The statement you're dealing with concerns linear algebra, specifically the relationship between linear polynomials (which we interpret as linear functionals or linear maps) on a vector space and their respective null spaces.

We can rephrase the problem as follows:

Let $g, f_1, f_2, \dots, f_r$ be linear polynomials (or linear functionals) on a vector space $V$, with respective null spaces $N$ and $N_1, N_2, \dots, N_r$. The claim is:

$g \text{ is a linear combination of } f_1, f_2, \dots, f_r \iff N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$

Proof Outline

Forward Direction:

If $g$ is a linear combination of $f_1, f_2, \dots, f_r$:

Assume that $g = a_1 f_1 + a_2 f_2 + \dots + a_r f_r$ for some scalars $a_1, a_2, \dots, a_r$. We need to show that $N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$.

1. If $v \in N_1 \cap N_2 \cap \dots \cap N_r$, this means that $f_i(v) = 0$ for all $i = 1, 2, \dots, r$, i.e., $v$ belongs to the null space of each of the functionals $f_1, f_2, \dots, f_r$.

2. Since $g(v) = a_1 f_1(v) + a_2 f_2(v) + \dots + a_r f_r(v)$, and each $f_i(v) = 0$, it follows that $g(v) = 0$.

3. Therefore, $v \in N$, meaning that any vector in $N_1 \cap N_2 \cap \dots \cap N_r$ also belongs to $N$. This proves that $N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$.

Reverse Direction:

If $N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$:

We want to show that $g$ is a linear combination of $f_1, f_2, \dots, f_r$. Assume $N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$.

1. Consider the quotient space $V / (N_1 \cap N_2 \cap \dots \cap N_r)$, where we identify vectors in $N_1 \cap N_2 \cap \dots \cap N_r$ with the zero vector. The functionals $g, f_1, f_2, \dots, f_r$ descend to well-defined linear functionals on this quotient space because their null spaces include $N_1 \cap N_2 \cap \dots \cap N_r$.

2. Since the null spaces of $f_1, f_2, \dots, f_r$ are larger than or equal to $N_1 \cap N_2 \cap \dots \cap N_r$, the images of these functionals in the quotient space span the same subspace of the dual space.

3. If $g$ vanishes on $N_1 \cap N_2 \cap \dots \cap N_r$, it must be expressible as a linear combination of $f_1, f_2, \dots, f_r$ on the quotient space, meaning there exist scalars $a_1, a_2, \dots, a_r$ such that $g = a_1 f_1 + a_2 f_2 + \dots + a_r f_r$.

4. Hence, $g$ is a linear combination of $f_1, f_2, \dots, f_r$.

Conclusion:

Thus, we have shown both directions:

$g \text{ is a linear combination of } f_1, f_2, \dots, f_r \iff N \supseteq N_1 \cap N_2 \cap \dots \cap N_r$

This completes the proof.

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Would you like more details on any part of this proof?

Here are some related questions to explore further:

1. What is the relationship between the null space of a functional and its kernel?
2. How does this theorem generalize to higher-dimensional spaces or more complex vector spaces?
3. Can we interpret the quotient space in terms of direct sums of null spaces?
4. What role does the dimension of $V$ play in these kinds of linear dependence arguments?
5. How can the Rank-Nullity Theorem help in analyzing the dimensions of the spaces involved?

Tip: In linear algebra, always try to interpret abstract vector spaces and functionals through concrete examples, such as $\mathbb{R}^n$, to gain a better intuitive understanding of the results.