Let W = {(f(x)∈ P2[R] : f '(x) + xf(0) = 0}

i) Prove that W is a subspace of P2[R].

ii) Find a basis for W.

Here's what I have so far:

i) I have to verify that the subspace (1) contains the zero space, (2) is closed under addition, and (3) is closed under scalar multiplication.

    (1) Contains the zero space
        f'(x) + xf(0) = 0
        f'(0) + (0)f(0) = 0
        0 = 0, 
        so 0 ∈ W

I'm stuck on numbers 2 and 3, because I'm not sure how to represent f '(x) + xf(0) = 0 with examples that I can just plug in and show that it is closed under addition and scalar multiplication.

I generally do these problems with vectors or the like - for example, if the question was

Let Let W = {(x, y, z)∈ R^3 : x + y - 2z = 0}

and I wanted to show if it was a subspace, I would just show from some vectors v1 = (a1, a2, a3) and v2 = (b1, b2, b3) that

(a1 + b1) + (a2 + b2) - 2(a3 + b3) = (a1 + a2 - 2a3) + (b1 + b2 - 2b3), thus is closed under addition


c(a1 + a2 - 2a3) = ca1 + ca2 - 3ca3, thus is closed under scalar multiplication.

and I would find the basis by just finding the kernel of x + y - 2z = 0

I'm not sure how to deal with the problem when it is converted into polynomials. Thanks.

  • 1
    $\begingroup$ You don't need to check that the point $0$ is in this set. You need to check the function $(x\mapsto 0)$ is in this set (that's the zero vector in this vector space). $\endgroup$
    – user123641
    Jul 21, 2014 at 3:51
  • $\begingroup$ Keep in mind that the 'basis vectors' of a polynomial ring are the monomials $1,x,x^2,\ldots$ themselves. So you could just as well write a polynomial $1+3x-x^2$ as the vector $(1,3,-2)$. $\endgroup$ Jul 21, 2014 at 3:52
  • $\begingroup$ Wait, because verified that the POINT 0 is in the set instead of the vector 0 is in the set, did I do the first step incorrectly? $\endgroup$
    – Ted
    Jul 21, 2014 at 4:12
  • $\begingroup$ Yeah, actually your proof of (1) needs revising. You don't replace $x$ by 0. You replace $f$ by the function $0$, and the desired equality follows for every $x$. $\endgroup$ Jul 21, 2014 at 5:47

2 Answers 2


Proving (2) goes like this: Suppose that $f$ and $g$ both belong to $W$. Ask yourself why $f+g$ also belongs to $W$. It's easy if you write out the equation that $f$ satisfies, namely $f'(x)+x f(0)=0$, then the equation that $g$ satisfies (same equation, with $f$ replaced by $g$) and then add the two equations and apply the distributive property to get stuff times $(f+g)$.

You will need to use the fact that $(f+g)(x)=f(x)+g(x)$, which is the definition of the sum of two functions, and also the fact that $f'+g'=(f+g)'$, which is an elementary property of the derivative.

Proving (3) is even easier.

  • $\begingroup$ What I did, which I'm not sure is correct, is h, g ∈ W h'(x) + x h(0) = 0 g'(x) + x g(0) = 0 h'(x) + g'(x) + xh(0) + xg(0) = (h'(x) + xh(0)) + (g'(x) + xg(0)), c(g'(x) + xg(0)) = cg'(x) + xg(0) $\endgroup$
    – Ted
    Jul 21, 2014 at 4:09
  • $\begingroup$ That's basically right except that you should justify the steps ($h'(x) + x h(0) = 0$ is an assumption, etc.) and you also need to clinch the argument: you've written down $$h'(x) + g'(x) + xh(0) + xg(0) = (h'(x) + xh(0)) + (g'(x) + xg(0)),$$ but the important thing is that this shows that $(h+g)'(x)+x(h+g)(0)=0$, and therefore $(h+g)$ satisfies condition (2). $\endgroup$ Jul 21, 2014 at 4:58
  • $\begingroup$ I think I got it now for the proof part. Thank you! $\endgroup$
    – Ted
    Jul 21, 2014 at 5:31

Let $\phi: P_2 \to P_2$, be given by $\phi(p)(t) = p'(t)+t p(0)$. It is easy to verify that $\phi$ is linear.

Then $W = \ker \phi$. Since the null space of a linear operator is a subspace, it follows that $W$ is a subspace.

If we use the basis $\{t \mapsto 1, t \mapsto t, t \mapsto t^2 \}$ for $P_2$, we see that $\phi$ has the matrix representation $\Phi (p_0,p_1,p_2)^T = (p_1, 2p_2+p_0,0)^T$. That is, $\Phi = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 0 & 0\end{bmatrix}$. It is straightforward to see that $\dim {\cal R} \Phi = 2$, hence $\dim \ker \Phi = 1$. It is also straightforward to see that $\Phi(2, 0, -1)^T = 0$, hence $\ker \Phi = \operatorname{sp} \{ (2, 0, -1)^T \}$.

Now use the basis above to find a basis for $W$.

  • $\begingroup$ I'm not sure how to arrived here: If we use the basis $\{t \mapsto 1, t \mapsto t, t \mapsto t^2 \}$ for $P_2$, we see that $\phi$ has the matrix representation $\Phi (p_0,p_1,p_2)^T = (p_1, 2p_2+p_0,0)^T$. $\endgroup$
    – Ted
    Jul 21, 2014 at 4:14
  • $\begingroup$ Suppose $p \in P_2$, then $p(t) = p_0+p_1 t + p_2 t^2$, and so $p'(t)+tp(0) = p_1+(p_0+2 p_2)t$. The representation of $p$ in the basis is $(p_0,p_1,p_2)^T$ and the representation of $t \mapsto p'(t)+tp(0)$ is $(p_1, p_0+2 p_2, 0)^T$. $\endgroup$
    – copper.hat
    Jul 21, 2014 at 5:40

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