# Abstract Nonsense

## Halmos Sections 34 and 35:Products and Polynomials

Point of post: In this post we complete the problems at the end of sections 34 and 35 of Halmos’s book.

1.

Problem: Let $S$

$\displaystyle S:\mathbb{C}[x]\to\mathbb{C}[x]:\sum_{j=0}^{n}a_j x^j\mapsto \sum_{j=0}^{n}\frac{a_jx^{j+1}}{j+1}$

and

$\displaystyle D:\mathbb{C}[x]\to\mathbb{C}[x]:\sum_{j=0}^{n}a_j x^j\mapsto \sum_{j=1}^{n}ja_j x^{j-1}$

compute $\left(S^nD^n\right)(p(x))$ and $\left(D^nS^n\right)(p(x))$ for an arbitrary $p\in\mathbb{C}[x]$.

Proof: Note that

$\displaystyle \left(SD\right)\left(\sum_{j=0}^{m}a_jx^j\right)=S\left(\sum_{j=1}^{m}j a_j x^{j-1}\right)=\sum_{j=1}^{m}a_j x^j$

and

$\displaystyle \left(DS\right)\left(\sum_{j=0}^{m}a_j x^j\right)=D\left(\sum_{j=0}^{m}\frac{a_j x^{j+1}}{j+1}\right)=\sum_{j=0}^{m}a_j x^j$

and so in particular in general

$\displaystyle \left(S^n D^n\right)\left(\sum_{j=0}^{m}a_j x^j\right)=\sum_{j=n}^{m}a_j x^j$

where the sum is the zero polynomial if $n>m$ and

$\displaystyle \left(D^n S^n\right)\left(\sum_{j=0}^{m}a_j x^j\right)=\sum_{j=0}^{m}a_j x^j$

2.

Problem: If $A$ and $B$ are linear transformations such that $AB-BA$ commutes with $A$ then

$A^kB-BA^k=kA^{k-1}\left(AB-BA\right)$

for every $k\in\mathbb{N}$.

Proof:We first note that by induction

$A^k\left(AB-BA\right)=\left(AB-BA\right)A^k$

which, upon expansion gives

$A^{k+1}B-A^kBA=ABA^k-BA^{k+1}$

and for our particular purposes

$A^{k+1}B-ABA^k=A^kBA-BA^{k+1}$

Thus, with this in hand we may proceed by induction. Note that $A^kB-BA^k=kA^{k-1}\left(AB-BA\right)$ holds trivially for $k=1$ and if we assume that it holds for $k$ we see that

\begin{aligned}(k+1)A^k\left(AB-BA\right) &= kA^k\left(AB-BA\right)+A^k\left(AB-BA\right)\\ &= A kA^{k-1}\left(AB-BA\right)+A^k\left(AB-BA\right)\\ &= A\left(A^kB-BA^k\right)+A^k\left(AB-BA\right)\\ &= A^{k+1}B-ABA^k+A^{k+1}B-A^kBA\\ &=A^kBA-BA^{k+1}+A^{k+1}B-A^kBA\\ &= A^{k+1}B-BA^{k+1}\end{aligned}

from where the conclusion follows.

3.

Problem: Suppose that $A:\mathbb{C}_n[x]\to\mathbb{C}_n[x]$ is given by $\left(A(p)\right)(x)=p(x+1)$. Prove that

$\displaystyle A=\sum_{k=0}^{n-1}\frac{D^k}{k!}$

Proof: Recall that $\mathbb{C}_n[x]$ has basis $\{x^0,\cdots,x^{n-1}\}$ and since a linear transformation is determined entirely on a basis it suffices to check that this equality is true on this particular basis. But,

$\displaystyle \sum_{k=0}^{n-1}\frac{D^k\left(x^{\ell}\right)}{k!}=\sum_{k=0}^{\ell}\frac{\ell(\ell-1)\cdots(\ell-k+1)x^{\ell-k}}{k!}=\sum_{k=0}^{\ell}{\ell \choose k}x^{\ell-k}=\left(x+1\right)^{\ell}$

where we’ve evidently used the Binomial Theorem in the last step.

From previous comment the problem follows.

4.

Problem:

a) If $A$ is an endomorphism on an $n$-dimensional $F$-space $\mathscr{V}$, then there exists a non-zero polynomial of degree less than or equal to $n^2$ which annihilates $A$.

Proof: This is a fairly crude estimate (since for all of us know that there is a $n$-degree polynomial which annihilates $A$, the characteristic polynomial). Regardless, we merely note that if there did not exist a non-zero polynomial $p$ with $\deg p=n^2$ which annihilates $A$ then for any $\alpha_0,\cdots,\alpha_{n^2}\in F$ we’d have that

$\alpha_0\mathbf{1}+\alpha_1 A+\cdots+\alpha_{n^2}A^{n^2}=\mathbf{0}\implies \alpha_0=\alpha_1=\cdots=\alpha_{n^2}=0$

which implies that $\text{End}\left(\mathscr{V}\right)$ has $n^2+1$ linearly independent vectors, which is clearly impossible.

5.

Problem: The product of linear transformations is defined only if they “match” in the following sense. Suppose that $\mathscr{U}$, $\mathscr{V}$, and $\mathscr{W}$ are all $F$-spaces and suppose that $A\in\text{Hom}\left(\mathscr{U},\mathscr{V}\right)$ and $B\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right)$. The product $BA$ is defined to be the linear transformation from $\mathscr{U}$ to $\mathscr{W}$ given by $B\circ A$. Interpret and prove as many of the five axioms for an associative unital algebra as you can.

Proof: The first axiom can be thought of as follows. If $A\in\text{Hom}\left(\mathscr{U},\mathscr{V}\right)$, $B\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right)$, and $C\in\text{Hom}\left(\mathscr{W},\mathscr{X}\right)$, then it’s true that $A(BC)=(AB)C$ since function composition  is associative. If $A\in\text{Hom}\left(\mathscr{U},\mathscr{V}\right)$ and $B,C\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right)$ then we can say that $A\left(B+C\right)=AB+AC$ since $A(B(x)+C(x))=A(B(x))+A(C(x))$. Similarly, the third axiom holds. The fourth one holds trivially. The fifth one doesn’t hold well at all. Namely, there is a left and right identity elements, namely $\text{id}_\mathscr{V}$ and $\text{id}_{\mathscr{U}}$ respectively but there is no ‘single’ identity element.

Reference:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

November 23, 2010 -