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Introduction and description of my problem

I have trouble when finding the matrix change of base $P$ that allows me to obtain the Jordan form from the matrix $A$, in other words, find $P$ that satisfies $J=P^{-1}AP \ $ where $J$ is the Jordan matrix for $A$.

$$A=\begin{pmatrix}{-1} & {0} & {1} & {0} \\ {-1} & {-1} & {1} & {1} \\ {-1} & {0} & {1} & {0} \\ {-1} & {0} & {2} & {-1}\end{pmatrix}$$

The characteristic polynomial of $A$ is $p_{A}(X)=X^2(X+1)^2$, so we have the eigenvalues $X=0$ and $X=1$.

What I have worked so far:

  1. Eigenvectors and Eigenspaces calculus

    • For $X=0$, we have got that the $dim(Ker(A))=1$ so we need to obtain $A^2$ and check for the dimension of the kernel of $A^2$.

$$A^2=\begin{pmatrix}{0} & {0} & {0} & {0} \\ {0} & {1} & {1} & {-2} \\ {0} & {0} & {0} & {0} \\ {0} & {0} & {-1} & {1}\end{pmatrix}$$

Since $dim(Ker(A^2))=2$, it matchs the multiplicity of our eigenvalue we now can find its eigenvectors.

We solve for $X$ the system of equations $A^{2}X=0 $, where $X=(x,y,z,t)^{T}$.

We get $\begin{cases}z=t \\ y=t \end{cases}$, $\ $so $Ker(A^2)=<(1,0,0,0),(0,1,1,1)>$

$\begin{array}{lllll} {n_{1}=dim(Ker(A))=1} & {p_{1}=n_{1}-n_{0}=1} & {q_{1}=p_{1}-p_{2}=0} & { \Rightarrow \text{no Jordan blocks of order }1} \\ {n_{2}=dim(Ker(A^{2}))=2} & {p_{2}=n_{2}-n_{1}=1} & {q_{2}=p_{2}=1} & {\Rightarrow 1\text{ Jordan block of order }2}\end{array}$


  • For $X=-1$, we calculate $(A+I)$:

$$(A+I)=\begin{pmatrix} {0} & {0} & {1} & {0} \\ {-1} & {0} & {1} & {1} \\ {-1} & {0} & {2} & {0} \\ {-1} & {0} & {2} & {0} \end{pmatrix}$$

Since $rank(A+I)=3$, we get that $dim(Ker(A+I))=1$ so we need to calculate $(A+I)^2$.

$$(A+I)^2=\begin{pmatrix} {-1} & {0} & {2} & {0} \\ {-2} & {0} & {3} & {0} \\ {-2} & {0} & {3} & {0} \\ {-2} & {0} & {3} & {0} \end{pmatrix}$$

Clearly $dim(Ker((A+I)^2))=2$ so it matches the multiplicity of the eigenvalue $X=-1$. Now we solve for the eigenvalues.

We get $\begin{cases} -x &+2z &=0 &\Rightarrow &x=2z\\ -2x&+3z&=0 \end{cases} \Big\} \ \Rightarrow \ z=0 \ \Rightarrow \ x=0 $

$$Ker((A+I)^2)=<(0,1,0,0),(0,0,0,1)>$$

As for the previous eigenvalue we get $1$ Jordan box of order $2$ for the eigenvalue $X=-1$.

  1. Jordan Matrix and arise of the problem

Therefore, we now that the Jordan matrix has to be: $$J=\left(\begin{array}{cc|cc} \bf 0 & \bf 1 & 0 & 0 \\ \bf 0 & \bf 0 & 0 & 0 \\ 0 & 0 & \bf -1 & \hline \bf 1 \\ 0 & 0 & \bf 0 & \bf -1 \\ \end{array}\right)$$

By the Jordan Decomposition Theorem we also now that $\mathbb{R}^4=Ker(A^2) \oplus Ker((A+I)^2)$, so we can construct an eigenvector base $B=\{(1,0,0,0),(0,1,1,1),(0,1,0,0),(0,0,0,1)\}$.

