A Proof of Brill’s Theorem via Directional Derivatives

Algebra, Linear algebra, Polynomials, ProofWiki 5 Minutes

Let K be the field of real numbers \mathbb{R} or the field of complex numbers \mathbb{C}. Given a vector v = (v_1, \ldots, v_n) \in K^n. The directional derivative operator with respect to the vector v, denoted by D_v, is a map from the ring K[x_1, \ldots, x_n] to itself, defined by

\displaystyle D_v(P) = \sum_{i=1}^n v_i \frac{\partial P}{\partial x_i},

where \frac{\partial P}{\partial x_i} is the formal partial derivative of the polynomial P with respect to the variable x_i.

Theorem 1. Let P, Q \in K[x_1, \ldots, x_n], constants a, b \in K, and a linear form L(x) = c_1x_1 + \ldots + c_nx_n. The operator D_v has the following properties.

(1) D_v(aP + bQ) = aD_v(P) + bD_v(Q).

(2) D_v(PQ) = D_v(P)Q + P D_v(Q).

(3) For any positive integer k, D_v(P^k) = k P^{k-1} D_v(P).

(4) D_v(L) = L(v).

Proof. (1) By the definition of the partial derivative of a polynomial, we have

\displaystyle \frac{\partial (aP + bQ)}{\partial x_i} = a\frac{\partial P}{\partial x_i} + b\frac{\partial Q}{\partial x_i},

thus we immediately obtain \displaystyle D_v(aP + bQ) = aD_v(P) + bD_v(Q).

(2) Since the formal partial derivative obeys the product rule

\displaystyle \frac{\partial (PQ)}{\partial x_i} = \frac{\partial P}{\partial x_i}Q + P\frac{\partial Q}{\partial x_i} we have

\displaystyle D_v(PQ) = \sum_{i=1}^n v_i \left( \frac{\partial P}{\partial x_i}Q + P\frac{\partial Q}{\partial x_i} \right)

\displaystyle = \left( \sum_{i=1}^n v_i \frac{\partial P}{\partial x_i} \right) Q + P \left( \sum_{i=1}^n v_i \frac{\partial Q}{\partial x_i} \right) = D_v(P)Q + P D_v(Q).

(3) We prove this by mathematical induction on k.

For k=1, the formula becomes D_v(P^1) = 1 \cdot P^0 \cdot D_v(P), which is trivially true.

Suppose the property holds for k-1, that is

\displaystyle D_v(P^{k-1}) = (k-1)P^{k-2}D_v(P).

Applying (2), we obtain

\displaystyle D_v(P^k) = D_v(P)P^{k-1} + P D_v(P^{k-1})

=D_v(P)P^{k-1} + P \big( (k-1)P^{k-2}D_v(P) \big)=k P^{k-1} D_v(P).

By the principle of mathematical induction, the property holds for all positive integers k.            

(4) For a linear form L(x) = c_1x_1 + \ldots + c_nx_n, we have

\displaystyle D_v(L) = \sum_{i=1}^n v_i \frac{\partial L}{\partial x_i} = \sum_{i=1}^n v_i c_i = L(v).

The theorem is completely proved. \Box

Theorem 2 (Brill). Suppose K is the field of real numbers \mathbb{R} or the field of complex numbers \mathbb{C}. Let L_1, L_2, \ldots, L_m be m non-zero linear forms in n (n>1) variables that are pairwise non-proportional over K. Then, for any integer k \ge m-1, the set of powers \{L_1^k, L_2^k, \ldots, L_m^k\} is linearly independent over K.

Proof. We prove the theorem by mathematical induction on the number of linear forms m. For m=1, we have 1 non-zero linear form L_1. For any integer k \ge 0, the polynomial L_1^k is not the zero polynomial, so the set consisting of only one element \{L_1^k\} is trivially linearly independent. Thus, the assertion holds for m=1.              

Now suppose the assertion holds for m-1 linear forms, for some integer m > 1. Consider m non-zero pairwise non-proportional linear forms L_1, \ldots, L_m and an exponent k \ge m-1.

Consider the linear relation

\displaystyle \sum_{i=1}^m c_i L_i^k = 0 (1)

where c_i \in K. We need to show that c_i = 0 for all i = 1, \ldots, m.              

Since L_m is not identically 0, the set of points x \in K^n such that L_m(x) = 0 is an (n-1)-dimensional vector subspace of the space K^n. At the same time, since L_i is not proportional to L_m for all i < m, the intersection of the two spaces L_i(x) = 0 and L_m(x) = 0 is an (n-2)-dimensional subspace.

The field K has characteristic 0 and is therefore an infinite field. A vector space over an infinite field cannot be the union of finitely many proper subspaces (see [1]). Therefore, there exists a vector v = (v_1, \ldots, v_n) \in K^n such that L_m(v) = 0 and \displaystyle L_i(v) \neq 0, \quad \forall i = 1, \ldots, m-1.

Consider the directional derivative operator with respect to the vector v defined as follows

\displaystyle D_v = \sum_{j=1}^n v_j \frac{\partial}{\partial x_j}.

When applying D_v to a linear form \displaystyle L(x) = a_1x_1 + \ldots + a_nx_n,

we have \displaystyle D_v(L) = a_1v_1 + \ldots + a_nv_n = L(v).

When applying this operator to the k-th power of L, we obtain

\displaystyle D_v(L^k) = k \cdot L^{k-1} \cdot D_v(L) = k \cdot L(v) \cdot L^{k-1}.

Applying the operator D_v to both sides of equation (1), we have

\displaystyle \sum_{i=1}^m c_i k L_i(v) L_i^{k-1} = 0.

Since L_m(v) = 0, the m-th term vanishes completely. Since m \ge 2, we have k \ge m-1 \ge 1, so we can divide both sides by k (which is valid since K has characteristic 0) to obtain

\displaystyle \sum_{i=1}^{m-1} \big(c_i L_i(v)\big) L_i^{k-1} = 0.

Since k \ge m-1, we deduce that k-1 \ge m-2. Applying the inductive hypothesis to m-1 linear forms with the exponent k-1, the set of polynomials \{L_1^{k-1}, \ldots, L_{m-1}^{k-1}\} is linearly independent. Therefore, all coefficients in the sum above must be zero. That is,

\displaystyle c_i L_i(v) = 0, \quad \forall i = 1, \ldots, m-1. But by the choice of the vector v, we already have L_i(v) \neq 0, so we must have \displaystyle c_i = 0, \quad \forall i = 1, \ldots, m-1. Substituting c_1 = \ldots = c_{m-1} = 0 back into (1), we have c_m L_m^k = 0. Since L_m is not zero, c_m = 0.            

Thus c_1 = c_2 = \ldots = c_m = 0, and \{L_1^k, L_2^k, \ldots, L_m^k\} is linearly independent. By the principle of mathematical induction, the theorem is proved. \Box

References

[1] https://nttuan.org/2010/02/04/finite-unions-of-proper-subspaces-over-an-infinite-field/

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