Find eigenspace for eigenvalues of a linear transformation $T(M)=-2M^t+M$

Question

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

Answer 1

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 $V^{S}_{-1}$.

Answer 2

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.

Answer 3

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 $n=4$.)