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Problem: Suppose that po, p1, ...., pm are polynomials in Pm(F) such that pj(2)=0 for every j. Prove that (p0, p1, ...., pm) is not linearly independent in Pm(F).
Not really sure where to start with this one guys. Any help would be much appreciated. THANKS.
letQuote:
Originally Posted by eg37se
let
and
then
and so
thus the columns of
and hence the rows
ofare linearly dependent, i.e. there exist scalars
which are not all 0, such that
where
is the
-th row of
this result is now equivalent
to what we're looking for:
I believe you that that works, but those are concepts we unfortunately have not covered yet. Therefore, I think I am going to try to use the Linear Dependence Lemma in my book which states that, (v1,...,vm) is linear dependent provided that:
vj is an element of span (v1, ...., v(j-1))
and
if the jth term is removed from v1,..., vn the span of the remaining list equals span (v1,...., vn)
Let me know if this looks like it could work. Thanks.
Consider the set. Clearly this is a subspace of
, and every
can be written as
. Consider the map
This is a surjective (i.e. onto) linear transformation, because to every, we have
. Moreover it is injective (i.e. one-to-one) because the inverse image of the zero polynomial is just the zero polynomial (check this). Hence
.
It follows that since you haveelements of
in your list, they cannot be linearly independent.
