Is canonical basis matrix A for $T(x,y)=(2ix+y,x)$ diagonalisable?

Question

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?

Answer 1

$$ \left( \begin{array}{rr} 0 & 1 \\ 1 & -i \end{array} \right) \left( \begin{array}{rr} 2i & 1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{rr} i & 1 \\ 1 & 0 \end{array} \right)= \left( \begin{array}{rr} i & 1 \\ 0 & i \end{array} \right) $$ 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 $

Answer 2

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.