Let g,f1,f2....,fr be linear polynomials on a vector space V with respective null spaces N,N1,...Nr .Then g is a linear combination of g is a linear combination  of f1,f2,...fr ⟺ N contains N1∩N2∩…∩Nr.

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.

Let g,f1,f2....,fr be linear polynomials on a vector space V with respective null spaces N,N1,...Nr. Then g is a linear combination of f1,f2,...fr ⟺ N contains N1∩N2∩…∩Nr.

Explore the relationship between linear combinations of linear polynomials and their respective null spaces in vector spaces. This solution provides a detailed proof and explanation suitable for college-level linear algebra students.

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