Determine if T can be diagonalized (in $\mathbb C$ and $\mathbb R $)

Question

Let $\ T: \mathbf R^3 \rightarrow \mathbf R^3 \ \ T(x,y,z) = (3x, 2y-5z, y-2z) $ for every $\ (x,y,z) \in \mathbf R^3 $

I need to find if T can be diagonalised So basically I know I'm looking for $\ Tv - I_3\lambda = 0 $

$\begin{bmatrix} 3 & 0 & 0 \\\ 0 & 2 & -5 \\\ 0 & 1 & - 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} = \begin{bmatrix} 3 - \lambda & 0 & 0 \\ 0 & 2-\lambda & -5 \\ 0 & 1 & -2 -\lambda \end{bmatrix}$

and the determinant of this matrix will be $\ (3- \lambda)(\lambda^2 + 1) =0 $ we have only one eigenvalue of $\mathbb R $ which is 3 and the eigen vector will be $\ Sp \{(1,0,0) \} $ so under $\mathbb R $ the $\ T$ matrx can not be diagonalized. and Under $\ \mathbb C $ we have another two eigen values, $\ i, -i $ so I get this linear equations which I'm not sure how to proceed from here:

$\ (3-i)x =0 \\ (2-i)y -5z = 0 \\ y + (-2-i)z = 0 $

Answer 1

There’s no compelling reason to actually compute a diagonalizing basis for $T$ in order to prove this. $T$ has three distinct eigenvalues, so its characteristic polynomial is also its minimal polynomial. Over $\mathbb C$, this obviously splits into the product of three distinct linear terms, hence $T$ is diagonalizable over $\mathbb C$, but the factor $x^2+1$ is irreducible over $\mathbb R$, therefore $T$ is not diagonalizable over that field.

If you do need to find eigenvectors for the two complex eigenvalues, proceed as you would if they were real: solve the system of linear equations. The presence of $i$ in the system doesn’t really change the solution method, but you’ll end up with complex solutions.

Answer 2

The matrix diagonalizes over $\mathbb{C}$ since it is $3\times 3$ and has three distinct eigenvalues.

As for your three equations, you have $x = 0$. Now, look at the other two by solving for $y$ in each. What do you see?