# Show that the polynomials $1$,$x$, $x^2$, ..., $x^n$ form a linearly independent set in $P_n$.

This is the solution according to my book.

Solution. For convenience, let us denote the polynomials as $$p_0=1,p_1=x,p_1=x^2,\dots,p_n=x^n$$ We must show that the only coefficients satisfying the vector equation \begin{equation}\tag{5} a_0 p_0 + a_1 p_1 + a_2 p_2 + \cdots + a_n p_n = 0 \end{equation} are $$a_0=a_1=a_2=\cdots=a_n=0$$ But $(5)$ is equivalent to the statement that \begin{equation}\tag{6} a_0+a_1 x+a_2 x^2+\cdots+a_n x^n=0 \end{equation} for all $x$ in $(-\infty,\infty)$, so we must show that this is true if and only if each coefficient in $(6)$ is zero. To see that this is so, recall from algebra that a nonzero polynomial of degree $n$ has at most $n$ distinct roots. That being the case, each coefficient in $(6)$ must be zero, for otherwise the left side of the equation would be a nonzero polynomial with infinitely many roots. Thus, $(5)$ has only the trivial solution.

(original here: http://i.stack.imgur.com/sp7vY.png)

I don't really understand the conclusion that it makes. Why would the left side of the equation be a non-zero polynomial with infinitely many roots? Doesn't the sentence preceding it say that a non-zero polynomial of degree $n$ has at most $n$ distinct roots? Does that not contradict the statement that it would have infinitely many roots? And how does this all end up meaning that each coefficient must be zero?

Can somebody clarify this a bit for me?

• They actually form a linearly independent set in $P_n$. Mar 19, 2016 at 7:44
• That's what I meant. Is there a way to edit that? @learner Mar 19, 2016 at 7:47

The degree of a (nonzero) polynomial is the highest power of the variable appearing which has a nonzero coefficient. In the expression on the left, we do not know the values of $a_0,\dots,a_n$, so the degree of the left side could be any value up to $n$, or it might be the zero polynomial.
The right side of the equation is the zero polynomial. The zero polynomial has infinitely many roots, since for every value of $x$, $0$ is still $0$. Due to the equality between the left and right sides, the left side also has this property.
The left side has infinitely many roots because we have $Q(x)=a_0+a_1x+\ldots+a_nx^n=0$ for all $x\in (-\infty,\infty)$, which means every real number is a root of the polynomial $Q(x)$.
So, since every non-zero polynomial of degree $n$ has at most $n$ roots, and the polynomial $Q(x)$, the only possibility is that $Q(x)$ is the zero polynomial, that is, the polynomial whose coefficients $a_i$ are all equal to zero.