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Problem of the Week #45 - April 8th, 2013

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: Let $V=C^{\infty}(\mathbb{R})$ denote the $\mathbb{C}$-vector space of continuous functions $f:\mathbb{R}\rightarrow\mathbb{C}$ which have continuous $n^{\text{th}}$ derivatives for all $n\geq 1$. Let $D\in\text{End}_{\mathbb{C}}(V)$ denote the derivative operator. View $V$ as a $\mathbb{C}[X]$-module, in the usual way, so that $X\cdot v=D(v)$ for all $v\in V$.


  1. Describe $\text{Ann}\{e^x\}$ (the annihilator of $e^x$), and prove the accuracy of your description.
  2. It is a basic result in analysis that for any $f\in V$, and any $\lambda\in\mathbb{C}$, $Df=\lambda\cdot f$ if and only if $f(x)=C\cdot e^{\lambda x}$ for some $C\in\mathbb{C}$. Now assume that $W$ is a subspace of $V$. Define \[ [D-\lambda]^{-1}(W)=\{f\in V: Df-\lambda f\in W\}.\] Prove that $\dim([D-\lambda]^{-1}W)\leq\dim(W)+1$.
  3. Consider a homogeneous linear differential equation with constant coefficients, of the following form: \[D^nf+a_{n-1}D^{n-1}f+\cdots+a_nDf+a_0f=0.\] Prove that the set of solutions $f\in V$ to this equation is a subspace of $V$ with dimension at most $n$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.

  1. Claim: $\text{Ann}\{e^x\}=(X-1)$, the ideal generated by $X-1$.
    Proof: Consider the element $X-1\in\mathbb{C}[X]$. Since $V$ is a $C[X]$-module, we see that$$(X-1)\cdot e^x = X\cdot e^x - 1\cdot e^x= De^x-e^x = e^x-e^x = 0.$$
    Therefore, $X-1\in\text{Ann}\{e^x\}$. Since $\mathbb{C}[X]$ is a PID, and $\text{Ann}\{e^x\}\neq (0)$, it follows that $\text{Ann}\{e^x\}$ is generated by a unique monic polynomial $P$. Since $X-1\in\text{Ann}\{e^x\}$, $P$ divides $X-1$. So, $P=1$ or $P=X-1$. If $P=1$, then $1\cdot e^x=e^x\neq 0$ for all $x$. Thus, $1\notin\text{Ann}\{e^x\}$ and $P\neq 1$. Therefore $P=X-1$, which implies that $(X-1)=\text{Ann}\{e^x\}$.$\hspace{1in}\mathbf{Q.E.D.}$
  2. Proof: We first show that $[D-\lambda]^{-1}(W)$ is a subspace of $V$. Let $f,g\in[D-\lambda]^{-1}(W)$. By definition, $w = Df-\lambda f\in W$ and $w^{\prime} = Dg-\lambda g\in W$. Furthermore,
    $$D(f+g)-\lambda(f+g)=Df+Dg-\lambda f-\lambda g = (Df-\lambda f)+(Dg-\lambda g) = w+w^{\prime}\in W.$$
    Therefore, $f+g\in[D-\lambda]^{-1}(W)$.\vskip0.5pc
    Similarly, let $f\in[D-\lambda]^{-1}(W)$ and $\alpha\in\mathbb{C}$. By definition, $w=Df-\lambda f\in W$. Thus,
    $$D(\alpha\cdot f)-\lambda\cdot(\alpha\cdot f)=\alpha\cdot Df-\alpha\cdot (\lambda\cdot f)=\alpha\cdot(Df-\lambda f)=\alpha w\in W.$$
    Therefore $\alpha\cdot f\in[D-\lambda]^{-1}(W)$.

    Now, let $\varphi:[D-\lambda]^{-1}(W)\rightarrow W$ be a map defined as $\phi( f)=Df-\lambda f$ for all $f\in[D-\lambda]^{-1}(W)$. We see that $\phi$ is a homomorphism of $\mathbb{C}$ vector spaces, based on calculations done above. Therefore, by the Rank-Nullity theorem,
    $$\dim([D-\lambda]^{-1}(W))=\dim(\text{Im}(\varphi))+\dim(\ker(\varphi)).$$
    $\text{Im}(\varphi)$ is a subspace of $W$; hence $\dim(\text{Im}(\varphi))\leq\dim(W)$. It is an analytic result that, for any $f\in V$ and $\lambda\in\mathbb{C}$, $Df=\lambda f$ iff $f(x)=C\cdot e^{\lambda x}$ for some $C\in\mathbb{C}$. That is,
    $$\ker(\varphi)=\{f\in[D-\lambda]^{-1}(W): Df-\lambda f=0\}$$
    is at most one-dimensional, so $\dim(\ker(\varphi))\leq 1$. Therefore, we now see that
    $$\dim([D-\lambda]^{-1}(W))=\dim(\text{Im}(\varphi))+\dim(\ker(\varphi))\leq\dim(W)+1.\hspace{1in}\mathbf{Q.E.D}$$
  3. Proof: Note that $D$ is a linear operator (follows a similar argument presented in part 2). If $f$ is a solution to the differential equation, then
    $$\begin{aligned} & D^nf+a_{n-1}D^{n-1}f+\cdots+a_1Df+a_0f = 0\\ \implies &Df-a_1^{-1}a_0f = -a_1^{-1}D^nf-a_1^{-1}a_{n-1}D^{n-1}f-\cdots-a_1^{-1}a_2D^2f.\end{aligned}$$
    Therefore, $f\in[D-\lambda]^{-1}(W)$ with $\lambda=a_1^{-1}a_0$ with $W$ being spanned by $\{D^2f,\ldots,D^nf\}$. This implies that the set of solutions to this linear equation is a subspace of $[D-\lambda]^{-1}(W)$. So the dimension of the solution set is at most the dimension of $[D-\lambda]^{-1}(W)$, which $n-1+1=n$. $\hspace{1in}\mathbf{Q.E.D.}$
 
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