<s>[INST] $\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ not diagonalizable I would like to ask you about this problem, that I encountered:
Show that there exists no matrix T such that $$T^{-1}\cdot 
 \left( \begin{array}{cc}
1 & 1  \\
0 & 1 \\
 \end{array} \right)\cdot T $$ is diagonal.
In other words our matrix let's call it A cannot be diagonalizable. (A being the matrix "in between the T's").
I saw the following: $$\left( \begin{array}{cc}
1 & 1  \\
0 & 1 \\
 \end{array} \right)=\left( \begin{array}{cc}
1 & 0  \\
0 & 1 \\
 \end{array} \right)+\left( \begin{array}{cc}
0 & 1  \\
0 & 0 \\
 \end{array} \right)$$
Let's denote them: $$A=D+N$$
Also easy to see is that $DN =ND$ and $N^{2}=0$.
It follows that
$(D+N)^{t}=D^{t}+tN = \text{Identity}^{t}+t\left( \begin{array}{cc}
0 & 1  \\
0 & 0 \\
 \end{array} \right)=\left( \begin{array}{cc}
1 & t  \\
0 & 1 \\
 \end{array} \right)$
Note: the algebraic expression was reduced to this, since all terms $N^2$ and higher are $0$, also $D=\text{Identity}$.
But I somehow fail to see why from here one can deduce (or not) that $A$ is not diagonalizable.
Any hint or help greatly appreciated!
Thanks
 [/INST] A different approach:
Your matrix has a single eigenvalue $\lambda=1$ (with multiplicity $2$) so if it was diagonalizable it would be similar to the identity, i.e.
$$P^{-1} \begin{pmatrix}
1 & 1  \\
0 & 1 \\
 \end{pmatrix} P=\begin{pmatrix}
1 & 0  \\
0 & 1 \\
 \end{pmatrix} \Rightarrow \\
\begin{pmatrix}
1 & 1  \\
0 & 1 \\
 \end{pmatrix}=P \begin{pmatrix}
1 & 0  \\
0 & 1 \\
 \end{pmatrix}P^{-1}=PP^{-1}=\begin{pmatrix}
1 & 0  \\
0 & 1 \\
 \end{pmatrix};$$
contradiction.
More generally the matrix 
$$
A= \begin{pmatrix}
1 & * &* &\ldots & * & *\\
0 & 1 &* &\ldots & * & *\\
0 & 0 &1 &\ldots & * & *\\
\vdots & \vdots &\vdots &\ddots & \vdots &\vdots \\
0 & 0 &0 &\ldots & 1 & * \\
0 & 0 &0 &\ldots & 0 & 1 \\
 \end{pmatrix}
$$
is diagonalizable iff $A=I$.
</s>