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!!
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.