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Find the Jordan Canonical Form of the following matrix $$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 & 1 & 1\\ \end{bmatrix}$$

My try: I go about finding the Jordan Basis for this matrix. It is clear that $1$ is the only eigenvalue of this matrix. So $$A-I=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1 & 1 & 0\\ \end{bmatrix}$$

Moreover Rank$(A-I)^2=1$. Moreover Rank$(A-I)^3=0$ . So We don't need to go further on evaluating the powers of matrices in our search for generalized eigenvectors. $$(A-I)^2=\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 4 & 0 & 0 & 0 & 0 & 0\\ \end{bmatrix}$$

Also $(A-I)^3=0$.

It is seen that the generalized eigenspace (say $U$) consists of

$U$=span$\{v_1=(0,1,0,0,0,0)^t,v_2=(0,0,1,0,0,0)^t,v_3=(0,0,0,1,0,0)^t,v_4=(0,0,0,0,1,0)^t,v_5=(0,0,0,0,0,1)^t\}$

Here I run into a little problem. Since $(A-I)v_i=v_5$ for each $i=1,2,3,4$, I have only five vectors with me for the Jordan Canonical Basis. Moreover I can't choose any other arbitrary vector linearly independent to these four simply because if $v$ were such a vector , then there is no $i \gt 0$ (and integer) such that $(A-I)v^{i}=0$

I have also another question here. Since $(A-I)^3=0$, why should I not choose general eigen vectors corresponding to $(A-I)^3$??

I am a little stuck here.

Thanks for the help!!!

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    $\begingroup$ $A-I$ is nilpotent, it will most certainly not have the same rank as its square! You'll find that $(A-I)^2$ has all its coefficients zero, except for its last line which equals $$(\;4\quad 0\quad 0\quad 0\quad 0\quad 0\;)$$ $\endgroup$ – Olivier Bégassat Jun 3 '15 at 11:00
  • $\begingroup$ @OlivierBégassat Thanks. Edited $\endgroup$ – tattwamasi amrutam Jun 3 '15 at 11:14
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From rank computations you find: there are four linearly independent eigenvectors ($rank(I-A)=2$), hence four Jordan blocks. The nilpotence index of $I-A$ is three, hence the largest Jordan block has size $3\times 3$. Thus, the other blocks are $1\times 1$ blocks.

You need a vector $v_6$ that satisfies $$ (I-A)^3 v_6=0, \ (I-A)^2v_6\ne0. $$ Take $$ v_6=\pmatrix{1&0&0&0&0&0}. $$ Then $(I-A)^2v_6, (I-A)v_6$ are in the Jordan basis too, which are $$ (I-A)^2v_6 = \pmatrix{0&0&0&0&0&1}^T=v_5, \\ (I-A)v_6 = \pmatrix{0&1&1&1&1&1}^T=:v_7. $$ Now you have three linearly independent vectors. Complement this set with three eigenvectors to form a basis, for instance $$ \pmatrix{0&1&-1&0&0&0}^T, \pmatrix{0&1&0&-1&0&0}^T, \pmatrix{0&1&0&0&-1&0}^T. $$ Arrange them in the right order to obtain the Jordan basis.

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  • $\begingroup$ what happens if I take $e_2$, $e_3$ and $e_4$ instead of the eigen vectors?? Can I take them and form a basis as well??? $\endgroup$ – tattwamasi amrutam Jun 3 '15 at 11:50
  • $\begingroup$ No, these are no eigenvectors, and all of them satisfy $(I-A)e_i=v_5$. The resulting matrix will not be in Jordan form (it would contain three $1$'s in a row). $\endgroup$ – daw Jun 3 '15 at 11:52

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