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I can't find the Reduced Row Echelon Form to find the number of pivots because I don't have numbers to work with. I know an upper bound for the rank is the smaller amongst $p$ and $n+1$. Any tips on how to approach this problem? Basically, I need to find the dimension of the column space.

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You need to choose a basis for your polynomial space (I recommend the standard basis $\{1,x,x^2,\ldots\}$), then represent each polynomial in terms of that basis. That is, polynomial space of degree at most $n$ is isomorphic with $\mathbb{R}^{n+1}$. Once you have your polynomials represented as $(n+1)$-tuples, then you can do row reduction as usual.

Example: $u=1+x, v=2+3x$. $E=\{1,x\}$ is the standard basis. Then $[u]_E=[1, 1]$ and $[v]_E=[2,3]$. We can now row reduce $\left[\begin{smallmatrix}1&1\\2&3\end{smallmatrix}\right]$ to get $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$. This has two pivots, so $\{u,v\}$ are independent.

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I don't understand how I would do row reduction in this case. My variables in the first row would still be of the form a11, a21, ..., ap1, etc. I was thinking of finding a basis for the column space in terms of the columns of the matrix. – ask Oct 18 '13 at 2:17
I should clarify. I don't actually have numbers in my coefficient matrix to work with. I'm supposed to find the rank for ANY set of p distinct polynomials of size at most n. That's where I'm stuck. I'm guessing the answer should be in terms of p and n since those are the only variables I have. – ask Oct 18 '13 at 2:21
All you can say in this case is the rank is (1) at most $p$, and (2) at most $n+1$. – vadim123 Oct 18 '13 at 2:23
Yeah...that's what I thought. Maybe that's about it. Thank you. – ask Oct 18 '13 at 2:24

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