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Just had an exam where the last question was:

Find a basis for the subset of $\mathbb{P}_3$ where $p(1) = 0$ for all $p$.

I answered $\{t,t^2-1,t^3-1\}$, but I'm not entirely confident in the answer. Did I think about the question in the wrong way?

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Your first polynomial, $t$, isn’t in the subspace: if if $p(t)=t$, then $p(1)=1$, not $0$. Did you mean $t-1$? That would work. – Brian M. Scott Nov 15 '12 at 20:58
Since you have only told us your answer, and haven't told us how you thought about the question, it is impossible to answer your query as to whether you thought about the question the wrong way. – Gerry Myerson Nov 16 '12 at 1:51

Another way to get to the answer:

$P_3=\{{ax^3+bx^2+cx+d:a,b,c,d{\rm\ in\ }{\bf R}\}}$. For $p(x)=ax^3+bx^2+cx+d$ in $P_3$, $p(1)=0$ is $$a+b+c+d=0$$ So, you have a "system" of one linear equation in 4 unknowns. Presumably, you have learned how to find a basis for the vector space of all solutions to such a system, or, to put it another way, a basis for the nullspace of the matrix $$\pmatrix{1&1&1&1\cr}$$ One such basis is $$\{{(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)\}}$$ which corresponds to the answer $$\{{x^3-x^2,x^3-x,x^3-1\}}$$ one of an infinity of correct answers to the question.

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I'm assuming by $\mathbb{P}_3$ you mean the vector space of polynomials of degree 3 or less. It has dimension $4$, and since you have one condition (namely $p(1)=0$), we expect the subspace to have dimension 3, one less.

So it is enough to find three linearly independent basis vectors. In this example, that means three linearly independent polynomials, all with $p(1)=0$.

In your answer ($\{t, t^2-1, t^3-1\}$), the first one must be wrong, since $t(1)=1$. If you replace $t$ by $t-1$, then you have three linearly independent polynomials, all satisfying $p(1)=0$.

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