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Chapter 2, Question 12 from Axler's "Linear Algebra Done Right":

Suppose that $p_{0}, p_{1}, \ldots , p_{m}$ are polynomials in $P_{m}(F)$ such that $p_{j}(2) = 0$ for each $j$. Prove that $(p_{0}, p_{1}, \ldots , p_{m})$ is not linearly independent in $P_{m}(F)$.

What I have worked out so far:

Since $p_{j}(2) = 0$ for all $j$, $x^{0} = 1 \notin$ span$(p_{0}, p_{1}, \ldots , p_{m})$. That is , $(p_{0}, p_{1}, \ldots , p_{m})$ does not span $P_{m}(F)$.

Thanks.

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That's it, you are finished. Your space $P_m$ has dimension $m+1$, so by a theorem you probably already have, any set of $m+1$ linearly independent vectors is a basis. Your $m+1$ vectors are (as you proved) not a basis, so they cannot be linearly independent. –  André Nicolas Oct 10 '12 at 5:31
    
So are you using the contrapositive of a proved theorem? That is: "any set of m+1 linearly independent vectors is a basis" is equivalent to "not basis, then any set of m+1 vectors is not linearly independent" –  mathnoob Oct 10 '12 at 5:37
    
I prefer to say "Let $V$ be a space of dimension $m+1$. If a set of $m+1$ vectors is not a basis, then the set is not linearly independent." And yes, it is the contrapositive of the standard theorem. I don't really distinguish between $A\rightarrow B$ and $\lnot B\rightarrow \lnot A$, since they "say" the same thing. –  André Nicolas Oct 10 '12 at 5:47
    
okay thank you. –  mathnoob Oct 10 '12 at 6:02
    
+1 for nicely put, explained and, eventually, solved question –  DonAntonio Oct 10 '12 at 12:41

1 Answer 1

up vote 1 down vote accepted

Here's another way. $p_j(2)=0$ implies $p_j(x)=(x-2)q_j(x)$ for some $q_j$ in $P_{m-1}(F)$. Now $P_{m-1}(F)$ has dimension $m$, so $\{{\,q_0,q_1,\dots,q_m\,\}}$ is too big to be linearly independent. Take any nontrivial linear dependence among the $q_j$ and multiply by $x-2$ to get a nontrivial linear dependence among the $p_j$.

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