Show that the set $\{u_1, u_2, u_3,\dots, u_k\}$ is linearly independent

Question

Let $\{u_1, u_2, u_3,\dots, u_k\}$ be real polynomials:

$u_1(x)=1$

$u_2(x)=1+x$

$u_3(x)=1+x+x^2$

$\dots$

$u_k(x)=1+x+\dots+x^{k−1}$

with $k$ a positive integer.

Show that the set $\{u_1, u_2, u_3,\dots, u_k\}$ is linearly independent.

I know that to be linearly independent, we must have

$u_1a_1+u_2a_2+\dots+u_ka_k = 0 $

if and only if $a_i=0$ for every $i$

But I am unable to develop this idea.

Answer 1

Insert now those polynomials:

$$ a_1+(1+x)a_2+(1+x+x^2)a_3+...+(1+x+x^2+...+x^{k-1})a_k=0$$

so $$(a_1+a_2+a_3+...+a_k)\cdot 1 +(a_2+a_3+...+a_k)x+ (a_3+a_4+...+a_k)x^2+...+(a_{k-1}+a_k)x^{k-2}+a_kx^{k-1}=0$$

so polynomial on left is identical to 0, so all it coefficents are 0:

$$a_1+a_2+a_3+...+a_k=0$$ $$a_2+a_3+...+a_k=0$$ $$ \vdots $$ $$ a_{k-1}+a_k =0$$ $$ a_k=0$$ so starting from down to up you see that all coefficients are 0 and we are done.

Answer 2

Proceed by induction on $k$. Any linear combination of the $u_i$ will be a polynomial in $x$ of degree at most $k-1$. If the linear combination is $0,$ what does that tell you about the coefficient of $u_k$ in your sum? Using the induction hypothesis, what further conclusion can you draw?

Answer 3

The matrix built with the coordinate vectors with respect to the basis $\{1,x,x^2,\dots,x^{k-1}\}$ of the vector space of polynomials of degree at most $k-1$ is $$ \begin{bmatrix} 1 & 1 & 1 & \dots & 1 & 1 \\ 0 & 1 & 1 & \dots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1 & 1 \\ 0 & 0 & 0 & \dots & 0 & 1 \end{bmatrix} $$ and this matrix has rank $k$.

Alternatively, rewrite the linear combination as $$ a_kx^{k-1}+(a_k+a_{k-1})x^{k-2}+\dots+(a_k+\dots+a_2)x+(a_k+\dots+a_1)1=0 $$ and finish.

Answer 4

Consider their Wronskian:

$$W(u_1,u_2,...,u_k)= \begin{vmatrix} u_1&u_2&\cdots&u_{k-1}&u_k \\ u_1'&u_2'&\cdots&u_{k-1}'&u_k' \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ u_1^{(k-2)}&u_2^{(k-2)}&\cdots&u_{k-1}^{(k-2)}&u_k^{(k-2)}\\ u_1^{(k-1)}&u_2^{(k-1)}&\cdots&u_{k-1}^{(k-1)}&u_k^{(k-1)} \end{vmatrix}=\\ \begin{vmatrix} \color{red}1&1+x&\cdots&1+x+\cdots x^{k-2}&1+x+\cdots x^{k-1} \\ 0&\color{red}{1!}&\cdots&1+2x+\cdots (k-2)x^{k-3}&1+2x+\cdots (k-1)x^{k-2} \\ \vdots&\vdots&\ddots&\vdots&\vdots \\ 0&0&\cdots&\color{red}{(k-2)!}&(k-2)!+(k-1)!x\\ 0&0&\cdots&0&\color{red}{(k-1)!} \end{vmatrix}=\\ \prod_{i=1}^{k-1} i!\ne 0.$$ Hence, the given functions are linearly independent.