The whole question is this:
$A=\begin{pmatrix}1 & 3\\ 0 & 6\end{pmatrix}$
$T:M_2(\mathbb{R})\rightarrow M_2(\mathbb{R})$ is a linear transformation defined by $T(M)=AM$
I need to show that T is diagonizable and find basis B of $M_2(\mathbb{R})$ such that $[T]_B$ is diagonal.
I've tried the following:
$P_A(t)=|tI-A|$ is the characteristic polynomial
$P_A(t)=(t-1)(t-6)$
Hence, the eigenvalues are $\lambda=1,6$
$V_\lambda = P(\lambda I-A)$, where $P$ is the solution space.
I got stuck here:
$V_1=(I-A)=\begin{pmatrix}0 & -3\\0 & -5\end{pmatrix}$ which leaves me the only solution of $\begin{pmatrix}0\\0\end{pmatrix}$. Seems to me like it's too bad for an eigenvector.
Best Answer
HINT
Note that a not trivial solution to $V_1x=0$ is $x=(1,0)$.
Anyway as noticed by David we need to consider the matrix associated to the given transformation $T(M)=AM$, that is with reference to the basis $E_1$, $E_2$, $E_3$, $E_4$
$E_1=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$
$E_2=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$
$E_3=\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}$
$E_4=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}$
$$T_A=\begin{pmatrix}1 & 0 &3&0\\ 0 & 1 &0&3\\ 0 & 0 &6&0\\ 0 & 0 &0&6 \end{pmatrix}$$