If I have the matrix:
$$M=\begin{pmatrix}-1 & 1 & 1 \\ -10 & 5 & 3 \\ -2 & 1 & 3 \end{pmatrix} $$
I know the matrix $M$ is triangularisable, but I am not sure how to find $A$ such that $M=P^{-1}AP$ to find the corresponding triangular matrix $A$.
So far I have found the eigenvalues are $2,3$, then once I do $M-3I$, I get that the basis for that matrix is $\begin{pmatrix}1 & 2 & 1 \end{pmatrix}^T$ and $\begin{pmatrix}1 & 3 & 1 \end{pmatrix}^T$. However I am not sure how to go any further? Can someone show me how? Thanks!
The vector $(1,2,2)$ is an eigenvector of $M$ corresponding to the eigenvalue $3$ and the vector $(0,-1,1)$ is an eigenvector of $M$ corresponding to the eigenvalue $2$. Now, consider the equation$$M.(x,y,z)=2(x,y,z)+(0,-1,1).$$One solution of this system is $(1,3,0)$. So, if $v_1=(1,2,2)$, $v_2=(0,-1,1)$, and $v_3=(1,3,0)$, you have $M.v_1=3v_1$, $M.v_2=2v_2$, and $M.v_3=2v_3+v_2$. So, take$$P=\begin{bmatrix}1&0&1\\2&-1&3\\2&1&0\end{bmatrix}$$(the columns of $P$ are the vectors $v_1$, $v_2$, and $v_3$). Then$$P^{-1}MP=\begin{bmatrix}3&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$