$T:f(x)\to f(x-1)+x^3f'''(x)/3$ Find the Jordan normal form and a Jordan basis for $T$.

Question

Let $T\in \mathcal{L}(\mathcal{P_3}(\mathbb{C})$ be the operator $$T:f(x)\to f(x-1)+\frac{x^3f'''(x)}{3}$$ Find the Jordan normal form and a Jordan basis for $T$.

Answer 1

If we take the canonical basis $1,x,x^2,x^3$, then $$ f(x)\longmapsto f(x-1) $$ maps $1\longmapsto 1$, $x\longmapsto x-1$, $x^2\longmapsto x^2-2x+1$, $x^3\longmapsto x^3-3x^2+3x-1$, so it has matrix $$ \begin{bmatrix} 1 &-1&1&-1\\ 0 &1&-2&3\\ 0 &0&1&-3\\ 0 &0&0&1\\ \end{bmatrix}. $$ And $f(x)\longmapsto x^3 f'''(x)/3$ maps $1$, $x$, and $x^2$ to $0$, and maps $x^3$ to $2x^3$. Thus its matrix is $$ \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&2 \end{bmatrix}. $$ Then matrix for $T$ is then the sum of those two, namely $$ T=\begin{bmatrix} 1 &-1&1&-1\\ 0 &1&-2&3\\ 0 &0&1&-3\\ 0 &0&0&3\\ \end{bmatrix}. $$ Since $T$ is triangular, we see immediately that the eigenvalues are $1$ (with multiplicity 3), and $3$.

Solving the two systems $T-\lambda I$ for $\lambda$ equal to $1$ and $3$ respectively, we find that the eigenvectors for $1$ are scalar multiples of $$v_1=\begin {bmatrix}1\\0\\0\\0\end {bmatrix} $$ (that is, the constant polynomials) and the eigenvectors for $3$ are scalar multiples of $$ v_4=\begin{bmatrix} -11\\12\\ -6\\ 4\\ \end{bmatrix}. $$ Since the algebraic multiplicity of $1$ is greater then the geometric one, to complete the basis we need to look at the generalized eigenvectors. That is, the kernel of $$(T-I)^2 =\begin{bmatrix} 0 &-1&1&-1\\ 0 &0&-2&3\\ 0 &0&0&-3\\ 0 &0&0&2\\ \end{bmatrix}^2 =\begin{bmatrix} 0 &0&2&-8\\ 0 &0&0&12\\ 0 &0&0&-6\\ 0 &0&0&4\\ \end{bmatrix} $$ and of $$(T-I)^3 =\begin{bmatrix} 0 &-1&1&-1\\ 0 &0&-2&3\\ 0 &0&0&-3\\ 0 &0&0&2\\ \end{bmatrix}^3 =\begin{bmatrix} 0 &0&0&-22\\ 0 &0&0&24\\ 0 &0&0&-12\\ 0 &0&0&8\\ \end{bmatrix}. $$ The kernel of $(T-I)^2$

We need to choose $v_2,v_3$ in such a way that $Tv_2=v_1+v_2$, $Tv_3=v_2+v_3$. These turn out to be $$ v_2=\begin{bmatrix}0\\-1\\0\\0\end{bmatrix},\ \ \ v_3=\begin{bmatrix}0\\1/2\\1/2\\0\end{bmatrix}. $$ Thus a Jordan basis for $T$ is given by $$ p_1(x)=1,\ \ \ p_2(x)=-x,\ \ \ p_3(x)=\frac{x+x^2}2,\ \ \ p_4(x)=4x^3-6x^2+12x-11, $$ and the Jordan form of $T$ is $$ \begin{bmatrix}1&1&0&0\\ 0&1&1&0\\ 0&0&1&0\\0&0&0&3\end{bmatrix}. $$