Prove that $A$ is similar to $\begin{bmatrix} \lambda & 1\\ 0 & \lambda \end{bmatrix}$

Question

I need some help with this one.

Let $A \in M_{2×2}(\mathbb{R})$ such that $A$ has exactly one eigenvalue $λ$, with $\gamma(\lambda)=1$,

then $A$ is similar to $\begin{bmatrix}
\lambda & 1\\
0 & \lambda
\end{bmatrix}$.

Here is what I know so far:

The geometric multiplicity of $λ$ is $1$, then we have one eigenvector, let's say $\Bbb v$.

We can maybe take another vector, $\Bbb v_1$, such that $(\Bbb {v,v_1})$ is a basis.

I do not know Jordan form, and I do know theorems like Cayley-Hamilton, if that helps.

Thanks a lot, I appreciate your time!

Answer 1

Outline: Pick any $\mathbf v_1\neq 0$ which is not an eigenvector.

Let $\mathbf v_2=A \mathbf v_1-\lambda \mathbf v_1$ Show $\mathbf v_2$ is an eigenvector and that $\mathbf v_1, \mathbf v_2$ are a basis.

Then consider how $A$ acts on that basis:

$$\begin{align}A \mathbf v_1&=\lambda \mathbf v_1+ \mathbf v_2\\A \mathbf v_2&=0 \mathbf v_1+\lambda \mathbf v_2\end{align}$$

Then $S=\begin{pmatrix}\mathbf v_2&\mathbf v_1\end{pmatrix}$ satisfies:

$$S^{-1}AS=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$$