How to show diagonalisability?

Question

Let $$A= \left[ \begin{array}{ c c } 3 & 2 \\ -1 & 1 \end{array} \right] $$ Let $f : ^2\mathbb{C}\rightarrow^2\mathbb{C}$ be the linear map given by $f(u) = Au$, $\forall u ∈ ^2\mathbb{C}$. Determine the eigenvalues of $f$. Show that $f$ is diagonalisable, and find a diagonal matrix $D$ and an invertible matrix $P$ (both with complex entries) such that $P^{−1}AP = D$.

For the eigenvalues, I got $\lambda_1=2+i$ and $\lambda_2=2-i$. How do I show that $f$ is diagonalisable? I also tried to find the eigenvecctors

For $\lambda_1=2+i$: I ended up with the equations $$(1-i)x+2y=0$$ $$-x+(-1-i)y=0$$ Solving this gave me $x,y=0$.

For $\lambda_2=2-i$: I ended up with the equations $$(1+i)x+2y=0$$ $$-x+(-1+i)y=0$$ Solving this gave me $x,y=0$.

Answer 1

$$\det(xI-A)=\begin{vmatrix}x-3&-2\\1&x-1\end{vmatrix}=x^2-4x+5=(x-2-i)(x-2+i)$$

For example, for

$$\lambda=2-i:\;\;-(1+i)x-2y=0\implies y=-\frac{1+i}2x$$

so an eigenvector is, for example,

$$\binom{\;\;1}{\!\!\!-\frac{1+i}2}$$

Do the other one similarly.

Being a $\;2\times 2\;$ matrix with two different eigenvalues it is automatically diagonalizable.

Answer 2

Every distinct eigenvalue has at least one eingenvector. Then this matrix has two linear independent eingenvectors because it has two distinct eigenvalues. That is a basis for $\mathbb{R}^2$.