How do we prove that $(x^4-6x^3), (x^3-6x^2), (x^2-6x)$, and $(x-6)$ are linearly independent?

Question

Consider the following problem from Axler's Linear Algebra Done Right

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

$P_4(F)$ is the set of all polynomials of degree $\leq 4$ with coefficients in field $\mathbb{F}$.

For any polynomial $p$ in $U$, we can write

$$p(x)=(x-6)q(x)=(ax^3+bx^2+cx+d)$$

where $q(x)\in P_3(\mathbb{F})$.

Thus, we have

$$p(x)=a(x^4-6x^3)+b(x^3-6x^2)+c(x^2-6x)+d(x-6)$$

Thus, the vectors $(x^4-6x^3), (x^3-6x^2), (x^2-6x)$, and $(x-6)$ span $U$.

How do we prove that these vectors are linearly independent (l.i.)?

Here is my attempt at proving this

If they are l.i. then

$$a(x^4-6x^3)+b(x^3-6x^2)+c(x^2-6x)+d(x-6)=0$$

for all $x$ only in the case where $a=b=c=d=0$.

Now, no matter what values of $a,b,c$, and $d$ we choose that are not all zero, when we make $x$ very large then the monomial term with the largest degree dominates everything else, and thus the expression is definitely not zero at such values of $x$.

Thus, the only way to make the expression zero for all $x$ is to choose $a=b=c=d=0$.

This seems correct. Is it?

What are some other ways to prove it?

Answer 1

As you mentioned above, we have to show the only solution for $a(x^4-6x^3)+b(x^3-6x^2)+c(x^2-6x)+d(x-6)=0$ is the trivial solution $a=b=c=d=0$. To show this notice that the following system of equations must be satisfied

$$\begin{cases} a = 0 \\ -6a+b =0 \\ -6b+c=0 \\ -6c+d=0 \\ -6d = 0 \end{cases}$$

think of the monomials $1,x,x^2,x^3,x^4$ as your basis vectors and their coefficients must be zero. It is easy from above equations to conclude that $a=0,b=0,c=0,d=0$ is the only possible solution and therefore the vectors $(x^4-6x^3),(x^3-6x^2),(x^2-6x),(x-6)$ are linearly independent.

Don't think about the polynomials as a function $f:\mathbb{F}\rightarrow \mathbb{F}$ of field $\mathbb{F}$ onto itself, think about them as a vector space. The space of polynomials with coefficients from $\mathbb{F}$ is an infinite dimensional space where the natural basis is $(1,x,x^2,x^3,\cdots,x^n,\cdots)$ and the coefficient of each $x^i$ term is the scalar of the vector $x^i$.

Answer 2

We have $$ \begin{pmatrix} x^4-6x^3 \\ x^3-6x^2 \\ x^2-6x \\ x-6 \\ -6 \end{pmatrix} =\begin{pmatrix} 1 & -6 & 0 & 0 & 0 \\ 0 & 1 & -6 & 0 & 0 \\ 0 & 0 & 1 & -6 & 0 \\ 0 & 0 & 0 & 1 & -6 \\ 0 & 0 & 0 & 0 & -6 \end{pmatrix} \begin{pmatrix} x^4 \\ x^3 \\ x^2 \\ x \\ 1 \end{pmatrix} $$ Since the matrix is clearly invertible, the polynomials on the left hand side are a basis for $P_4$ and so are linearly independent.

Answer 3

An alternative approach, as hinted at by @K.K.McDonald, is to consider the standard basis for the polynomial space $P_4(F) $, which is $\{1, x, x^2, x^3, x^4\} $. We can then construct a matrix with column vectors corresponding to the coordinate vectors of the polynomials $(x^4 - 6x^3 ), ( x^3 - 6x^2),( x^2 - 6x)$, and $(x - 6)$

$$ \begin{pmatrix} -6 & 0 & 0 & 0 \\ 1 & -6 & 0 & 0 \\ 0 & 1 & -6 & 0 \\ 0 & 0 & 1 & -6 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

Upon inspection, it becomes clear that these column vectors cannot be expressed as linear combinations of one another. In other words, they are linearly independent.