Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given a subspace of all polynomials $f(t)$ in $\mathbf{P}_2$ such that $f(1)=0$. I know that a basis for this space is $1-t$, $1-t^{2}$, and when I look at it, it makes perfect sense as to why. I was just wondering what is a systematic way of finding it, without eyeballing. Thanks!

share|improve this question
add comment

1 Answer

up vote 10 down vote accepted

Note that there is no such thing as the basis for this subspace: there are only two vector spaces in the entire universe that have just one basis: the zero vector space (okay, this may count as more than one vector space: the zero vector space over any field), and the one dimensional vector space over a field of $2$ elements. This is neither, so there is no such thing as "the" basis of the subspace: the subspace has lots of different bases, this is just one of them.

Now, as to this particular problem. Added: You don't actually need the following paragraph, but it gives you a way to know ahead of time how "big" a basis you are looking for, which may be useful in other situations.

The subspace you want is the nullspace of the linear tranformation $\mathbf{P}_2\to\mathbb{R}$ given by "evaluation at $1$." By the Rank-Nullity Theorem, you know the nullspace has dimension $2$, so you are looking for two linearly independent polynomials that are $0$ at $1$. Added. To see how the Rank-Nullity Theorem comes into play, remember that when you interpret the Rank-Nullity Theorem in terms of linear transformations it tells you that if $T\colon\mathbf{V}\to\mathbf{W}$ is a linear transformation, then $$\dim(\mathbf{V}) = \mathrm{rank}(T) + \mathrm{nullity}(T) = \dim(\mathrm{Im}(T)) + \dim(\mathrm{ker}(T)).$$ Here, $\mathbf{V}=\mathbf{P}_2$, $\mathbf{W}=\mathbb{R}$, and $T$ is "evaluation at $1$". Since $\dim(\mathbb{R}) = 1$, the only two possibilities for the rank of $T$ are $0$ and $1$; but the rank of $T$ is not $0$, because $T$ is not constant $0$ (it maps the polynomial $x$ to $1\neq 0$). So the rank is $1$, and since $\dim(\mathbf{P}_2) = 3$, that means the nullity is $2$. So the subspace you are looking for has dimension $2$.

A way to think about the Rank-Nullity Theorem is like the "law of conservation of matter": dimensions are not created nor destroyed, they are just transformed. If you have $\mathbf{V}$ of dimension $n$, then if you add the dimensions you "transform" to $\mathbf{0}$ and the dimensions you get in the image, you should get all the dimensions (they add up to $n$).

And now, finally, how to get a basis:

A polynomial is $0$ at $1$ if and only if it is divisible by $t-1$. So you are looking for a basis for the polynomials that are multiples of $t-1$, that is, of the form $(t-1)q(t)$, where $q(t)$ is a polynomial of degree $0$ or $1$ (cannot be degree two, since you are in $\mathbf{P}_2$). So if you pick any basis for the subspace of polynomials of degree at most $1$, say $1$ and $t+1$, then you get a basis for your subspace by considering the products $(t-1)1$ and $(t-1)(t+1)$: just take your arbitrary $(t-1)q(t)$, express $q(t)$ in terms of the vectors you took, $1$ and $t+1$, and this gives you how to express $(t-1)q(t)$ in terms of $(t-1)1$ and $(t-1)(t+1)$.

You could also have gotten a basis by taking $1$ and $t$, to get $t-1$ and $t^2-t$; or any basis for the subspace of polynomials of degree at most $1$ will do.

share|improve this answer
Arturo, thanks a lot for your detailed reply. I will go through it right now. Could you please explain though how you are using rank-nullity with respect to this problem? I am used to using that theorem with matrices, but now that we moved on, it often feels counter-intuitive. I can see that the Image of the transformation is 1, since we are mapping to one number in $\mathbb{R}$, but that's about it. Thanks! –  user7016 Feb 13 '11 at 6:29
@Kira: Added a bunch of stuff; I hope it is useful. –  Arturo Magidin Feb 13 '11 at 6:36
Thanks, this is great! –  user7016 Feb 13 '11 at 6:51
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.