In each of the cases below, decide whether the linear operator $T : \mathbb K^n \to \mathbb K^n$ given by its matrix $[T]_B$ is diagonalizable. If so, calculate an eigenvector basis and its diagonal shape.
a)$\begin{pmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{pmatrix}$ ; and $\mathbb K = \mathbb R, \mathbb C$
b)$\begin{pmatrix} -4 & -1 \\ 4 & 0 \end{pmatrix}$ ; and $\mathbb K = \mathbb R$
My attempt
a)$det \alpha = det \begin{pmatrix} 6 - \lambda & -3 & -2 \\ 4 & -1 -\lambda & -2 \\ 10 & -5 & -3 -\lambda\end{pmatrix} = 0 \to -\lambda^3 + 2 \lambda^2 - \lambda + 2$,
so
$\lambda_1 = 2, \lambda_2 = i , \lambda_3 =-i $
$\lambda_1 = 2$
$\begin{pmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{pmatrix}\begin{pmatrix} a \\ b \\ c \end{pmatrix} = 2 \begin{pmatrix} a \\ b \\ c \end{pmatrix} \to \begin{pmatrix} 6a -3b -2c \\ 4a -b -2c \\10a -5b - 3c \end{pmatrix} = \begin{pmatrix} 2a \\ 2b \\ 2c \end{pmatrix} $
$6a -3b -2c = 2a;4a -b -2c = 2b; 10a -5b - 3c = 2c \to c = 2a, b = 0$
so
$v_1 = (1,0,2)$
$\lambda_2 = i$
$\begin{pmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{pmatrix}\begin{pmatrix} d \\ e \\ f \end{pmatrix} = i \begin{pmatrix} d \\ e \\ f \end{pmatrix} \to \begin{pmatrix} 6d -3e -2f \\ 4d -e -2f \\10d -5e - 3f \end{pmatrix} = \begin{pmatrix} d i \\ e i \\ f i \end{pmatrix} \to d = e, f = \frac{3}{2}-\frac{i}{2}d$
$v_2 = (1,1,\frac{3}{2}-\frac{i}{2})$
$\lambda_3 = -i$
$\begin{pmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{pmatrix}\begin{pmatrix} w \\ y \\ z \end{pmatrix} = -i \begin{pmatrix} w \\ y \\ z \end{pmatrix} \to \begin{pmatrix} 6w -3y -2z \\ 4w -y -2z \\10w -5y - 3z \end{pmatrix} = \begin{pmatrix} w i \\ y i \\ z i \end{pmatrix} \to w = y, z = \frac{3}{2}+\frac{i}{2}w$
$v_3 = (1,1,\frac{3}{2}-\frac{i}{2})$
Now we have
$\beta = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 2 & \frac{3}{2}-\frac{i}{2} & \frac{3}{2}+\frac{i}{2} \end{pmatrix}$
now, getting $\beta^{-1}$ and doing $\beta^{-1} \alpha \beta$ we get the diagonalization right?
b)$det \begin{pmatrix} -4 -\lambda & -1 \\ 4 & 0 -\lambda \end{pmatrix} = 0 \to \lambda^2 + 4\lambda + 4 = (x+2)^2 \to \lambda = -2$
$\begin{pmatrix} -4 & -1 \\ 4 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = -2 \begin{pmatrix} a \\ b \end{pmatrix} \to \begin{pmatrix} -4a - b \\ 4a \end{pmatrix}\begin{pmatrix} -2a \\ -2b \end{pmatrix} \to b = -2a \to v=(1,-2)$ is not diagonalizable ?
Not sure if b) is right? a) is not right, but is the procedure correct?
Thanks.
(a)$v_3 \text{ should be } (1,1,3/2+i/2) $, which is the conjugate vector of the one found for $i.$ You don't have to calculate the diagonal form. The theorem guarantees it will be$$ \text{diag}(2,i,-i)$$(b) Yes, the multiplicity of the eigenvalue is 2 and the dimension of the eigenspace is 1 so the matrix is not diagonalizable.