# Basis and Dimension of a Subpace.

1. May 23, 2012

### Hiche

1. The problem statement, all variables and given/known data

V = {p(x) belongs to P3 such that p'(1) + p'(-1) = 0}

2. Relevant equations

...

3. The attempt at a solution

Okay, so finding the first derivative of p(x) = ax^3 + bx^2 + cx + d and plugging in the values 1 and -1 (to find p'(1) and p'(-1)), we get c = -3a. Does this make the basis of V = {x^3 - 3x, x^2, 1}. And the dimension hence is 3? I'm fairly new to the subject.

Also, when wanting to prove a transformation is one-to-one, does finding that the nullspace of T = {0} (i.e, the homogenous system has only one trivial solution which is 0) suffice?

Last edited: May 23, 2012
2. May 25, 2012

### Kreizhn

You are exactly correct: Indeed, we can think of $1,x,x^2,x^3$ as basis vectors, and the requirement on V is that c=-3a as you have shown. Hence
$$V = \{ ax^3 + bx^2 -3ax + d: a,b,d \in F \}$$
where F is whatever field in which you are working. Thus there are only three free parameters, so V is three dimensional. You already gave an explicit basis, so everything is fine there.

You are also correct about showing one-to-one ness: one need only show that the nullspace is trivial. Indeed, having trivial kernel and being injective are equivalent for linear operators on vector spaces.

Let $T: V \to W$ be a linear map on vector spaces V and W. Suppose first that T is injective and let $x \in \ker T$ so that Tx=0. But T0 = 0 so we have that $Tx = T0$ and by injectivity this implies that x = 0. Thus the kernel is always trivial.

Conversely, assume that the kernel is always trivial and consider Tx=Ty. Since the function is linear, we must have that Tx-Ty = T(x-y) = 0 so that $x-y \in \ker T$. Since the kernel is trivial, this means that x-y=0, so in particular x=y. Thus the map is injective.

3. May 25, 2012

### Hiche

A (very slightly) different way of looking at it is that the condition f'(1)+ f'(-1)= 0 leads to, as you say, c= -3a and so any such polynomial may be written $ax^3+ bx^3+ cx+ d= ax^3+ bx^2- 3ax+ d$$a(x^3- 3x)+ bx^2+ d$ so that any such polynomial can be written as a linear combination of $x^3- 3x$, $x^2$ and $1$. Those are the "basis vectors".