How to find Jordan normal form and basis? I have a matrix:
$$A = \begin{pmatrix}-1&5&4\\ 4&-6&-6\\ -8&16&14 \end{pmatrix}$$
simplifying:
$$A = \begin{pmatrix}-1&5&4\\ 2&-3&-3\\ -4&8&7 \end{pmatrix}$$
What steps should I reproduce to find Jordan form and Jordan basis, I am really really confuswd without good step by step example!
What I really understand that I have to find eigenvalues, so I did:
$$\det|A-I\lambda| = \begin{vmatrix}-1 - \lambda&5&4\\ 2&-3-\lambda&-3\\ -4&8&7-\lambda \end{vmatrix}  $$
so I found cubic equation which is:
$$-\lambda^3+3\lambda^2-5\lambda+3 = 0$$
$$-(\lambda-1)^3-2(\lambda-1)$$
So as I can judge Jordan normal form will have three eigenvalues that equal to $\lambda = 1$ on its main diagonal, below main diagonal all the values should be "$0$". But I bet it is not all. I do not know how to proceed (If I am right at all), and do not know how to find Jordan's basis after :(
 A: I didn't find a satisfying answer writting about both algorithm to find basis and Jordan normal form and why this algorithm works, so I decide to write one by myself.

Suppose $f$ is a linear operator, whose characteristic polynomial factors into degree-one polynomial (repetition allowed), and its matrix representation under some basis is $A$. The algorithm to find Jordan normal form is as following:

*

*Calculate the characteristic polynomial and all eigenvalues of $f$, denoted as $\lambda_1,...,\lambda_n$.

*Calculate geometric multiplicity of each $\lambda_i$. The number of Jordan blocks of eigenvalue $\lambda$ is the geometric multiplicity of $\lambda$. The number of Jordan blocks of eigenvalue $\lambda$ of size $k$ is equal to $$2\cdot \dim(\ker(\lambda\cdot Id-f)^k)-\dim(\ker(\lambda\cdot Id-f)^{k-1})-\dim(\ker(\lambda\cdot Id-f)^{k+1}).$$
By these two information above, we can already determine the Jordan normal form.

Actually, there are more information given by algebraic multiplicity and minimal polynomial:
the Jordan block of the largest size of eigenvalue $\lambda$ is equal to the power of $(x-\lambda)$ in minimal polynomial of $\lambda\cdot Id-f$. And the sum of sizes of all Jordan blocks of eigenvalue $\lambda$ is equal to the algebraic multiplicity of $\lambda$.
To calculate the basis corresponding to the Jordan normal form, we need more work: from the calculation above, we can know that there are $m$ Jordan blocks of eigenvalue $\lambda$ and their sizes are $k_1,...,k_m$. Find a basis for $\ker(\lambda\cdot Id-f)=V_{\lambda}$, denoted as $b_1,...,b_m$. For each $b_i$, we recursively calculate $b_i^{j}$:

*

*Let $b_i^{0}=b_i$,

*Suppose we have calculated $b_i^{j}$, let $b_i^{j+1}$ be a vector s.t. $(\lambda\cdot Id-f)\cdot b_i^{j+1}=b_i^j$.

Such process will terminate at some $l$. Suppose we have got $b_i^{0},...,b_i^{l}$. We will see that they are linear independent and they are a basis corresponding to a Jordan block $J_l(\lambda)$. Perform this algorithm on each eigenvalue $\lambda$ and each eigenvector $b_i$. Put them together, we'll get a basis corresponding to the Jordan normal form.

