Using span of T to find Jordan

Question

Looking for help on all parts of the question below:

Let $U$ be the complex vector space of polynomials of at most degree 6. Define $D, T: U \rightarrow U$ by $D(f) = f'$ and $T = D^2 + D^3$ Set $V =$ Span $\left\{\, T^n(x^6) \mid n \geq 0 \,\right\}$ and $W =$ Span $\left\{\,T^n(x^5) \mid n \geq 0\,\right\}$
Questions

a)If $A \subset U$, define the span of $A$.

b) Prove that $T(V) \subset V$

c) Find bases of $V $ of $W$, with respect to whose union the matrix $T|_V$ is a Jordan block.

d) Prove $U = V \oplus W$

e) Deduce the Jordan normal form of T

My attempt

a) The span of $A$ is defined to be the set of all finite linear combinations of $A$ (Now I suspect $A$ is an infinite set so I imagine this doesn't hold?)

b) My understanding is that $V$ is equal to the set polynomials degree 4 or less, whilst $W$ is equal to the set of polynomials degree $3$ or less? If $v \in V$ then $v$ is a polynomial of deg 4 or less then $T(v)$ is a polynomial degree $2$ or less.

c) I tried to write out $T(e_i)$ for each of the basis vectors of $\mathbb{C}[x]$, thus got such a matrix: $\begin{bmatrix}0 & 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 0 \\ 6 & 6 & 0&0&0 \\ 0 & 24 & 12& 0 & 0\end{bmatrix}$ with a corresponding matrix for $W$ but this doesn't feel right either.

d) No idea how to prove this

e) Not really sure either, i guess i just take the block sums together, and then i have a lower triangular form so are all the eigen values equal to $0$? But i have no idea about the nullities.

Answer 1

Some hints :

a) as stated in the comment, your definition is correct.

b) $v \in V$ if and only if it exists a polynom $P_v$ such that $v = P_v(T)(x^6)$. Now let's take $u\in T(V)$. You have $u = T(v)$ for some $v \in V$. So $u = T(P_v(T))(x^6) = (X\cdot P_v)(x^6 ) = Q(x^6) \in V$.

Also dont forget that $T^0(x^6) = x^6 \in V$ and $T^0(x^5) = x^5 \in W$.

c) You have $deg(T^0(x^6)) = 6$, $deg(T^1(x^6)) = 4 $, $deg(T^2(x^6)) = 2$, d$eg(T^3(x^6)) = 0$ so they are linearly independent.

And since $T^4(x^6) = 0 $, $\{ T^i(x^6) | 0\le i \le 3 \}$ is a basis of $V$. You can do the same for $W$.

$T|_V$ will have a nice Jordan block form with respect to the union of the two basis.

d) Degrees arguments can easily prove that $V \cap W = \emptyset $. Dimensions arguments can prove the rest.

e) It's up to you. =)

Answer 2

a) The span of a (possibly infinite) est $A$ of vectors is the set of vectors that can be obtained as a linear combination of vectors of$~A$; such a linear combination is a finite sum $\sum_{i=1}^mc_ia_i$ with $m\in\Bbb N$, $c_i\in\Bbb C$ and $a_i\in A$ for all $i$. For the question above the consideration of infinite sets is not really relevant, since although in $T^n(x^6)$ of $T^n(x^5)$ the exponent can take infinitely many values, one only gets a nonzero vector in finitely many cases (for $0\leq n\leq3$ in the former case and for $0\leq n\leq2$ in the latter).

b) It is obvious that $V$ is closed under the operation of $T$, since the set $S=\left\{\, T^n(x^6) \mid n \geq 0 \,\right\}$ of vectors (polynomials) that spans $V$ is closed under this operation. Elements of $V$ are linear combinations $\sum_{i=1}^mc_ia_i$ with $a_i\in S$, and the image by $T$ of this combination is $T(\sum_{i=1}^mc_ia_i)=\sum_{i=1}^mc_iT(a_i)$ which is a linear combination of elements $T(a_i)\in S$, and therefore a vector in$~V$. Concretely each $a_i$ is some $T^n(x^6)$, and then $T(a_i)=T(T^n(x^6))=T^{n+1}(x^6)\in S$.

c) I am unsure what the question really asks for. Certainly the family $[ x^6, T(x^6), T^2(x^6), T^3(x^6) ]$ is a basis for$~V$ relative to which the matrix of $T|_V$ is lower triangular with as only nonzero values entries $1$ directly below the main diagonal, which is the form of a Jordan block (or its transpose, depending on your convention) of size$~4$ for $\lambda=0$. Explicitly the family is $[ x^6, 30x^4+120x^3, 360x^2+1440x+720, 720 ]$.

d) A similar basis for $W$ is $[ x^5, T(x^5), T^2(x^5) ]=[x^5,20x^3+60x^2, 120x+240]$, and its union with the basis of $V$ is still linearly independent (since all polynomials have distinct degrees), and $V+W$ is a direct sum. Its dimension is $4+3=7=\dim U$ so the direct sum equals the whole space$~U$.

e) For $T|_W$ we get a Jordan block of size$~3$ for $\lambda=0$, and by the direct sum of d), the two blocks together form a Jordan form for$~T$. Since this form is unique up to permutation of the blocks, we may call it the Jordan normal form of$~T$: $$ \pmatrix{ 0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&0\\ } $$