Let $T: \Bbb M_{3x3}(\Bbb R) \rightarrow \Bbb M_{3x3}(\Bbb R)$ be a linear transformation such that $T(B) = AB$ where:
$$ A= \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{pmatrix} $$
now let $W = \{B|AB=BA\}$ be the group of all matrices that commute with $A$.
I want to
- find the minimal polyonimal for $T|_w$.
- prove that $W$ is the direct sum of two spaces $W = W_1 \oplus W_2$
my idea so far was:
Let P be any matrix:$$ P =\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} $$ then we could look at the transformation as $\Bbb R^9 \rightarrow \Bbb R^9$ like that: $$ T\begin{pmatrix} a \\ d \\ g \\ b \\ e \\ h \\ c \\ f \\ i \\ \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{pmatrix} & & \\ & \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{pmatrix} & \\ & & \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \\ \end{pmatrix} \\ \end{pmatrix} \begin{pmatrix} a \\ d \\ g \\ b \\ e \\ h \\ c \\ f \\ i \\ \end{pmatrix} $$
Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \\ \end{pmatrix} $$
which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$
my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...
Can you help me to better understand the restriction? Thanks..
$W$ is a vector space and not a group. Note that $T=A\otimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)\subset W$. Moreover $spectrum(T)=\{0,0,0,3,3,3,3,3,3\}$; cf.
https://en.wikipedia.org/wiki/Kronecker_product
We may assume that $A=diag(0,3,3)$ -because $(P\otimes I)(A\otimes I)(P\otimes I)^{-1}=(PAP^{-1})\otimes I$-.
Thus $W=\{diag(a,R);a\in\mathbb{R},R\in M_2\}$ has dimension $5$. Moreover $W=W_1\oplus W_2$ where $W_1=span(E_{1,1}),W_2=span(E_{2,2},E_{2,3},E_{3,2},E_{3,3})$ and $T_{|W_1}=0,T_{|W_2}=3I$.
Finally, $T_{|W}$ has $2$ eigenvalues $0,3$ and $spectrum(T_{|W})=\{0,3,3,3,3\}$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.