So, I am given the matrix $$A =\begin{pmatrix}
0 & 1 & 0 & 0 & \\
-1 & 2 & 0 & 0 & \\
-2 & 2 & 1 & 0 & \\
0 & 1 & 0 & -1
\end{pmatrix}$$ which I need to put into Jordan canonical form (something like $A = P^{-1} J P$, where $J$ is the Jordan matrix). I have calculated the characteristic polynomial to be $(\lambda-1)^3(\lambda+1) = 0$, which implies there are two eigenvalues, $\lambda_1 = -1$, with multiplicity of one, and $\lambda_2 = 1$, with multiplicity of three.
Calculating the eigenvectors, I get that there is one eigenvector for $\lambda_1$ (call it $\mathbf{v_1} = (0, 0, 0, 1)^t$) and that $\lambda_2$ has two eigenvectors (call them $\mathbf{v_2} = (0, 0, 1, 0)^t$ and $\mathbf{v_3} = (2, 2, 0, 1)^t$).
So far I understand that I need to find one additional eigenvector since the dimension of $A$ is four, but I have run into trouble time and again while trying to compute it.
Since $\lambda_1$ has one eigenvector, I understand it will have a Jordan block of size one and that $\lambda_2$ will have a Jordan block of size two.
So, $$J = \begin{pmatrix}
-1 & 0 & 0 & 0 & \\
0 & 1 & 1 & 0 & \\
0 & 0 & 1 & 0 & \\
0 & 0 & 0 & 1
\end{pmatrix}$$
How do I complete $P$ using generalized eigenvectors, though? Thanks in advance 🙂
EDIT: I looked up the definition of generalized eigenvectors and got that $(A-I)^2 \mathbf{v_4} = 0$ is spanned by the set $\{(-1, 1, 0, 0)^t, (4, 0, 0, 1)^t\}$. Of these two, I concluded that since only $(4, 0, 0, 1)^t$ is linearly independent relative to $\mathbf{v_1},\mathbf{v_2}$, I should use it as a seed to generate $\mathbf{v_5} = (A-I)\mathbf{v_4} = (-4,-4,-8,-2)^t$. Why is it not possible to use $\mathbf{v_3}$ as a column in $P$, though? What is the rationale behind having to compute $\mathbf{v_5}$?
Best Answer
The first important thing to remember is that the Jordan form is only unique up to rearranging the blocks which appear in it. You have found the eigenvalues and the sizes of the blocks and written them in a specific order given by your $J$. Now, you want to find $P$ such that $P^{-1}AP = J$. Let's call the columns of $J$ by the names $u_1,u_2,u_3,u_4$. Then the following must be true:
Let's see what this means.
Since you don't want to "try" all possible vectors $v$ in the eigenspace and see if the equation $(A - I)u_3 = v$ has a solution, you work backwards. You find an element $u_3 \in \ker (A - I)^2$ (not all elements will work because you want at the end a basis) and then set $u_2 = (A - I)u_3$.