Diagonalizable operator $D$ on $R^3$ and a nilpotent operator $N$ Let T be the linear operator on R3 which is represented by the matrix
$\begin{pmatrix}
3  &1 &-1  \\
 2 &2& -1  \\
 2 &2& 0  \\
\end{pmatrix}$
in the standard ordered basis. Show that there is a diagonalizable operator $D$
on $R^3$ and a nilpotent operator $N$ on $R^3$ such that $T = D + N$ and $DN = ND$.
Find the matrices of $D$ and $N$ in the standard basis.
We have the relationships
$D=c_1E_1+...+c_kE_k$
and
$N=(T-c_1I)E_1+...+(T-c_kI)E_k$
but I do not know how to use it. Can anyone explain? In this case the characteristic polynomial is $(x-1)(x-2)^2$. I do not know if I should find the matrix of E_i, also I did not understand either if In this formula should I consider the repetition of eigenvalue?
 A: The minimal polynomial for $T $ is $p (x)=(x-1)(x-2)^2$. Define  $$\left\{ 
\begin{array}{c}f_1 (x):=\dfrac{p (x)}{x-1}=(x-2)^2\\f_2 (x):=\dfrac{p (x)}{(x-2)^2}=(x-1)\end{array}\right.. $$ Since  $f_1$ and $f_2$ are relatively prime, you can find  $g_1$ and $g_2$ such that $f_1 g_1+f_2g_2=1$(use Euclidean algorithm). Now define  $E_i=f_i (T)g_i (T),i\in\{1,2\}.$ Finally take $D=E_1+2E_2$ and $N=(T-I)E_1+(T-2I)E_2$.
A: Steps:


*

*Find a Jordan basis for $T$

*Express $T$ as a matrix in terms of this Jordan Basis (i.e. in Jordan Normal Form).

*Separate this matrix into a sum of two matrices: a diagonal matrix consisting of all the diagonal entries, and an upper triangular matrix consisting of the off-diagonal $1$s. The former is diagonal, the latter is nilpotent, and their product commutes.

*If you want, find the standard basis representations for these operators, to present the solution in the same form as you were presented the problem.


Let me know if you need more help than this.
A: The Jordan normal form of your matrix will be $J=\begin{pmatrix}1&0&0\\0&2&1\\0&0&2\end{pmatrix}$.  That is, there is a basis $\{v_1,v_2,v_3\}$ rel which the matrix for $T$ is the above matrix. 
Let $P$ be the matrix whose columns are the basis vectors.   Then we have $P^{-1}TP=J$.
But $J=\begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}+\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}= D'+N'$.
Now just solve for $T$.  That is, $$T=PJP^{-1}=P(D'+N')P^{-1}=PD'P^{-1}+PN'P^{-1}=D+N$$.
Finally you need the Jordan basis (a basis of generalized eigenvectors), $\{v_1,v_2,v_3\}$ to find $N$ and $D$ explicitly. 
To this end, I have computed the following Jordan basis:  $\{(1,0,2),(-1,0,1),(1,1,2)\}$.
