Example of real matrices $2\times 2$ and $3\times 3$ that are not similars to a diagonal matrix.
I find that $A =\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} $ then i suppose that its similar to some diagonal matrix and found $D = \begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix} $ = $2I$ which is not the case.
But I can't find a real matrix $3 \times 3$ that is not similar to a diagonal matrix. Some help for this please.
Since the eigenvalues of $\begin{bmatrix}a&b\\0&c\end{bmatrix}$ are $a$ and $c$, and having $n$ distinct eigenvalues for an $n$-dimensional transformation implies the transformation is diagonalizable, you'll need to look at cases where $a=c$.
Then, say, we could try $\begin{bmatrix}1&b\\0&1\end{bmatrix}$, which just has $1$ as an eigenvalue. If $b=0$ this is obviously diagonalizable, so say $b\neq 0$.
The equation $\begin{bmatrix}1&b\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x+by\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$ would imply $by=0$, hence $y=0$. Then the only eigenvectors possible are $\begin{bmatrix}x\\0\end{bmatrix}$, which only furnishes a $1$ dimensional eigenspace for the value $1$.
Keeping in mind that diagonalizable transformations must have a basis consisting of eigenvectors, we can see that is not possible here, so the matrix isn't diagonalizable.
Try to do something similar for $3\times 3$ matrices. Using just $1$ and $0$ and keeping the matrices as simple as possible will allow you to argue in a simliar way.