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:
-
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 $X=-1$.
- 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!
Best Answer
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 $A\cdot(0,1,1,1)=(1,1,1,1)$. 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 $(A+\operatorname{Id})\cdot(0,0,0,1)=(0,1,0,0)$. 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$$P^{-1}\cdot A\cdot P=\begin{bmatrix}0&1&0&0\\0&0&0&0\\0&0&-1&1\\0&0&0&-1\end{bmatrix}.$$