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Given $$ \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2\\ \end{bmatrix}$$am I need to find a Jordan Decomposition of this matrix. For this purpose I am trying to follow the following theorem:

Theorem 1

If $N \in L(V)$ is nilpotent, then there exist vectors $v_1$,$v_2$,...$v_k$ $\in V$ such that

(i)$\{v_1,Nv_1,..N^{m(v_1)}v_1,...,v_k,Nv_k,....,N^{m(v_k)}v_k\}$ is a basis of $V$

(ii) $\{N^{m(v_1)}v_1,....,N^{m(v_k)}v_k\}$ is a basis of null $N$.

Then I am trying to find generalized eigenspaces corresponding to the eigen values $1$(say $U_1$) and $2$ (say $U_2$). My calculations show that $$U_1=span\{(1,0,0)^t,(0,1,0)^t\}$$ and $$U_2=span\{(0,0,1)^t\}$$

Then $(T-\lambda_i)|_{U_i}$ is nilpotent for $i=1,2$.-Then I want to use Theorem $1$ to conclude. But I am going wrong somewhere.

Thanks for the help!!

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    $\begingroup$ Your matrix is in Jordan form . $\endgroup$ – Emilio Novati Jun 2 '15 at 12:30
  • $\begingroup$ I should get the same martrix as well when I try to find the decomposition then $\endgroup$ – tattwamasi amrutam Jun 2 '15 at 12:33
  • $\begingroup$ Obviously! See my answer. $\endgroup$ – Emilio Novati Jun 2 '15 at 12:43
  • $\begingroup$ you don't recognize that your matrix is in jordan canonical form? $\endgroup$ – abel Jun 2 '15 at 12:58
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Your job is correct but useless because the result is trivial and immediate from the start. The matrix $$ \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2\\ \end{bmatrix}$$

is in Jordan form and we see immediately that it has eigenvalues $2$ and $1$ with algebraic multiplicity $2$ so you have: $$ \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2\\ \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 2\\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix} $$ The generlized eigenvectors are the columns of the identity matrix.

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  • $\begingroup$ I have a few questions. I got the identity matrix. Why are you multiplying it with the given matrix?? Do I need to do the same always?? $\endgroup$ – tattwamasi amrutam Jun 2 '15 at 13:00
  • $\begingroup$ The Jordan decomposition of a matrix $A$ is $A=SJS^{-1}$ where $S$ is a matrix that has as columns the (generalized) eigenvalue. See en.wikipedia.org/wiki/Jordan_normal_form. $\endgroup$ – Emilio Novati Jun 2 '15 at 13:05
  • $\begingroup$ Got it. Thank you..... $\endgroup$ – tattwamasi amrutam Jun 2 '15 at 13:17

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