Find a matrix in Jordan normal form that is similar to $$A=\begin{pmatrix} 0 & 4 & 2 \\ -3 & 8 & 3 \\ 4 & -8 & -2 \\ \end{pmatrix}$$
The characteristic equation of $A$ is $(\lambda-2)^3=0$, hence, $\lambda=2$ is a eigenvalue of algebraic multiplicity three. So eigenvector corresponds to $\lambda=2$, $$(A-2I)=\begin{pmatrix} -2 & 4 & 2 \\ -3 & 6 & 3 \\ 4 & -8 & -4 \\ \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$ $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\left\{\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}2\\1\\0\end{pmatrix} \right\}$$For one more eigenvector I tried to get one general eigenvector. $$(A-2I)=\begin{pmatrix} -2 & 4 & 2 \\ -3 & 6 & 3 \\ 4 & -8 & -4 \\ \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\0\\1\end{pmatrix}\text{ or }\begin{pmatrix}2\\1\\0\end{pmatrix}$$ But both of them are inconsistent system and I stuck. Then how to get complete eigenspace to find $P$ such that $A=PJP^{-1}$
If you want systematic approach, you need to know bit more about Jordanization. For normal eigenvalues with single or multiple eigenvectors (not generalized eigenvectors), you just put the eigenvalue in the diagonal matrix (n times, where n is the multiplicity of that eigenvalue) and put your correspondong eigenvectors in the corresponding columns in your P matrix.
For eigenvalues with generalized eigenvectors it's a little more than that. First of all, eigenvector means the vectors which are in the null space (i.e. kernel) of $(A-\lambda I)$, and lets call this matrix $W_1 = (A-\lambda I)$. Then generalized eigenvector means it is not in the $ker(W_1)$ but in $ker(W_i)$ with $i>1$ where $W_i = (A-\lambda I)^i$, and the $i$ shows the order of the generalized eigenvector. You can say that generalized eigenvector of order 1 is normal eigenvector. Before we proceed, you need to find the dimension of all $ker(W_i)$ with $i$ starting from one. And these dimensions is monotonically increasing with increasing $i$ until $i=m$ where m is the power of the term corresponding to this eigenvalue in the minimal polynomial and all the higher kernels are the same with the dimension equal to the multiplicity of the eigenvalue. All the lower kernel spaces are contained (strictly inside with non-zero difference subspace) in the higher kernel spaces until m. To illustrate, let us assume we have an eigenvalue with multiplicity 5 and dimension of the kernels are like these: $dim(ker(W_1)) = 2$, $dim(ker(W_2))=4$ and $dim(ker(W_3))=5$. (For general case we first need to find these dimensions). We can visualize this (kernel spaces) as:
Now, if we take a vector which is in $ker(W_i) \setminus ker(W_{i-1})$ and multiply it with $(A-\lambda I)$ we get a vector which is in $ker(W_{i-1})$, because if $(A-\lambda_i)^iv=0$ and $u = (A-\lambda_i)v$ then $(A-\lambda_i)^{i-1}u=(A-\lambda_i)^{i}v=0$, meaning $u \in ker(W_{i-1})$. If you continue with multiplying $(A-\lambda I)$, you'll get a chain of vectors ending at a certain eigenvector at the innermost space. For example, in the figure you take vector $f_{31}$ in the outermost space and find chain $f_{31}$-$f_{21}$-$f_{11}$, where $f_{21} = (A-\lambda I)f_{31}$, $f_{11} = (A-\lambda I)f_{21}$ and $f_{11}$ is a normal eigenvector.
Now, we know that the number of Jordan blocks is equal to the dimension of normal eigenvector space and that the dimesnion difference between kernels is non-decreasing; therefore, we must have n Jordan blocks and n chain of vectors for each of them. And the sizes of the Jordan blocks are in accordance with the dimension of the chains. This is a fact, we do not prrove anything here, just for clarification.
Thus, as a systematic approach for finding column vectors for P matrix corrresponding to the Jordan blocks of a certain "abnormal" eigenvalue is as follows:
You take a vector (or lin.ind. vectors if there are many) at the outermost space and find a chain of vectors like in the figure. When finished, you go to the lower disc and look for independent vectors if any, and this way find a chain for each lin.ind. normal eigenvector. To be more clear, lets follow our example on the figure.
Edit: We have two normal lin.ind. eigenvectors, that is the dimension of the first disc is 2. Therefore, we have two Jordan blocks and we should be able to find two chains for these two Jordan blocks. Previously I said we start from the outermost disc but this is due to simplicity. Actually, what we should do is the following, starting from normal eigenvectors we should find chain (we peviously stated what is the relation between vectors in a chain) for each eigenvector with maximum possible length. For our first eigenvector $f_{11}$ (it is up to you, you can choose either of two) we can find two more vectors in the second and the third discs respectively to construct our first chain $f_{31}$-$f_{21}$-$f_{11}$ and these vectors satisfy $f_{21} = (A-\lambda I)f_{31}$, $f_{11} = (A-\lambda I)f_{21}$. For our second eigenvector $f_{12}$, we can find a vector in second disc $f_{22}$ which is lin.ind. of $f_{21}$ and satisfy $f_{12}=(A-\lambda I)f_{22}$. However, for this second chain we cannot find a vector in the third disc that is linearly independent of $f_{31}$ since the dimension of that disc is one (i.e $ dim(ker(W_3)-ker(W_2)) = 1$). So we have to cut this chain here, with two vectors only. And that's it, we are done. When I said we start from outermost disc, I meant it is easy as when you multiply vector with $(A-\lambda I)$ you go to inner disc (get a vector in inner disc).
