Proposition: If a polynomial $a_0+a_1z+\dots +a_nz^n=0$ for all $z\in F$, then $a_0=a_1=\dots=a_n=0$.
Where $F$ denotes the field of real or complex numbers.
Proof: Let $P_n(F)$ denote the vector space of all polynomials in $F$. We prove that $a_0,z,z^2,\dots,z^n$ is a basis of $P_n(F)$.
Clearly $a_0,z,z^2,\dots,z^n$ spans $P_n(F)$ by definition of $P_n(F)$. We just need to show that this list is linearly independent. To do this, it suffices to show that $z^j\not\in$ span$(a_0,z,\dots,z^{j-1})$.
Suppose there exist scalars $a_1,\dots,a_{j-1}\in F$ such that $z^j=a_0+a_1z+\dots+a_{j-1}a^{j-1}$. Then differentiating both sides $j$ times shows that the L.H.S is non-zero whereas the R.H.S is zero. A contradiction. Thus, the list $a_0,z,z^2,\dots,z^n$ is linearly independent.
Because the list $a_0,z,z^2,\dots,z^n$ is linearly independent, a linear combination of this equals zero if and only if $a_0=a_1\dots=a_n=0$. Hence, if a polynomial is the zero function, then all the coefficients are zero.
Is this proof correct?
Best Answer
Your argument is fine, but it is unnecessarily complicated. There is no need to use linear independence.
if you are going to use differentiation anyway, just evaluate at zero; then differentiate, and repeat. In each iteration you get respectively that $a_0=0$, $a_1=0$, etc.
but there is no need to use differentiation, either. Taking $z=0$, you get $a_0=0$. Now $$ 0=a_1z+\cdots+a_nz^n=z (a_1+a_2z+\cdots+a_nz^{n-1}). $$ Since this equality holds for all $z$, we get that the polynomial inside the brackets is zero for all nonzero $z$, and hence for all $z$ by continuity. Now we can iterate the argument.
a third argument can be obtained using linear algebra. If we consider the $n$ equations $$ a_0+a_1\,k+a_2\,k ^2+\cdots+a_n\,k ^n=0,\qquad\qquad k=1,\ldots,n, $$ this is a homogeneous system of linear equations with matrix $[k^{j-1}]_{k,j}$. This is a Vandermonde matrix and hence its determinant is nonzero. Which means that the only solution to the system is $a_0=\cdots=a_n=0$.