I have the following matrix $M$: \begin{pmatrix} 4&-1&1\\ 2&1&2\\ 1&-1&4 \end{pmatrix}
I have to find a matrix $C$ such that $C^{-1}MC$ is in Jordan normal form.
I have found that the characteristic polynomial is $p(x)=-(x-3)^3$.
So the eigenvalue is $3$ and base eigenvectors are $\left(\begin{smallmatrix}1\\1\\0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}-1\\0\\1\end{smallmatrix}\right)$.
So $C = \left(\begin{smallmatrix}1&-1&?\\1&0&?\\0&1&?\end{smallmatrix}\right)$
Is this correct? And how do i find the last vector? Thank you.
Actually, no $C$ of that type will work.
The eigenspace corresponding to the eigenvalue $3$ is$$\langle(1,1,0),(-1,0,1)\rangle=\{(x,x+z,z)\mid x,y\in\Bbb R\}.$$Now, let us see when is it that the system$$M.(a,b,c)=3(a,b,c)+(x,x+z,z)$$has a solution. The system is$$\left\{\begin{array}{l}a-b+c=x\\2a-2b+2c=x+z\\a-b+c=z.\end{array}\right.\tag1$$Clearly, it has a solution if and only if $x=z$. Take, say, $x=z=1$; in other words, take the vector $(1,2,1)$. Then a solution of the system $(1)$ is $(1,0,0)$. In other words,$$M.(1,0,0)=3(1,0,0)+(1,2,1).$$Take an eigenvector $v_1$ of $M$ corresponding to the eigenvalue $3$ such that $V_1$ and $(1,2,1)$ are linearly independent; for instance, take $v_1=(0,1,1)$. Now, let $v_2=(1,2,1)$, and let $v_3=(1,0,0)$. Then $v_1$, $v_2$, and $v_3$ are linearly independent, $M.v_1=3v_1$, $M.v_2=3v_2$, and $M.v_3=v_2+3v_3$. So,$$C=\begin{bmatrix}0&1&1\\1&2&0\\1&1&0\end{bmatrix}$$will work:$$C^{-1}.M.C=\begin{bmatrix}3&0&0\\0&3&1\\0&0&3\end{bmatrix}.$$