I have the matrix $A=\begin{pmatrix}0& 0& 1\\1& 0& -3\\0& 1& 3\\\end{pmatrix}$ to put into Jordan normal form. I have looked at various websites but they all seem to use different methods and I'm getting very confused. I have that the eigenvalues are $1$ (multiplicity $3$) and I tried to row reduce $(A-I)$ to get the eigenvectors but I got the identity matrix implying that the eigenvectors are $0$. The textbook I have shows an example which suggests that I should calculate $(A-I)^2$, then choose a vector $v$ (at random?) so that $v$ is not in the nullspace of $(A-I)^2$ but I don't follow what happens after that.
Thank you!
The matrix's characteristic polynomial is
$$\;\begin{vmatrix}t&0&\!\!-1\\\!\!-1&t&3\\0&\!\!-1&t-3\end{vmatrix}=(t-1)^3\;$$
Calculating the eigenspace ofr the unique eigenvalue $\;\lambda=1\;$ (btw, we already know the matrix isn't diagonalizable. Why?), inputting $\;t=1\;$ and forming the associated homogeneous system:
$$\begin{cases}&\;\;\;x&&-z=0\\&-x+&y+&\;3z=0\end{cases}\;\;\implies x=z\;,\;\;y=-2z$$
and thus the eigenspace is of dimension one, which gives he JNF (why in the above we don't need the third equation?)
$$\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$$