For our eigenvalue under consideration we have two Jordan blocks and two chains respectively:
$J = \begin{bmatrix} \ddots \\ && \lambda && 1 && 0 && \\ && 0 && \lambda && 1 && \\ && 0 && 0 && \lambda \\ && && && && \lambda && 1 \\ && && && && 0 && \lambda \\ && && && && && && \ddots\end{bmatrix}$, $P = \begin{bmatrix} \cdots && f_{11} && f_{21} && f_{31} && f_{12} && f_{22} && \cdots\end{bmatrix}$.
So, I hope you see the obvious systematic pattern here. Note that you may not change the order of the vectors in the chain. So you have flexibility for choosing certain vectors but the restriction is they should have chain pattern, they should be linearly independent and you must use all the vectors for certain eigenvalue (the chains should be as long as possible).
Edit2:
For your example, we have two normal eigenvectors, i.e. dimension of first disc is 2. As the multiplicity of the eigenvalue is 3, we need one more eigenvector which is second order generalized eigenvector and should be in the $ker((A-2I)^2)$ and should be linearly independent of the previous two. To visualize, we have:
So here the constraints on vectors to be satisfied are:
1) $f_1$, $f_2$ and $f_3$ should be linearly independent
2) $f_3$ should be in the set $ker((A-2I)^2) \setminus ker(A-2I)$. That is, $(A- 2I)^2f_3=0$, but $(A-2I)f_3 \neq 0$
3) $f_1 = (A-2I)f_3$.
So, choose any combination of vectors $f_1$, $f_2$, $f_3$ satisfying the above requirements. Then your Jordan form and P matrix is:
$J = \begin{bmatrix} 2 && 1 && 0 \\ 0 && 2 && 0 \\ 0 && 0 && 2 \end{bmatrix}$, $\hspace{1cm}$ $P = \begin{bmatrix} f_1 && f_3 && f_2\end{bmatrix}$.
First Method
Let us first start from inner disc. You can choose your $f_1$ and $f_2$ any two linearly independent vectors from $ker(A-2I) = span(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0\end{bmatrix})$. Let us choose $f_1 = \begin{bmatrix} 2 \\ 1 \\ 0\end{bmatrix}$, and $f_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$. As we said previously, we should find chain for each eigenvector with maximal length. For $f_1$ we can find one more vector in second disc to construct chain with length 2. But for $f_2$ we cannot, because dimension of second disc is 1. So second chain is going to have length 1. Now $f_3$ should satisfy constraints (2) and (3) stated above.
$f_1 = (A-2I)f_3 = \begin{bmatrix} 2 \\ 1 \\ 0\end{bmatrix} = \begin{bmatrix} -2 && 4 && 2 \\ -3 && 6 && 3 \\ 4 && -8 && -4\end{bmatrix}f_3$
But as you see the above equation does not have any solution. So we should choose $f_1$ so that the above equation has a solution for $f_3$. From the above equation due to the form of $(A-2I)$, $f_1$ should be of the form $k \begin{bmatrix} 1 \\ 1.5 \\ -2 \end{bmatrix}$ so that we have a solution for $f_3$. Then we can take our $f_1 = \begin{bmatrix} 1 \\ 1.5 \\ -2 \end{bmatrix} $ and $f_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$. Note that $f_1$ and $f_2$ again span the eigenspace (ker(A-2I)), but now our $f_1$ is so that we can find $f_3$ so that $f_1 = (A-2I)f_3$.
$f_1 = (A-2I)f_3 = \begin{bmatrix} 1 \\ 1.5 \\ -2 \end{bmatrix} = \begin{bmatrix} -2 && 4 && 2 \\ -3 && 6 && 3 \\ 4 && -8 && -4\end{bmatrix}f_3$.
We can take $f_3 = \begin{bmatrix} 2.5 \\ 1 \\ 1\end{bmatrix}$, then $f_3$ satisfies all the requirements: $f_3 \in ker{((A-2I)^2)}$, $f_3 \notin ker(A-2I)$, and $f_1=(A-2I)f_3$. Then our P matrix is:
$P = \begin{bmatrix} 1 && 2.5 && 1 \\ 1.5 && 1 && 0 \\ -2 && 1 && 1\end{bmatrix}$
We can check that:
$PJP^{-1}=A$.
Second Method
Now lets start from outer disc. Lets find $f_3$ that satisfies constraint (2) stated above. So $(A-2I)f_3 \neq 0$, but $(A-2I)^2f_3=0$. It means $f_3$ is not in the first disc, but it is in the second disc. As $(A-2I)^2 = 0_{3x3}$, any vector in $R^3$ is in its null space. So we should choose a vector for which $(A-2I)f_3 \neq 0$. For example a simple choice $f_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ satisfies the requirement. Now lets find $f_1$ so that $f_1 = (A-2I)f_3$:
$f_1 = (A-2I)f_3 = \begin{bmatrix} -2 && 4 && 2 \\ -3 && 6 && 3 \\ 4 && -8 && -4\end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ -4\end{bmatrix}$.
So now we have our first chain $f_1$, $f_3$. Since the dimension of the second disc is one, we do not search for another linearly independent vector there and go down to inner disc. The inner disc has dimension two and therefore we can find another vector there linearly independent of $f_1$. We can choose a simple one $f_2 = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix}$, so that it is independent of the other two vectors and is in $ker(A-2I)$ (inner disc). So our $f_1$, $f_2$ and $f_3$ satisfy the above constraints (1), (2) and (3), and we are done. The corresponding P matrix is:
$P = \begin{bmatrix} 2 && 0 && 1 \\ 3 && 0 && 1 \\ -4 && 1 && -1 \end{bmatrix}$
and to check:
$A = PJP^{-1}$.
From these two methods you can see that starting from outer disc is relatively easy, because you just multiply by (A-2I) to find the vector in inner disc, as I said before.