Let $$T = \left(\begin{array}{ccc} 0&-1&-1\\-3&-1&-2\\7&5&6 \end{array}\right)$$ Find the Jordan Form of T, then a basis relative when T is in Jordan Form.
I've found that the characteristic polynomial $f(x)=$ minimal polynomial $p(x)=(x-1)(x-2)^2$, which gives me the eigenvalues of $\lambda=1,2$.
To find the Jordan Form, I looked at the dimensions of $Ker(T-I)$ and $Ker(T-2I)$ to figure out how many Jordan Blocks and their size. So I have determined that the Jordan Form is $$J_T = \left(\begin{array}{ccc} 1&0&0\\0&2&0\\0&1&2 \end{array}\right)$$
Is this correct? If so, how do I continue with this to find the basis of T in jordan form?
Apart from finding the dimensions of $\ker(T-I)$ and $\ker(T-2I)^2$ you need to find a basis for them: an eigenvector for the first, a chain for the second, $(T-2I)v=w$, $(T-2I)w=0$. $$\ker(T-I)\mapsto \begin{pmatrix}0\\1\\-1\end{pmatrix}$$ $$\ker(T-2I)^2\mapsto\begin{pmatrix}1\\1\\-3\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}$$
So taking the change of basis matrix $P=\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&-3\end{pmatrix}$ gives $$P^{-1}AP=J=\begin{pmatrix}1&0&0\\0&2&0\\0&1&2\end{pmatrix}$$