $\lambda_1, \ldots, \lambda_k$ and corresponding eigenspaces $V^T_{\lambda_1}, \ldots, V^T_{\lambda_k}$ and $p(X) = a_0 + \ldots + a_nX^n$ is any polynomial with $a_i \in \mathbb{F}$ then $p(T)$ is also diagonalizable with eigenvalues $p(\lambda_1), \ldots, p(\lambda_k)$ and the same eigenspaces. More precisely,

$$ V^{p(T)}_{p(\lambda_i)} = \bigcup_{\{1 \leq j \leq k \, | \, p(\lambda_j) = p(\lambda_i) \} } V^T_{\lambda_j}. $$

In your case, $S$ satisfies $S^2 = \mathrm{id}$ and so it is diagonalizable with eigenvalues $\lambda_1 = 1$ and $\lambda_2 = -1$. We have

$$ V^S_{1} = \{ A \, | \, S(A) = A^t = A \} = \mathrm{span} \{ \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right), \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right), \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) \} $$

and

$$ V^S_{-1} = \{ A \, | \, S(A) = A^t = -A \} = \mathrm{span} \{ \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right) \}. $$

Thus, $T = -2S + \mathrm{id} = p(S)$ (with $p(X) = -2X + 1$) is diagonalizable with eigenvalues $p(\lambda_1) = p(1) = - 1$ and $p(\lambda_2) = p(-1) = 3$ and eigenspaces $V^{S}_1$ and $V^{S}_{-1}$.

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  • $\begingroup$ Thank you, but why the eigenvalue $\lambda=1$ has an eigenspace of three vectors and the other eigenvalue only one vector? $\endgroup$ – Alan Nov 7 '15 at 15:42
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    $\begingroup$ Well, there are three linearly independent $2 \times 2$ symmetric matrices and one linearly independent $2 \times 2$ anti-symmetric matrix. $\endgroup$ – levap Nov 7 '15 at 15:43
  • $\begingroup$ I see, that's the case by definition. Thank you. $\endgroup$ – Alan Nov 7 '15 at 15:44
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    $\begingroup$ Since $T$ is diagonalizable and you know the dimensions of the eigenspaces and the eigenvalues, you can immediately write the characteristic polynomial: $(x + 1)^3(x - 3)$. $\endgroup$ – levap Nov 7 '15 at 15:46
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    $\begingroup$ Forgot that, thank you so much. $\endgroup$ – Alan Nov 7 '15 at 15:47
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You want $M \in V$ such that $T(M)=-M$ and $T(M)=3M$. Let us consider the first equation. \begin{align*} T(M) & = -M\\ M-2M^t & = -M\\ M & = M^t. \end{align*} Thus $$V_{\lambda=-1}=\{M \in V\, | \, M=M^t\}.$$ or simply said: set of all $4 \times 4$ symmetric matrices.

Likewise you will get $$V_{\lambda=3}=\{M \in V\, | \, M=-M^t\}.$$ or simply said: set of all $4 \times 4$ anti-symmetric matrices.

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  • $\begingroup$ Thank you, but how do I translate it to numbers? And the two missing vectors, I just add the standard basis vectors? $\endgroup$ – Alan Nov 7 '15 at 15:21
  • $\begingroup$ I understand from your answer that the dimension of each eigenvector is... 2? $\endgroup$ – Alan Nov 7 '15 at 15:24
  • $\begingroup$ I may found it, since A is symmetric and anti symmetric at the same time it has to be that A=0? $\endgroup$ – Alan Nov 7 '15 at 15:33
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Every matrix $M$ can be written as a symmetric and an antisymmetric part $$ M = M_s + M_a \\ M_s = \frac{1}{2}(M+M^t),\;\; M_a = \frac{1}{2}(M-M^t). $$ That is $M_s^t=M_s$ and $M_a^t=-M_a$. Then $$ TM_s = -2M_s+M_s = -M_s \\ TM_a = 2M_a+M_a = 3M_a. $$ So you have a basis of eignevectors for $T$ with eigenvalues $3$ and $-1$. That means $T$ has minimal polynomial $(\lambda+1)(\lambda-3)=\lambda^{2}-2\lambda-3$.

The symmetric matrices are determined by arbitrary numbers along and above the diagonal, which means the dimension is $(n^{2}-n)/2+n=n(n-1)/2+n$. The antisymmetric matrices have zeros along the diagonal, and the numbers above the diagonal can be arbitrary, which gives dimension $(n^{2}-n)/2=n(n-1)/2$. (You are working with $n=4$.)

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  • $\begingroup$ So you're saying I can assign any number I want to find the "original matrix" as long as it fits the criteria? $\endgroup$ – Alan Nov 7 '15 at 15:39
  • $\begingroup$ @Alan : I'm saying that the space of $n\times n$ matrices has a basis of symmetric and antisymmetric matrices. The symmetric matrices are determined by the numbers on and above the diagonal. The antisymmetric matrices are determined by the numbers above the diagonal because the diagonal entries must be $0$. So that's how you know the dimensions of these two subspaces of matrices. And, these two subspaces span the full space of matrices. The symmetric matrices are all eigenvectors of $T$ with eigenvalue $-1$ and the antisymmetric are eigenvectors with eigenvalue $3$. $\endgroup$ – DisintegratingByParts Nov 7 '15 at 19:47

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