Diagonalize $\left(\begin{smallmatrix} 1 & 3 \\ 0 & 6 \end{smallmatrix}\right)$ 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.
 A: 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 $E_1$, $E_2$, $E_3$, $E_4$


*

*$E_1=\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}$

*$E_2=\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$

*$E_3=\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}$

*$E_4=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}$
$$T_A=\begin{pmatrix}1 & 0 &3&0\\ 0 & 1 &0&3\\ 0 & 0 &6&0\\ 0 & 0 &0&6 \end{pmatrix}$$
A: Based on the title, it seems that you think the problem is asking you to diagonalize  $A=\begin{pmatrix} 1 & 3 \\ 0 & 6 \end{pmatrix}$. Not so! $A$ is not the matrix representing $T$!  Since $M_2(\Bbb R)$ has dimension $4$, the matrix representing $T$ is a certain $4\times 4$ matrix.
Not that you should try to write down that matrix, instead you should do the problem "abstractly", looking for a basis of $M_2(\Bbb R)$ consisting of eigenvectors of $T$. (Note that an eigenvector for  $T$ is a $2\times 2$ matrix.)
So I'd call this a "problem" as opposed to an "exercise", meaning that there's something you have to figure out; it's not just a routine application of techniques covered in class and in the book. In fact once you've diagonalized $A$ it follows that you can diagonalize $T$, but you have to figure out how that  works - how does diagonalizing $A$ lead to a diagonalization of $T$?
Regarding diagonalizing $A$, which again is not the problem: This should be routine. You're stuck on finding a basis for the nullspace of $I-A$; there's an algorithm for finding a basis for the nullspace of a matrix that you should simply apply here! When you say it looks like the only solution to $(I-A)v=0$ is $v=(0,0)$ you're ignoring that algorithm.
Now regarding the actual problem, which asks you to diagonalize  $T$: I'd split the part you have to figure out into two pieces:
(i) Suppose first that $A$ is actually diagonal, and show that it follows that $T$ is diagonalizable. Hint: If  $A$ is diagonal then every matrix  in the standard basis for $M_2(\Bbb R)$ is an eigenvector of $T$, with eigenvalue ...
(ii) Now show how just diagonalizing $A$ shows that $T$ is diagonalizable.
Hint: Say $T=T_A$, and given a  diagonal matrix $D$ define $T_D:M_2(\Bbb R)\to M_2(\Bbb R)$ by $T_DM=DM$. Now if you've diagonalized  $A$ you have a diagonal  matrix  $D$ and an invertible matrix $P$ such that $A=PDP^{-1}$. Your  solution to (i)  gives you a complete set of eigenvectors  for $T_D$, and since $$T_AM=P^{-1}T_D(PM)$$ this allows you to calculate a complete set of eigenvectors for $T_A$. (Hint: Show that the eigenvectors for $T_A$ are precisely the matrices $P^{-1}M$, where $M$ is an eigenvector for $T_D$.)
