Finding the Jordan Form of a matrix... I know that this type of question has been asked on here before but I am still having a hard understanding what is going on. The text that I am learning from is "Linear Algebra Done Right by Sheldon Axler" and I don't think it covers this topic very well. Does anyone have any suggestions? I am studying for my linear comp at the end of January.
Given the matrix
$$A = \left(\begin{matrix}0&-1&-1\\-3&-1&-2\\7&5&6\end{matrix}\right).$$
Find the Jordan form $J$ and an invertible matrix $Q$ such that $A = QJQ^{-1}$.
I know that I want to start by finding the eigenvalues of this matrix. I ended up getting det$(A - \lambda I) = (\lambda + 2)(\lambda62 + 3\lambda - 2) \implies \lambda = -2, \frac{-3 -\sqrt{17}}{2}, \frac{\sqrt{17} - 3}{2}$.
Now, from what I have read on this website, I know now that I want to find the null space of $A - \lambda I$ for each eigenvalue. Then my invertible matrix $Q$ will have entries whose columns are these vectors. Why does one do this? Is there a "nicer" way to go about doing this other than direct computation?
This problem is from a previous linear comp at my university, so I can't imagine that they would have wanted me to the direct calculation.
Hopefully, my questions and explanation of the problem were clear. Thanks for any help in advance.
 A: Eigenvalues are $2,2,1$, which simplifies the calculation a lot.
A: The characteristic polynomial is
$$
    p(\lambda)=\lambda^3-5\lambda^2+8\lambda-4=(\lambda-1)(\lambda-2)^2.
$$
The vectors involved in the Jordan blocks associated with eigenvalue $2$ is the column space of
$$
           A-I = \begin{pmatrix}-1 & -1 & -1 \\
                                -3 & -2 & -2 \\
                                 7 &  5 &  5\end{pmatrix}.
$$
The column space of $A-I$ is two-dimensional, which gives $\mbox{dim}(\mbox{ker}(A-I))=1$, and it is obvious that $\mbox{ker}(A-I)$ is spanned by
$$
     \begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}.
$$
(It is obvious because the last two columns of $A-I$ are identical.) Then
\begin{align}
     (A-2I)(A-I)&=\begin{pmatrix}-2 & -1 & -1 \\
                                -3 & -3 & -2 \\
                                 7 &  5 &  4\end{pmatrix}.
\begin{pmatrix}-1 & -1 & -1 \\
                                -3 & -2 & -2 \\
                                 7 &  5 &  5\end{pmatrix} \\
  &= \begin{pmatrix}-2 & -1 & -1 \\
        -2 & -1 & -1 \\
        6 & 3 & 3\end{pmatrix}
\end{align}
Because $(A-2I)^2(A-I)=0$, you have
\begin{align}
        (A-2I)\begin{pmatrix} -1 \\ -2 \\ 5\end{pmatrix}&=\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} \\
       (A-2I)\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} &= 0.
\end{align}
The Jordan form is
$$
          J = \begin{pmatrix} 1 & 0 & 0 \\
                    0 & 2 & 1 \\
                    0 & 0 & 2\end{pmatrix}
$$
and the transition matrix $Q$ is
$$
            Q = \begin{pmatrix} 0 & -1 & -1 \\
                                1 & -1 & -2 \\
                                -1 &  3 & 5
                \end{pmatrix}.
$$
