Is canonical basis matrix A for $T(x,y)=(2ix+y,x)$ diagonalisable? I'm working from Kaye and Wilson's Linear Algebra textbook. Exercise 13.2 asks if the matrix $A$ with respect to the usual basis of $\mathbb{C}^2$ for the linear map $T:\mathbb{C}^2\to \mathbb{C}^2$ given by $T(x,y)=(2ix+y,x)$ is diagonalisable.
My attempt:
The usual basis for $\mathbb{C}^2$ is $\{(1,0), (0,1)\}$.
So I got $A=\big(\begin{smallmatrix}
  2i & 1\\
  1 & 0
\end{smallmatrix}\big)$.
Now $\chi_A(x)=(x-i)^2$ so $A$ has only 1 eigenvalue, namely $\lambda =i$ and I found that $Av=\lambda v$ implies $v=t(1,-i), t\in\mathbb{R}$.
I'm not sure how to progress. I know that $A$ being diagonalisable means there is some invertible $P$ such that $P^{-1}AP$ is diagonal, right?
 A: Notice that $A$ is diagonalisable iff $a(\lambda_k)=g(\lambda_k)$, where $a(\lambda_k)$ denote the algebraic multiplicity of $\lambda_k$, in this case $a(i)=\color{blue}{2}$ because  $\chi_{A}(x)=(x-i)^{\color{blue}{2}}$ and $g(\lambda_k)$ is the geometric multiplicity of $\lambda_k$, in this case $g(i)=\dim E_i=1$ with $E_i={\rm span}\{(i,1)\}$. Thus, $A$ is not diagonalisable in the usual sense.
Very close to the previous argument, notice that since $A\in M_{2}(\bf C)$ then $2$ independent eigenvectors diagonalize $A$, but in this case we have only the eigenvector $(i, 1)$. Therefore, $A$ can not be diagonalisable.
NB: The Jordan decomposition, is of course possible in this case. But I suppose that you're working with the diagonalisation in the usual sense of eigenvectors non generalized.
A: $$
\left(
\begin{array}{rr}
0 & 1 \\
1 & -i
\end{array}
\right)
\left(
\begin{array}{rr}
2i & 1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
i & 1 \\
1 & 0
\end{array}
\right)=
\left(
\begin{array}{rr}
i & 1 \\
0 & i
\end{array}
\right)
$$
which is, well, partway to diagonal.  Call it $J = \left(
\begin{array}{rr}
i & 1 \\
0 & i
\end{array}
\right) $
The specific diagonal element does not much matter.  Can we find another matrix $P$   so that $P^{-1}  J P = D$ is diagonal? If so, the diagonal entry would remain $i.$  Also, we can multiply  $P$   by a constant, possibly complex, so that $\det P = 1.$  Thus if
$P = \left(
\begin{array}{rr}
a & b \\
c & d
\end{array}
\right) $    with $ad-bc = 1,$  we get $P^{-1}= \left(
\begin{array}{rr}
d & -b \\
-c & a
\end{array}
\right)$   and we are trying to solve
$$
???  \; \;
 \left(
\begin{array}{rr}
d & -b \\
-c & a
\end{array}
\right) 
\left(
\begin{array}{rr}
i & 1 \\
0 & i
\end{array}
\right) 
 \left(
\begin{array}{rr}
a & b \\
c & d
\end{array}
\right) =
\left(
\begin{array}{rr}
i & 0 \\
0 & i
\end{array}
\right) \; \; ???
$$
Assume there is such a matrix $P.$
Multiplying the left side while using $ad-bc=1$  leads to
$$
??? \; \;
\left(
\begin{array}{rr}
i +cd & d^2 \\
-c^2 & i-cd
\end{array}
\right)=
\left(
\begin{array}{rr}
i & 0 \\
0 & i
\end{array}
\right) \; \; ???
$$
Clearly $d=c=0.$  However, this contradicts $ad-bc = 1 $
