If we know the eigenvalues of a matrix $A$, and the minimal polynom $m_t(a)$, how do we find the Jordan form of $A$?

Question

We have just learned the Jordan Form of a matrix, and I have to admit that I did not understand the algorithm.

Given $A = \begin{pmatrix} 1 & 1 & 1 & -1 \\ 0 & 2 & 1 & -1 \\ 1 & -1 & 2 & -1 \\ 1 & -1 & 0 & 1 \end{pmatrix} $, find the Jordan Form $J(A)$ of the matrix.

So what I did so far:

(I) Calculate the polynomial: $P_A(\lambda) = (\lambda - 1)^2(\lambda -2)^2$.

(II) Calculate the minimum polynomial: $m_A(\lambda) = P_A(\lambda) =(\lambda - 1)^2(\lambda -2)^2 $

But I am stuck now, how do we exactly calculate the Jordan Form of $A$?

And an extra question that has been confusing me. In this case, does $A$ have $4$ eigenvalues or $2$ eigenvalues?

Answer 1

Since the minimal polynomial has degree $4$ which is the same order of the matrix, you know that $A$'s smith normal form is $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & m_A(\lambda )\end{pmatrix}$.

Therefore the elementary divisors (I'm not sure this is the correct term in english) are $(\lambda -1)^2$ and $(\lambda -2)^2$.

Theory tells you that one jordan block is $\color{grey}{(\lambda -1)^2\to} \begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ and the other is $\color{grey}{(\lambda -2)^2\to} \begin{pmatrix} 2 &1\\ 0 & 2\end{pmatrix}$.

Therefore a possible JNF for $A$ is $\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 2 &1\\ 0 & 0 & 0 & 2\end{pmatrix}$.

Regarding the extra question, it seems to be asking about the geometric multiplicity. In this case it has two eingenvalues and not four.