The statement you're describing can be understood and proven by using the properties of linear maps and the basis of vector spaces. Here's the explanation and the proof in a structured way:

Problem Setup:

1. Finite-dimensional vector space $V$:

   • The vector space $V$ is finite-dimensional over the field $K$.
   • The vectors $v_1, v_2, \dots, v_n$ form a basis for $V$.

2. Vector space $W$:

   • $W$ is another vector space over the same field $K$.
   • The vectors $w_1, w_2, \dots, w_n$ are arbitrary vectors in $W$.

3. Objective:

   • We want to prove the existence of a unique linear map $f: V \to W$ such that: $f(v_i) = w_i \quad \text{for all} \quad i = 1, 2, \dots, n.$

Existence and Uniqueness of the Linear Map:

Existence:

• Let $\{ v_1, v_2, \dots, v_n \}$ be a basis for the vector space $V$.

• Any vector $v \in V$ can be written uniquely as a linear combination of the basis vectors: $v = a_1 v_1 + a_2 v_2 + \dots + a_n v_n,$ where $a_1, a_2, \dots, a_n$ are scalars from $K$.

• Now, define the map $f: V \to W$ on the basis vectors $\{ v_1, v_2, \dots, v_n \}$ as follows: $f(v_i) = w_i \quad \text{for all} \quad i = 1, 2, \dots, n.$

• For any arbitrary vector $v \in V$, given that $v = a_1 v_1 + a_2 v_2 + \dots + a_n v_n$, we define: $f(v) = f(a_1 v_1 + a_2 v_2 + \dots + a_n v_n) = a_1 f(v_1) + a_2 f(v_2) + \dots + a_n f(v_n).$ Using the fact that $f(v_i) = w_i$, we have: $f(v) = a_1 w_1 + a_2 w_2 + \dots + a_n w_n.$ Therefore, $f$ is a linear map because it respects both scalar multiplication and addition.

Uniqueness:

• Suppose there is another linear map $f' : V \to W$ that satisfies $f'(v_i) = w_i$ for all $i = 1, 2, \dots, n$.

• Consider any vector $v = a_1 v_1 + a_2 v_2 + \dots + a_n v_n \in V$. By linearity: $f'(v) = a_1 f'(v_1) + a_2 f'(v_2) + \dots + a_n f'(v_n) = a_1 w_1 + a_2 w_2 + \dots + a_n w_n.$ But this is exactly what $f(v)$ is by the definition of $f$. Hence, $f'(v) = f(v)$ for all $v \in V$.

• Therefore, $f' = f$, which proves that $f$ is the unique linear map that satisfies $f(v_i) = w_i$ for all $i$.

Conclusion:

There exists a unique linear map $f: V \to W$ such that $f(v_i) = w_i$ for all $i = 1, 2, \dots, n$.

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Would you like any further clarifications or details on the proof? Here are some related questions that might interest you:

1. How does the concept of a linear transformation work in higher-dimensional vector spaces?
2. What is the kernel of a linear map, and how can it be determined?
3. How do coordinate systems relate to bases in vector spaces?
4. What is the significance of the rank of a linear map in this context?
5. How does this proof extend to infinite-dimensional vector spaces, if at all?

Tip: To fully grasp linear maps, it helps to practice applying them to simple examples of vector spaces, such as $\mathbb{R}^2$ and $\mathbb{R}^3$, to see how the map operates on combinations of basis vectors.