Given the following matrix: $$ A = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 1 & 2 \end{pmatrix} $$ We find its eigenvalues, and eigenvectors: $$ |A - \lambda \mathbb{I}| = (3 - \lambda)^3 = 0 $$ We get the eigenvalue: $\lambda = 3$, with geometrical multiplicity $k = 3$. $$ \dim(\ker(A - 3 \mathbb I)) = 2 \\ \; \\ \ker(A - 3 \mathbb I) = \text{span} \left \{ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right \} $$ So the matrix is not diagonalizable. So we try to find its Jordan normal form. But I have some doubts. First, how would you know that the Jordan form is like: $$ \begin{pmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} $$ or like this: $$ \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{pmatrix} $$ how can I know?
And, how would I find the change of basis matrix $P$? I've tried with: $$ (A - 3 \mathbb{I})^2 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \\ \; \\\ \ker(A - 3 \mathbb{I})^2 = \mathbb{R}^3 = \text{span} \left \{ \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right \} $$ Would this three last vectors be the columns of the matrix $P$?
Could someone explain me this in depth, and how would you relate this vectors with the Jordan form (in a similar way, you can relate the vectors of a diagonal matrix with the basis of its eigenspace).
Both of the matrices that you mentioned are Jordan normal forms of $A$.
Let $v_1=(1,0,0)$ and let $v_2=(0,1,1)$. Then $A.v_1=3v_1$ and $A.v_2=3v_2$. Besides, if $(x,y,z)\in\Bbb R^3$, then$$A.(x,y,z)-3(x,y,z)=(0,y-z,y-z).$$Clearly, if you take, say $y=1$ and $z=0$, then $A.(x,y,z)-3(x,y,z)=(0,1,1)=v_2$. So, let $v_3=(0,1,0)$. Then $A.v_3=3v_3+v_2$, and therefore, if you take$$P=\begin{bmatrix}1&0&0\\0&1&1\\0&1&0\end{bmatrix}$$(the columns of $P$ are $v_1$, $v_2$, and $v_3$), then$$P^{-1}.A.P=\begin{bmatrix}3&0&0\\0&3&1\\0&0&3\end{bmatrix}.$$