Prove that $A$ is not similar to any diagonal matrix, but is similar to a lower triangular matrix. 
Show that A is not similar to a diagonal matrix, but is similar to a triangular matrix of the form $\begin{bmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{bmatrix}$.

$$
A = 
\begin{bmatrix}
2 & -1 \\
0 & 2 \\
\end{bmatrix}
$$
The book from which I'm self-studying doesn't really tell me when a matrix is diagonalisable, so, I resorted to Wikipedia and found this:

An $n×n$matrix $A$ over a field $F$is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to $n$

So, the eigenvalues of $A$ is
$$
det 
\begin{bmatrix}
\lambda -2 & 1 \\
0 & \lambda -2 \\
\end{bmatrix} = 0 \\
(\lambda -2)^2 = 0$$
Thus, eigenvalues = 2,2.
Eigenvectors:
$$
\begin{bmatrix}
2 & -1 \\
0 & 2 \\
\end{bmatrix} \times
\begin{bmatrix}
x_1 \\
x_2 \\
\end{bmatrix} 
= 
\begin{bmatrix}
2x_1 \\
2x_2 \\
\end{bmatrix}\\
2x_1 -x_2 = 2x_2 \implies x_2 = 0 \\
2x_2 = 2x_2
$$
Thus, the eigenspace = $(x,0)$, the basis is $(1,0)$, thus the dimension is 1. The dimension of eigenspace is less than the dimension of the matrix, therefore, $A$ is not diagonalisable.
But I'm not sure if I'm right, because the statement of Wikipedia is $A \iff B$, and in our case $B$ is false, does that imply $A$ is also false?
The second part of the question says that we have to prove that $A$ is similar to $\begin{bmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{bmatrix}$. How can I show that? I have two options:

*

*To find a $C$ such that $\begin{bmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{bmatrix}= C^{-1} A ~C$, or

*To show that $\begin{bmatrix} \lambda & 0 \\ 1 & \lambda \\ \end{bmatrix}$ represents the same linear transformation. (By now you might have got an impression that my concepts are very fragile in this topic)

Going by first option, I assume there exists such a $C$ and I assumed it to be $\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$, and proceeded thus
$$
\begin{bmatrix}
 \lambda & 0 \\ 
  1 & \lambda \\ 
\end{bmatrix}= C^{-1} A ~C $$
$$
\begin{bmatrix} 
a & b \\
c & d 
\end{bmatrix} \times 
\begin{bmatrix} \lambda & 0 \\
  1 & \lambda 
 \end{bmatrix}
= 
\begin{bmatrix}
2 & -1 \\
0 & 2 
\end{bmatrix} \times
\begin{bmatrix} 
a & b \\
c & d \\
\end{bmatrix}
$$
I will get four equations by equating element-wise.
I need a detailed explanation for solving this problem.
 A: You calculated that the matrix is not diagonalizable, and it has one real eigenvalue with algebraic multiplicity of 2 and geometric multiplicity of 1, so by the Jordan's theorem its canonical jordan form is $\begin{bmatrix} 2 & 0 \\ 1 & 2 \\ \end{bmatrix}$, or $\begin{bmatrix} 2 & 1 \\ 0 & 2 \\ \end{bmatrix}$. It depends on how you choose your basis for the change but it does not matter because one is the transpose of the other
A: you don't have enough eigenvectors, so something less must be used. We have $(A-2I)^2 = 0.$   We take any column vector  that is not an eigenvector, say
$u=\left(\begin{array}{c}0 \\1 \\\end{array}\right).$
We know $(A-2I)^2 u = 0.$
Calculate $v = (A-2I)u.$   We have $(A-2I) v = (A-2I)^2 u = 0$
so that  this $v$   will be a genuine eigenvector.  Here
$v=\left(\begin{array}{c}-1 \\0 \\\end{array}\right).$
We place the vectors as the columns of a matrix $P.$  For lower triangular  the order is $(u,v)$  so
$$P=\left(\begin{array}{cc}
0&-1 \\
1&0 \\
\end{array}\right).$$
Since $\det P = 1$  the usual recipe says $ P^{-1} = -P.$  Put another way, $P^2 = - I$
Finally
$$J= P^{-1}AP =
\left(\begin{array}{cc}
2&0 \\
1&2 \\
\end{array}\right).$$
