Finding if a linear transformation is diagonalisable

Question

Hi i am having some trouble tackling this question for my exam revision.

Let $M_{(2,2)}(\mathbb{R})$ denote the vector spce of 2x2 matrices over the real numbers. Also, let A denote the matrix $$\begin{bmatrix}2&\lambda\\1&0\end{bmatrix}$$

where $\lambda$ is a real number. Consider the map

$$\psi : V\rightarrow V, \psi(X)=AX-XA. $$

Compute the eigenvalues for $\psi$. For which values of $\lambda$ is $\psi$ diagonalisable?(check the dimension of the kernel of $\psi$.)

It is clear to see that $\psi$ is a linear transformation and hence it is diagonalisable if it has two distinct real eigenvalues.

The problem i am having is how do I find the eigen values of this linear transformation. I have tried finding the eigen values for the matrix $A$ and obtain $\lambda > -1$ but i do not think this is correct as I haven't considered the kernel or the actual function itself at all.

Secondly i tried using the hint and firstly i obtained the kernel

$$AV - VA=0$$ $$\begin{bmatrix}2&\lambda\\1&0\end{bmatrix} \begin{bmatrix}v_1&v_2\\v_3&v_4\end{bmatrix}-\begin{bmatrix}v_1&v_2\\v_3&v_4\end{bmatrix}\begin{bmatrix}2&\lambda\\1&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$ rearranging I obtained $$\begin{bmatrix}-v_2+\lambda v_3&2v_2+\lambda v_4-\lambda v_1\\v_1-2v_3-v_4&v_2-\lambda v_3\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$$ from this i obtained $\lambda=\frac{v_2}{v_3}$ so the ker($\psi$)=$c\begin{bmatrix}2&\frac{v_2}{v_3}\\1&0\end{bmatrix}$ for any $c,v_2,v_3 \epsilon\mathbb{R}$ i am stuck on how am i supposed to compute the eigen values ? and i can see that the $dim(ker)=1$ using the rank nullity theorem this means the $dim(im(\psi))$=1

Answer 1

I suppose there is confusion. First the vector space used is espace of size 4 which is the canonical basis {E11 E12 E21 E22}. When evaluating \ psi (Eij) we obten the matrix of \ psi relativly to the canonical basis Eij. Then its characteristic polynomial il is X ^ 2 (X ^ 2-4 (\ lambda +1)). So the value 0 is an eigenvalue of multiplicity 2 and the dimension of associated eigensubspace is 2, and {(1,0,0,1), (2, \ lambda, 1,0)} is a basis. Other roots of characteristic polynmial are simple but must be real (to end what are the eigenvalues) becaus the vector space is real. So we need \ lambda greater than or equal to -1, and in this cas the transformation is diagonalisable. Note if you remplace R with the complex number C, \psi is diaginalisable for all \lambda.

Answer 2

$\psi$ is a linear transformation between $M_{2\times 2} \rightarrow M_{2\times 2},$ so you need to find a matrix representation $[\psi]$ of $\psi$ and compute the eigenvalues and eigenvectors of $[\psi]$. Since $M_{2\times 2}$ is 4-dimensional, $[\psi]$ will be a $4 \times 4$ matrix. It will be diagonalizable if and only if it has 4 linearly independent eigenvectors. It is not necessary that it has 4 distinct eigenvalues. It may have only 1 eigenvalue with an eigenspace of dimension 4, for example.