For the operator $T:\mathbb{R}[x]_3\to \mathbb{R}[x]_3$ defined by $T(f(x))=f(x)+xf''(x)$.Find a Jordan canonical from $J$ of $T$ and a basis $\beta$ for $\mathbb{R}[x]_3$ such that $J=[J]^{\beta}_{\beta}$
**Can any body tell me please is my solution right or wrong? Thank you **
The basis for $\mathbb{R}[x]_3$ is $\{1,x,x^2,x^3\}$
So..
$T(1)=1\\ T(x)=x+0=x\\ T(x^2)=x^2+2x=2x+x^2\\ T(x^3)=x^3+6x^2=6x^2+x^3$
so.. $A=[T]=\begin{bmatrix} 1 &0 &0 &0 \\0 &1 &2 &0 \\0 &0 &1 &6 \\0 &0 &0 &1 \end{bmatrix}$
Hence the characteristic equation is $P_T(x)=(x-1)^4$
And $(A-I)=\begin{bmatrix} 0 &0 &0 &0 \\0 &0 &2 &0 \\0 &0 &0 &6 \\0 &0 &0 &0 \end{bmatrix}$
and $(A-I)^3=0$
Then $d_1=\dim N(A-I)=2\\ d_2=\dim N((A-I)^2)=3\\ d_3=\dim N((A-I)^3)=4$
We have $N(A-I)=\{(a,b,c,d)|c=d=0\}$ and $N(A-I)^2=\{(a,b,c,d)|d=0\}$
hence , We may choose $(0,0,0,1) $ to start our first chain , we get $(A-I)(0,0,0,1)=(0,0,6,0)$ and $(A-I)^2=(0,12,0,0)$ Now, in $N(A-I)$ We already have $(0,0,12,0)$ . Hence to complete basis of $N(A-I)$ we take $(1,0,0,0)$ . Thus Jordan canonical basis is $\{((0,0,0,1),(0,0,6,0),(0,12,0,0,),(1,0,0,0)\}$ which is $\{x^3,6x^2,12x,1\}$
$f(x^3)=x^3+6x^2\\ f(6x^2)=6x^2+12x\ f(12x)=12x\\ f(1)=1$
Hence Jordan matrix $\begin{bmatrix} 1 &0 &0 &0 \\1 &1 &0 &0 \\0 &1 &1 &0 \\0 &0 &0 &1 \end{bmatrix}$