Let $U =\{ p\in \mathbb{P_4}(\mathbb{R}): p(6) = 0 \}$. Find a Basis of $U$.

Question

LADR 3rd ed. S.Axler|Section 2.C|problem 4. page.48

Let $U = \{ p\in \mathbb{P_4}(\mathbb{R}): p(6) = 0 \}$.

I take the general form of a polynomial of degree 4 : $p(x)=e+dx^1+cx^2+bx^3+ax^4$ and then I tried to narrow the form down to one that represents polynomials of degree 4 with the property that p(6)=Zero

$0=e+d6^1+c6^2+b6^3+a6^4$
$e=-d6^1-c6^2-b6^3-a6^4$

$p(x)=e+dx^1+cx^2+bx^3+ax^4$ turns into: $p(x)=d6^1-c6^2-b6^3-a6^4+dx^1+cx^2+bx^3+ax^4$ $p(x)=d(x-6)+c(x^2-6^2)+b(x^3-6^3)+a(x^4-6^4)$

According to a worked-out solution I found on the net, $\{(x-6),(x^2-6^2),(x^3-6^3),(x^4-6^4)\}$ is indeed supposed to be a basis for the subspace described above.

What I fail to understand is how $\{(x-6),(x^2-6^2),(x^3-6^3),(x^4-6^4)\}$ can be linearly independent and thus a basis for the subspace(assuming the answer is correct).

$0=p(\textbf{6})=d(\textbf{6}-6)+c(\textbf{6}^2-6^2)+b(\textbf{6}^3-6^3)+a(\textbf{6}^4-6^4)$

If x=6, then p(x) is zero, but the coefficients do not have to be zero.

Thank you.

Answer 1

Vectors $\{(x-6),(x^2-6^2),(x^3-6^3),(x^4-6^4)\}$ are linearly independent simply because by definition

$$c_1(x-6)+c_2(x^2-6^2)+c_3(x^3-6^3)+c_4(x^4-6^4)=0\iff c_i=0$$

exactly as for the standard basis $1,x,x^2,x^3,x^4$.

Note that you can't plug in a particular value for x to check linear independence since $1,x, x^2,x^3,x^4$ represent invariable elements/vectors of the vector space $\mathbb{P_4}$.

Answer 2

You just need to prove that $\{x^i-6^i: i=1,\ldots,4\}$ is a linearly independent set. Suppose that $\alpha_1,\ldots,\alpha_4$ are real numbers such that $$ \sum_{i=i}^4 \alpha_i (x^i-6^i) = \alpha_4x^4+\alpha_3x^3+\alpha_2x^2+\alpha_1x-\left(\sum_{i=1}^4 6^i\right) $$ is the constant polynomial $0$. Therefore $$ \alpha_1=\cdots=\alpha_4=0. $$