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Determine the Jordan Canonical Form of the following matrix: $$A=\begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 4\\ \end{bmatrix}$$

I am trying to determine the Jordan Basis first. For that purpose I am trying to find out the generalized Eigenvectors of this matrix.

Corresponding to $1$, Let $U_1$ be the generalized eigenspace. My calculations show that $$U_1=span\{(1,0,0)^t\}$$ and $U_2$ be the corresponding generalized eigenspace for $4$. I found out $$U_2=span\{(1,0,-9)^t,(0,1,6)^t\}$$All I need to do now is find the Jordan basis. Since $(A-\lambda_i I)|_{U_i }$ is nilpotent, all I need to do is find the basis for each such $i$.

I am confused from here on what to take as the jordan basis. I am sure that $(1,0,0)^t$ will feature as the first column. I am not sure about the other two.

Thanks for the help!!

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    $\begingroup$ Is $U_2$ an eigenspace? In other words, are all elements of $U_2$ eigenvectors, or is that true only for a $1$d subspace of $U_2$? $\endgroup$ – Travis Jun 3 '15 at 6:17
  • $\begingroup$ It is a generalized eigenspace. All the elements are not necessarily eigen vectors $\endgroup$ – tattwamasi amrutam Jun 3 '15 at 6:18
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    $\begingroup$ @tattwamasiamrutam Yes, that's precisely what I'm saying, and precisely why I gave the hint in my first comment. $\endgroup$ – Travis Jun 3 '15 at 7:39
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    $\begingroup$ @tattwamasiamrutam No, the eigenspace is one-dimensional - see my answer. Note: I didn't say the generalised eigenspace is one-dimensional, it is indeed two-dimensional. But this does not determine the Jordan form because it only tells you that the total size of the blocks for $\lambda=4$ is $2$, it could still be a single block or two $1\times1$ blocks. $\endgroup$ – David Jun 3 '15 at 7:48
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    $\begingroup$ @tattwamasiamrutam I just found a large number of typos in my previous comment and have fixed them, perhaps you need to read it again. Sorry ;-) $\endgroup$ – David Jun 3 '15 at 7:50
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First compute \begin{align*} A-4\,I &= \begin{bmatrix}-3&2&3\\ 0&0&5\\ 0&0&0\end{bmatrix} & (A-4\,I)^2 &= \begin{bmatrix}9&-6&1\\ 0&0&0\\ 0&0&0\end{bmatrix} & (A-4\,I)^3 &= \begin{bmatrix}-27&18&-3\\ 0&0&0\\ 0&0&0\end{bmatrix} \end{align*} so $\DeclareMathOperator{rank}{rank}$ \begin{align*} \rank\left(A-4\,I\right) &=2 & \rank\left((A-4\,I)^2\right) &=1 & \rank\left((A-4\,I)^3\right) &=1 \end{align*}

This tells us that the smallest value of $k$ so that $\DeclareMathOperator{null}{null}\null\left((A-4\,I)^k\right)\DeclareMathOperator{Span}{Span}$ stabilizes is $k=2$. This is called the index of nilpotency for the eigenvalue $\lambda=4$.

Now, note that $$ \null\left(A-4\,I\right)=\Span\left\{ v_1= \begin{bmatrix}2\\3\\0\end{bmatrix} \right\} $$ We wish to extend this basis for $\null\left(A-4\,I\right)$ to a basis $\left\{v_1,v_2\right\}$ for $\null\left((A-4\,I)^2\right)$ such that $(A-4\,I)v_2=v_1$. That is, $v_2$ must satisfy \begin{align*} (A-4\,I)^2 v_2 &= \vec 0 & (A-4\,I)v_2 &= v_1 \end{align*} One checks that $$ v_2=\begin{bmatrix}-1/15\\0\\3/5\end{bmatrix} $$ satisfies these equations.

Hence our Jordan form is $A=PJP^{-1}$ where \begin{align*} P&=\begin{bmatrix}1&2&-1/15\\0&3&0\\ 0&0&3/5\end{bmatrix} & J&=\begin{bmatrix}1&0&0\\0&4&1\\0&0&4\end{bmatrix} \end{align*}

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  • $\begingroup$ This is exactly the procedure I wanted to follow. I was confused over taking the basis elements since the the generalized eigenspace has dimension $2$. Thanks $\endgroup$ – tattwamasi amrutam Jun 3 '15 at 7:41
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    $\begingroup$ Short cut: as $\lambda=4$ has algebraic multiplicity $2$ and geometric multiplicity $1$, you can say in advance that the kernels will stabilise at $(A-4I)^2$, and you don't need to spend any effort calculating $(A-4I)^3$. Not that's it's very much effort ;-) $\endgroup$ – David Jun 3 '15 at 7:54
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    $\begingroup$ @David True, though I wanted to give some insight into the general procedure. $\endgroup$ – Brian Fitzpatrick Jun 3 '15 at 7:57
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To find $E_4$ we have $$A-4I=\pmatrix{-3&2&3\cr 0&0&5\cr0&0&0\cr}\ ,$$ which is already in echelon form; the solution is $$x_3=0\ ,\quad x_2=t\ ,\quad x_1=\tfrac23t\ ,$$ so the eigenspace $$E_4=\left\{t\pmatrix{\tfrac23\cr1\cr0\cr}\ \bigg|\ t\in{\Bbb R}\right\}$$ is one-dimensional. So the Jordan basis vectors will form one chain for $\lambda=4$, and (obviously) one chain for $\lambda=1$. So there is one Jordan block for each eigenvalue, and a Jordan form is $$J=\pmatrix{1&0&0\cr0&4&1\cr0&0&4\cr}\ .$$

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