Prove that $A$ is similar to $\begin{bmatrix} \lambda & 1\\ 0 & \lambda \end{bmatrix}$ 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!
 A: 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:
$$S^{-1}AS=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$$
A: Well, start with your eigenvector $v$. Then take a second vector $w$ to get a basis $\{v, w\}$. We have $Av = \lambda v$ and $Aw = xv + yw$ for some unknown scalars $x, y$. Thus, with respect to this basis, we have
$$[A] = \begin{bmatrix} \lambda & x \\ 0 & y \end{bmatrix}$$
A common principle here is that because $y$ is on the diagonal like this, $A - y \mathrm{Id}$ is singular. So $y$ is an eigenvalue, which means $y = \lambda$. Thus
$$[A] = \begin{bmatrix} \lambda & x \\ 0 & \lambda \end{bmatrix}$$
And now it's just a matter of playing around with things to get $x = 1$. Well first, can you see that $x \neq 0$ given these hypotheses?
Then the next step is to just kind of play around with a change of basis: $\{v, av + bw\}$ until you can get $x = 1$. Try simple cases like $b = 1$ or $a = 0$ first.
Edit: use a computer algebra system if it helps:
$$\begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix} \begin{bmatrix} \lambda & x \\ 0 & \lambda \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix}^{-1} = \;?$$
A: Since $\gamma(\lambda) = 1$, it follows that the minimal polynomial of $A$ is $p_A(t) = (t-\lambda)^2$, hence thanks to Jordan canonical form theory, $A$ is similar to $\begin{bmatrix}\lambda & 1 \\ 0 & \lambda\end{bmatrix}$
