Proof that if a polynomial is the zero function then all the coefficients are zero.

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?

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 $$a_0=\cdots=a_n=0$$.

Your proof is essentially correct, but you must replace $$a_0$$ by $$1$$ because it is possible that $$a_0 = 0$$ in which case $$a_0,z,z^2,\dots,z^n$$ is not a basis. Also you should not write

Suppose there exist scalars $$a_1,\dots,a_{j-1}\in F$$ such that $$z^j=a_0+a_1z+\dots+a_{j-1}a^{j-1}$$.

1. The $$a_k$$ could be confused with the given coefficients of the polynomial in your question. But this is just a question of presentation.

2. The $$a_0$$ here is not the same as the $$a_0$$ in your "basis" $$a_0,z,z^2,\dots,z^n$$ (but as I said, you should use $$1$$ here).

3. You must suppose that there exist $$b_0,\ldots,b_n \in F$$ not all zero such that $$b_01 + b_1z + \ldots + b_nz^n = 0$$ for all $$z \in F$$. Then let $$j$$ be the largest index such that $$b_j \ne 0$$. You get $$z^j = \frac{b_0}{b_j} + \frac{b_1}{b_j}z + \ldots +\frac{b_{j-1}}{b_j}z^{j-1} .$$ This allows to proceed as you did.

I think there is a much simpler proof. Assume that $$\sum_{k=0}^n a_k z^k = 0.$$ Differentiating $$n$$ times, we get $$n!a_n = 0$$, so $$a_n = 0$$. Repeat this process to find that $$a_i = 0$$ for $$i = 0,1,\dots,n$$. Therefore, $$f(z) = 0 + 0z + \dots + 0z^n$$ is clearly the additive identity polynomial function in a field. Note that this also proves that $$(1,z,z^2,\dots,z^n)$$ form a basis for all polynomials of degree less than or equal to $$n$$, however the converse was unnecessary.

The fundamental theorem of algebra is often stated as: an $$n$$-th degree polynomial over $$\Bbb C$$ has exactly $$n$$-roots.

The easy direction is that it can have at most $$n$$. That's also called the factor theorem: if $$p(a)=0$$, then $$(z-a)\mid p(z)$$.

Now to your problem: if a polynomial equals zero, then in particular it has more than $$n$$ roots. This implies that it's the zero polynomial (by the factor theorem).