How to prove whether a set of polynomials is linearly independent?

Question

$P_4(\mathbb{R})$ is the set of polynomials with degree at most $4$ with real coefficients.

$U=\{p \in P_4(\mathbb{R})∣p(3)=0\}$

A basis of $U$ would be $\{(x−3)^4,(x−3)^3,(x−3)^2,(x−3)\}$

But how do we prove, that the set of polynomial is linearly independent?

If we build the linear combination: $a\cdot(x−3)^4 + b\cdot(x−3)^3 + c\cdot(x−3)^2 + d\cdot(x−3) = 0$

Then we see that $a=b=c=d=0$. However, in order to be linearly independent, $a=b=c=d=0$ has to be the only solution for the equation. But this is not the case, since $x=3$ is also a solution $a\cdot(3−3)^4 + b\cdot(3−3)^3 + c\cdot(3−3)^2 + d\cdot(3−3) = 0$.

Could anyone tell me please, where is my mistake?

Thanks

Answer 1

.Ah,no. It's a minor confusion, but worth clearing up all right.

See, as values, it is true that $a(3-3)^4 + b(3-3)^3 + c(3-3)^2 + d(3-3) = 0$. However, this does not mean that $a(x-3)^4 + b(x-3)^3 + c(x-3)^2 +d(x-3)=0$ as polynomials.

Equality as polynomials, means that the two polynomials must evaluate to the same quantity at every point. That is, $p \equiv q$ as polynomials if for all $x$, $p(x)=q(x)$.

You have only checked this for one point, $x=3$. There are many points left. As it turns out, $a(x-3)^4 + b(x-3)^3 + c(x-3)^2 +d(x-3)=0$ as polynomials, if and only if $a=b=c=d=0$. this can be proved by appropriate substitution of certain $x$, and then finding $a,b,c,d$ (or by differentiation).

I hope this clears the confusion.

Answer 2

The zero vector in your vector space is the zero polynomial, not the number $0$. Thus you need to find coefficients $a,b,c,d$ such that the polynomial evaluates to $0$ for all $x$, not just a particular $x$ such as your $x=3$. And only $a=b=c=d=0$ fits that bill.