Now, we wonder why this works. We'll firstly look at a special case -- nilpotent operator (i.e., an operator $f$ s.t. there exists $n\in\mathbb{N}^{>0}$ s.t. $f^n=0$).
$\textbf{Definition 1.}$ Suppose $V$ is a vector space and $f:V\to V$ is an linear operator. A $\textbf{cyclic subspace}$ associated to $f$ and $v$ is the subspace of $V$ spanned by $\{v,f(v),f^{2}(v),...\}$.
$\textbf{Claim 1.}$ Suppose $f$ is a nilpotent operator and $v\in V$. If $k$ is the least positive number s.t. $f^k=0$, then $\{v, f(v),...,f^{k-1}(v)\}$ is a basis of $Z(f,v)$.
$\textbf{Claim 2.}$ If $f$ is a nilpotent operator, then there exists $v_1,...,v_n\in V$ s.t. $$V=Z(f,v_1)\oplus Z(f,v_2)\oplus\dots\oplus Z(f,v_n).$$
Note that if we take $\{v_1,f(v_1),...,f^{k_1-1}(v_1),...,v_n,f(v_n),...,f^{k_n-1}(v_n)\}$ as  basis of $V$, then the matrix representation of $f$ under this basis is
\begin{pmatrix} J_{k_1}(0) & &\\
                         & \ddots & \\
                         &   & J_{k_n}(0).\\              
   \end{pmatrix}
Now, look at an arbitrary linear operator $f:V\to V$. We have the following lemma.
$\textbf{Lemma 1.}$ (Fitting's Lemma) If $V$ is a finite-dimensional vector space and $f:V\to V$ is a linear operator. Then, $V_N=\cup_{k=1}^{\infty} \ker(f^k)$ and $V_I=\cap_{k=1}^{\infty} \ker(f^k)$ are two $f$-invariant subspaces s.t. $f\upharpoonright_{V_N}$ is nilpotent and $f\upharpoonright_{V_I}$ is invertible. $V=V_N\oplus V_I$. Moreover, if $U$ and $W$ are two $f$-invariant subspaces of $V$ s.t. $V=U\oplus W$, $f\upharpoonright_U$ is nilpotent and $f\upharpoonright_W$ is invertible, then $U=V_N$ and $W=V_I$.
Once we have Lemma 1, we can use induction to prove the existence of Jordan normal form. However, we need to be careful that it is possible for $V_N$ to be zeros space. Therefore, for an arbitrary $f$, we use Fitting's lemma for $\lambda\cdot Id-f$, where $\lambda$ is an eigenvalue of $f$. Then, the kernel of $\lambda\cdot Id-f$ is nontrivial. Thus, we can apply Fitting's lemma and induction hypothesis on $V_I$ of it. This is also why we need to require the characteristic polynomial of $f$ to factor into product of linear forms.
(I'll write about this point in a bit more detailed way when I have time.)
A: I have characteristic equation
$-\lambda^3 + 7\lambda^2 -16\lambda + 12 = -(\lambda - 2)^2(\lambda-3)$
lets get the easy ones out of the way.
$\lambda = 3$
$\pmatrix{-4&5&4\\4&-9&-6\\-8&16&11}\mathbf v_1 = 0\\
\mathbf v_1 = \pmatrix{3\\-4\\8}$
$\lambda = 2$
$\pmatrix{-5&5&4\\4&-8&-6\\-8&16&12}\mathbf v_2 = 0\\
\mathbf v_2 = \pmatrix{4\\-4\\8}$
Curious.
When you have $Av = 0$ you can simplify $A$ with elementary row operations to find $v.$
Now for the odd man...
$A \mathbf v_3 = 2v_3 + v_2\\
(A - 2I)v_3 =v_2$
If you perform any row operations, that is the equivalent to multiply on the left some matrix, you must also apply the same operations to $\mathbf v_2$
$\pmatrix{-3&5&4\\4&-8&-6\\-8&16&12}v_3 =\pmatrix{4\\-4\\8}\\
\pmatrix{-3&5&4\\0&-4&-2\\0&0&0}v_3 =\pmatrix{4\\4\\0}\\
\mathbf v_3 = \pmatrix{-3\\-1\\0}$
$\pmatrix{-1&5&4\\4&-6&-6\\-8&16&14}\pmatrix{3&4&-3\\-4&-4&-1\\8&8&0} = \pmatrix{3&4&-3\\-4&-4&-1\\8&8&0}\pmatrix{3\\&2&1\\&&2}$
Update:
Since I had a doubled eigenvalue and only found one vector for that eigenvalue, I took a bit of a short cut.  I knew that I had this Jordan block, and 
$A[v_2,v_3] = [v_2,v_3]\begin{bmatrix} 2&1\\&2\end{bmatrix} = [2v_2,v_2+v_3]$
and solved for that.
There is another way to do it, and it might line up with your notes better.
If you have a defective eigenvalue.  You can find the generalized eigenvector this way.
(It starts out the same)
We have found $\mathbf v_2$ as an eigenvector.
$(A-\lambda I)\mathbf v_3 = \mathbf v_2\\
(A-\lambda I)^2\mathbf v_3 =  (A-\lambda I) \mathbf v_2$
Since $\bf v_2$ is an eigenvector $(A-\lambda I) \mathbf v_2 = \mathbf 0$
$(A-\lambda I)^2\mathbf v_3 = \mathbf 0$
And in this problem
$(A-\lambda I)^2 = \pmatrix{-3&9&6\\4&-12&-8\\-8&24&16}$
$\mathbf v_3 = \pmatrix{-3\\-1\\0}$ is in the kernel of $(A-\lambda I)^2$ 