If I construct the change of basis matrix from the canonic base $(\mathscr{C})$ to base $B$ which I denote by $(Id)_{B \mathscr{C}}$, that should be: $$(Id)_{B \mathscr{C}}=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ \end{pmatrix} \qquad \qquad (Id)_{\mathscr{C}B}=\left((Id)_{B \mathscr{C}}\right)^{-1}= \begin{pmatrix} {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {1} & {-1} & {0} \\ {0} & {0} & {-1} & {1} \\ \end{pmatrix}$$

When I do $$(Id)_{\mathscr{C}B} \cdot A \cdot (Id)_{B \mathscr{C}},$$ shouldn't this matrix be $J$?

However, it turns out that $$(Id)_{\mathscr{C}B} \cdot A \cdot (Id)_{B \mathscr{C}}= \begin{pmatrix} {-1} & {1} & {0} & {0} \\ {-1} & {1} & {0} & {0} \\ {0} & {0} & {-1} & {1} \\ {0} & {0} & {0} & {-1} \end{pmatrix} \neq \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix}=J$$

I have to be doing something wrong... Can you please help me finding the matrix $P$ such that $P^{-1}\cdot A \cdot P=J$. Thanks in advance!

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First of all, $\ker A=\bigl\langle(1,1,1,1)\bigr\rangle$, $\ker A^2=\bigl\langle(1,1,1,1),(0,1,1,1)\bigr\rangle$ and furthermore $A.(0,1,1,1)=(1,1,1,1)$. On the other hand, $\ker(A+\operatorname{Id})=\bigl\langle(0,1,0,0)\bigr\rangle$, $\ker(A+\operatorname{Id})^2=\bigl\langle(0,1,0,0),(0,0,0,1)\bigr\rangle$ and furthermore $(A+\operatorname{Id}).(0,0,0,1)=(0,1,0,0)$. So, take$$P=\begin{bmatrix}1&0&0&0\\1&1&1&0\\1&1&0&0\\1&1&0&1\end{bmatrix}$$and then$$P^{-1}.A.P=\begin{bmatrix}0&1&0&0\\0&0&0&0\\0&0&-1&1\\0&0&0&-1\end{bmatrix}.$$

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  • $\begingroup$ That's the method I like; meanwhile, I put a working 0-1 matrix, 4 by 4, at your answer math.stackexchange.com/questions/3391472/… $\endgroup$ – Will Jagy Oct 16 at 23:45
  • $\begingroup$ The vector $(0,1,1,1)$ does not satisfy that $A^2 \cdot (0,1,1,1)^T=(0,0,0,0)$ so why is it in the $Ker(A^2))$? $\endgroup$ – Rober Oct 17 at 9:22
  • $\begingroup$ You are wrong. Since$$A^2=\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1\end{bmatrix},$$we do have $A^2.(0,1,1,1)^T=(0,0,0,0)$. $\endgroup$ – José Carlos Santos Oct 17 at 10:09
  • $\begingroup$ The first vector of the change-of-base matrix will have to be a vector $v$ such that $A.v=0$. The second one will have to be a vector $w$ such that $A.w=v$. $\endgroup$ – José Carlos Santos Oct 17 at 10:29
  • $\begingroup$ This is just a correction of my previous comment where I didn't express well, but your answer that still holds for it: I corrected the matrix $A^2$,but I keep getting a matrix that isn't $J$, because I get that the eigenvectors that generate $Ker(A^2)$ are $\{(1,0,0,0),(0,1,1,1)\}$ instead of $\{(1,1,1,1),(0,1,1,1)\}$, which are the right ones (as you showed in your previous answer). What I do not understand is why it doesn't work with $(1,0,0,0)$ and it works with $(1,1,1,1)$. Isn't it sufficient that $(1,0,0,0) \in Ker(A^2)$ to be a vector that gives the change of base matrix? $\endgroup$ – Rober Oct 17 at 10:39

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