MHB Jordan-Basis for Matrix: Need Help!

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The discussion centers on finding a Jordan basis for a given 4x4 matrix with a quadruple eigenvalue of 1. The user has identified one eigenvector and attempted to find additional vectors but encountered inconsistencies. They derived the canonical Jordan form, which includes three blocks, and outlined the necessary conditions for constructing the Jordan basis. Ultimately, they successfully identified the Jordan basis vectors and the transition matrix, confirming that the transformation satisfies the Jordan canonical form. The thread highlights the process of deriving a Jordan basis and the importance of compatibility conditions in linear algebra.
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Unsolved question posted by gotmejerry on MHF (April 8th 2013, 01:06 PM).

Hi,
I couldn't find a Jordan-basis for this matrix:


png.latex


I have found an eigenvector: v1=[3 5 0 0] then an other vector v2 for (A-I)v2=v1, then I tried to compute (A-I)v3=v2 but that was inconsistent so I think I need 2 other eigenvectors, but I don't know how to find them.

Thanks for your help!
 
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Eigenvalues of $A=\begin{bmatrix}{1}&{0}&{3}&{0}\\{0}&{1}&{5}&{0}\\{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{1}\end{bmatrix}\in\mathbb{R}^{4\times 4}$: $$\det (A-\lambda I)=(1-\lambda)^4=0\Rightarrow \lambda=1 \quad(\mbox{cuadruple)}$$ On the other hand, we have: $$\dim \ker (A-I)=4-\mbox{rank} \begin{bmatrix}{0}&{0}&{3}&{0}\\{0}&{0}&{5}&{0}\\{0}&{0}&{0}&{0}\\{0}&{0}&{0}&{0}\end{bmatrix}=4-1=3$$ This means that the canonical form of Jordan $J$ has three blocks: $$J=\begin{bmatrix}{1}&{1}&{0}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{1}\end{bmatrix}$$ We have to find o basis $B=\{e_1,e_2,e_3,e_4\}$ of $\mathbb{R}^4$ satisfying $$\left \{ \begin{matrix} (A-I)e_1=0\\ (A- I)e_2=e_1\\(A-I)e_3=0\\(A-I)e_4=0 \end{matrix}\right.$$ The first and the second systems have the form $(A-I)x=h$, that is: $$\begin {bmatrix}{0}&{0}&{3}&{0}\\{0}&{0}&{5}&{0}\\{0}&{0}&{0}&{0}\\ {0}&{0}&{0}&{0}\end{bmatrix} \begin {bmatrix}{x_1}\\{x_2}\\{x_3}\\{x_4}\end{bmatrix} = \begin {bmatrix}{h_1}\\{h_2}\\{h_3}\\{h_4}\end{bmatrix} \Leftrightarrow \left \{ \begin{matrix} 3x_3=h_1\\5x_3=h_2\\0=h_3\\0=h_4 \end{matrix}\right.\quad (1)$$ Easily we get the general solution ($GS$) of $(1)$ and the compatibility conditions ($CC$): $$GS\;\left \{ \begin{matrix} x_1=\alpha\\x_2=\beta\\x_3=\frac{h_1}{3}\\x_4= \gamma \end{matrix}\right.\quad CC\;\left \{ \begin{matrix} 5h_1-3h_2=0\\h_3=0\\h_4=0 \end{matrix}\right.$$ Vector $e_1$. In this case $h_1=h_2=h_3=h_4=0$ and the general solution of $(1)$ is $e_1=(\alpha,\beta,0,\gamma).$ This vector $x$ will be as $h$ in the next system so, we impose $CC$: $5\alpha-3\beta=0,0=0,\gamma=0$. We choose $\alpha=3,\beta=5,\gamma=0$ so, $\boxed{e_1=(3,5,0,0)}$.

Vector $e_2$. In this case $h_1=3,h_2=5,h_3=h_4=0$ and the general solution of $(1)$ is $e_2=(\alpha,\beta,1,\gamma).$ We simply choose $\alpha=0,\beta=0,\gamma=0$ so, $\boxed{e_2=(0,0,1,0)}$.

Vectors $e_3$ and $e_4$. These are eigenvectors associated to the eigenvalue $1$, That is, solutions of the system $(A-I)x=0\equiv x_3=0$. We choose $e_3$ and $e_4$ linearly independent such that $\{e_1,e_3,e_4\}$ form a basis of $\ker (A-I)$ for example $\boxed{e_3=(0,1,0,0)}$ and $\boxed{e_4=(0,0,0,1)}$. As a consequence:

$(i)$ Jordan basis for $A$: $B_J=\{e_1,e_2,e_3,e_4\}$

$(ii)$ Transition matrix: $P=\begin{bmatrix}{3}&{0}&{0}&{0}\\{5}&{0}&{1}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{0}&{1}\end{bmatrix}$

$(iii)$ Canonical form of Jordan: $J=\begin{bmatrix}{1}&{1}&{0}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{1}&{0}\\{0}&{0}&{0}&{1}\end{bmatrix}$

P.S. Easily verified, $P^{-1}AP=J$ or equivalently $AP=PJ$
 
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