To prove that the set $S \cup \{ \beta \}$ is linearly independent, we will follow these steps:

Given:

• $S$ is a linearly independent subset of a vector space $V$.
• $\beta$ is a vector in $V$ that is not spanned by $S$, meaning that $\beta$ is not a linear combination of the vectors in $S$.

Goal:

Prove that the set $S \cup \{ \beta \}$ is linearly independent.

Proof:

We want to show that the set $S \cup \{ \beta \}$ is linearly independent. According to the definition of linear independence, a set of vectors is linearly independent if the only solution to the equation

$c_1 v_1 + c_2 v_2 + \dots + c_n v_n = 0$

is $c_1 = c_2 = \dots = c_n = 0$.

Here, let $S = \{ v_1, v_2, \dots, v_k \}$ be a linearly independent set, and $\beta$ is a vector in $V$. We need to show that the only solution to

$c_1 v_1 + c_2 v_2 + \dots + c_k v_k + c_{\beta} \beta = 0$

is $c_1 = c_2 = \dots = c_k = c_{\beta} = 0$.

Step 1: Assume a linear combination equals zero

Suppose there exist scalars $c_1, c_2, \dots, c_k, c_{\beta} \in \mathbb{F}$ such that

$c_1 v_1 + c_2 v_2 + \dots + c_k v_k + c_{\beta} \beta = 0$

We need to prove that this implies $c_1 = c_2 = \dots = c_k = c_{\beta} = 0$.

Step 2: Consider the case where $c_{\beta} \neq 0$

If $c_{\beta} \neq 0$, then we can solve for $\beta$ in terms of the other vectors in the set $S$:

$\beta = - \frac{c_1}{c_{\beta}} v_1 - \frac{c_2}{c_{\beta}} v_2 - \dots - \frac{c_k}{c_{\beta}} v_k$

This means that $\beta$ is expressed as a linear combination of the vectors $v_1, v_2, \dots, v_k$. However, this contradicts the assumption that $\beta$ is not spanned by $S$, since we just found a linear combination of vectors in $S$ that equals $\beta$. Therefore, we must have $c_{\beta} = 0$.

Step 3: Conclusion from $c_{\beta} = 0$

Now, we are left with the equation

$c_1 v_1 + c_2 v_2 + \dots + c_k v_k = 0$

Since $S = \{ v_1, v_2, \dots, v_k \}$ is linearly independent by assumption, the only solution to this equation is

$c_1 = c_2 = \dots = c_k = 0$

Final Conclusion:

Thus, the only solution to the equation

$c_1 v_1 + c_2 v_2 + \dots + c_k v_k + c_{\beta} \beta = 0$

is $c_1 = c_2 = \dots = c_k = c_{\beta} = 0$. Therefore, the set $S \cup \{ \beta \}$ is linearly independent.

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Would you like further clarification or more details on any part of the proof?

Here are some related questions to help deepen your understanding:

1. What is the definition of linear independence and how is it applied in vector spaces?
2. How do you prove that a set of vectors is linearly independent?
3. Can a linearly independent set become linearly dependent if you add a new vector from the span of the set?
4. How does the concept of a basis relate to linear independence?
5. What is the difference between a linearly independent and a linearly dependent set?

Tip: When adding a vector to a linearly independent set, check whether the new vector can be written as a linear combination of the existing vectors. If it can, the set becomes dependent.