# prove list of polynomials is a spanning list

Question:

Suppose $p_0,p_1,..,p_m \in P(F)$ are such that each $p_j$ has degree $j$. Prove that $p_0,p_1,..,p_m$ is a basis of $P_m(F)$, where $P_m(F)$ is the set of all polynomials with coefficients in $F$ and degree at most $m$.

Is this enough to prove the above statement? $\rightarrow$

Since each $p_i \in \{p_0,\dots,p_m\}$ is of degree $i$, that implies each $p_i$ is linearly independent.

Since $p_0...p_m$ is a linearly independent list of length $m+1$, it spans $P_m$, because the dimension of $P_m$ is of length $m+1$, and a linearly independent list of same dimension is a spanning list.

Your proof concerning the linear independenece of the list $p_0,p_1,\dots,p_m$ is incorrect .
What you need to do is to assume that for some $c_0,c_1,\dots c_m\in\mathbf{F}$ we have $q = \sum_{i=0}^{m}c_ip_i = 0$ and then use this to prove that $c_1 = c_2 = \dots = c_m = 0$ which you can do by repeatedly differentiating $q$ a total of $n+1$ times and each time infering that $c_i = 0$ after each differentiation.
Or alternatively you could do that by proving that $p_0$ alone is linearly independent and $p_j\not\in\operatorname{span}(p_0,p_1,\dots,p_{j-1}),\forall j\in\{1,2,\dots,m\}$.
Having proved linear independence then assuming that you know that $\dim\mathcal{P}_m(\mathbf{F}) = m+1$ you can infer that the given list is indeed a basis.
• Sheldon Axler "Linear Algebra Done Right". So the proof is wrong because I have not shown they are linearly independent ? So, for your differentiation method, are the constants $0$ because if $q=ax=0$, then $q'=a=0=>a=0$? So this shows linear independence of the $p_i$, and a linearly independent list of length $m+1$ spans $P_m(F)$ because their dimensions are the same? Is that enough to say proof finished? – Frank May 24 '18 at 8:12
• Yes that is the essence of proving linear indepenence using differentiation method and yes once you have established linear independence it is sufficient to use this together with the the fact $\dim\mathcal{P}_m(\mathbf{F}) = m+1$ to conclue that the list of polynomials in question do indeed span $\mathcal{P}_m(\mathbf{F})$. – Atif Farooq May 24 '18 at 9:09