Suppose V is the real vector space of real-valued functions with a derivative. What is the dimension of the null space of the linear map

$$Tf = x\frac{df}{dx} - 4f\;\;?$$

The basis I found for the null space is $\{x^4\}$, which gives dimension 1, but the hint for problem was that the dimension is not 1. Why is that so?


You can solve the equation $x{dy\over dx}=4y$ using separation of variables: $$\tag{1} {1\over y}\,dy={4\over x}\,dx. $$ We want to avoid division by zero; so, in solving $(1)$, we consider the cases where $x<0$ and $x>0$. Integration of $(1)$ shows that for $k$ a constant, $y=k x^4$ is a solution for both the domain $x>0$ and the domain $x<0$.

One may then verify that the solutions of $(1)$ over $\Bbb R$ have the form $$\tag{2} f(x)=\cases{cx^4,&$x\ge0$\cr d x^4,&$x\le 0$ } $$ where $c$ and $d$ are constants (in particular, one can (and should) show such an $f$ is differentiable at $0$ with $f'(0)=0$).

Note that any such $f$ is a linear combination of the independent functions obtained by taking $(c=1, d=0)$ and $(c=0,d=1)$ in $(2)$.

  • 1
    $\begingroup$ I think this is correct and I'm upvoting it, yet it'd perhaps be a little clearer if we write $$f_1(x):=\begin{cases}x^2&\,\text{if}\;\; x>0\\0&\,\text{if}\;\;x\leq 0\end{cases}$$and likewise for the other possibility $\endgroup$ – DonAntonio Dec 18 '12 at 14:17

We have a simple linear homogeneneous differential equation of order $\,1\,$ to solve

$$Tf=0\Longrightarrow x\frac{df}{dx}=4f\Longrightarrow \int\frac{df}{f}=4\int\frac{dx}{x}\Longrightarrow$$

$$\log|f|=4\log|x|+C\Longrightarrow f=kx^4\,\,,\,k=\,\text{a constant}$$

and yes: you're answer is correct.

Whose hint was that $\,\dim\ker T\neq 1\,$ ?

  • 3
    $\begingroup$ Wouldn't you have $|f|=kx^4, k\ge 0$? This gives another solution: $f(x)=\cases{x^4,& $x\ge0$\cr -x^4,& $x\le0$}$ . $\endgroup$ – David Mitra Dec 18 '12 at 13:18
  • $\begingroup$ It was a hint from the professor as part of some review questions for an exam. $\endgroup$ – user53593 Dec 18 '12 at 13:21
  • $\begingroup$ That's true, David. Good point. Yet both cases are contained in the same $\,1-$dimensional subspace $\,Span\{x^4\}\,$... $\endgroup$ – DonAntonio Dec 18 '12 at 13:44
  • $\begingroup$ The $f$ in my comment certainly isn't in the span of $\{x^4\}$... $\endgroup$ – David Mitra Dec 18 '12 at 13:45
  • $\begingroup$ why is $f$ not in the span of $\{x^4\}$? $\endgroup$ – user53593 Dec 18 '12 at 14:13

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