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I am trying to find the dimension of the vector space ($P_2$ means polynomials of degree at most $2$) $$V = \{p(x) \in P_2 \mid xp'(x) = p(x)\}.$$ However, I don't even know how to start...

(Additionally, the question asks to find the basis for $V$, which I don't know how to do, but I think I might if someone tells me how to understand the above first...)

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Suggestions: Write down what a typical element of $P_2$ looks like. Write down what its derivative is, and multiply by $x$. Using these formulas, write down and analyze the equation $xp'(x)=p(x)$. You will be able to explicitly describe what $V$ is, leading to both the dimension and a basis. – Jonas Meyer May 28 '12 at 23:09
up vote 4 down vote accepted

Consider the linear transformation $T\colon P_2\to P_2$ given by $T(p) = xp' - p$. You should verify that this is indeed a linear transformation.

For example, if $p(x) = 3x^2 - 2x$, then $p'(x) = 6x-2$, so $xp'(x) = x(6x-2) = 6x^2-2x$. Therefore, $$T(p) = xp' - p = (6x^2-2x)-(3x^2-2x) = 6x^2-2x-3x^2+2x = 3x^2.$$

The vector space you want is nothing more than the nullspace of this linear transformation. So you can compute the nullspace with any method you may already know (e.g., find a basis for $P_2$, find the coordinate matrix for $T$ relative to that basis, and then find the nullspace of the matrix), and use it to find its dimension.

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Taking an arbitrary element $p(x)=ax^2+bx+c\in V$, we see it has to satisfy the special condition $xp'(x)=p(x)$, so $x(2ax+b)=2ax^2+bx+0=ax^2+bx+c$.

Then immediately $c=0$, and $2a=a$ implies that $a=0$.

So the only things in $V$ are of the form $bx$ for any $b\in F$.

I think you can work out a basis :)

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So, $dim (V) = 1$ and Basis = {bx} where b∈R. Correct? – MathMathCookie May 29 '12 at 2:25
@MathMathCookie: Yes, the dimension is $1$, but no, that is not the basis, that is $V$. An example of a basis is $\{x\}$. If you meant to say that for each $b\in\mathbb R$, $\{bx\}$ is a basis, that is close but not quite right, because $\{bx\}$ is not a basis for $V$ when $b=0$. – Jonas Meyer May 29 '12 at 3:03
@JonasMeyer Oops, thanks! – MathMathCookie May 29 '12 at 18:44

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