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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?

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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 $(2)$.

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    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 possibility2012-12-18
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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 $\,\dim\ker T\neq 1\,$ ?

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    Read David's answer, @user53593: I think that's right on the money.2012-12-18