Let $A$ be a $3\times 3$ matrix defined by $$A=\begin{pmatrix} 3&-1& 1\\2 & 0 &1\\1&-1&2\end{pmatrix}$$
- Find the eigenvalues as well as the eigenspaces related to it.
- Prove that the matrix $A$ has a triangular form $$T=\begin{pmatrix} \lambda_1&m&0\\0 &\lambda_1 &0\\0&0&\lambda_2\end{pmatrix}$$ where $m$ is a constant to be determined.
My Attempt The eigenvalues are $\lambda_1=2$, and $\lambda_2=1$, with $\lambda_1$ having a double multiplicity (i.e $P(\lambda)=-(1-\lambda)(2-\lambda)^2$).
The spaces are $$V_{\lambda_1}=Vect\{(1,1,0)^{\top}\}$$ $$V_{\lambda_2}=Vect\{(0,1,1)^{\top}\}$$ In all of my attempts, I tried to complete the basis with a canonical vector of $\mathbb R^3$. I tried $(1,0,0)^{\top}$, $(0,0,1)^{\top}$, and $(1,1,1)^{\top}$.
My problem is that at every single attempt, I get these two triangular matrices, $$\begin{pmatrix} 2&0&1\\0 & 1 &0\\0&0&2\end{pmatrix}, \hspace{0.5cm} \begin{pmatrix} 2&0&1\\0 & 1 &1\\0&0&2\end{pmatrix} $$
I really don't understand what's wrong with my attempts, I thought I might've miscalculated the inverse, but turns out it's correct, with every basis completing. Probably that's the problem too.
Any help or guidance is much appreciated.
let $\lambda_{1}=2$,find a basis $\{f_{1}=(0,1,1);f_{2}\}$ for $V_{\lambda_{1}}^{*}=Ker(T-\lambda_{1})^{2}$,let $f_{3}=(1,1,0)$ ,wrt basis $\{f_{1};f_{2};f_{3}\}$ the matrix has the desired form. Remark that you can always pick $m=1$ by replacing $f_{2}$ by $t f_{2}+ f_{1}$ for a suitable choice of $t$.