MHB Jordan-Basis for Matrix: Need Help!

[Fernando Revilla]

Unsolved question posted by gotmejerry on MHF (April 8th 2013, 01:06 PM).

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

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!

 

[Fernando Revilla]

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 $AP=PJ$