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