To prove that the mapping $\varphi : K^n \rightarrow V$ defined by $\varphi(\lambda_1, \ldots, \lambda_n) = \lambda_1 v_1 + \ldots + \lambda_n v_n$ is surjective, we need to show that for every element $v \in V$, there exists a tuple $(\lambda_1, \ldots, \lambda_n) \in K^n$ such that $v = \lambda_1 v_1 + \ldots + \lambda_n v_n.$

Assumptions

Let $V$ be a vector space over a field $K$, and suppose $\{v_1, \ldots, v_n\}$ is a basis for $V$.

Proof

1. Basis Property: Since $\{v_1, \ldots, v_n\}$ is a basis for $V$, every vector $v \in V$ can be written as a linear combination of the basis vectors. This means that for any $v \in V$, there exist scalars $\lambda_1, \ldots, \lambda_n \in K$ such that $v = \lambda_1 v_1 + \ldots + \lambda_n v_n.$

2. Existence of Pre-image: By the definition of the map $\varphi$, the tuple $(\lambda_1, \ldots, \lambda_n) \in K^n$ is mapped to $v$ under $\varphi$. Since we can find such a tuple for any $v \in V$, we have shown that every vector in $V$ has a pre-image in $K^n$.

3. Conclusion: Therefore, $\varphi$ is surjective, as every element in $V$ can be expressed as an image of some element in $K^n$ under the map $\varphi$.

Thus, $\varphi$ is surjective.

Would you like a more detailed explanation, or do you have further questions?

Here are some related questions that expand upon this concept:

1. What are the implications if $\{v_1, \ldots, v_n\}$ is not a basis for $V$?
2. How does this proof change if $V$ has infinite dimension?
3. Can we show that $\varphi$ is also injective if $V$ has dimension $n$?
4. How would this proof differ if $K$ were not a field?
5. What are some practical applications of surjective linear maps in linear algebra?

Tip: In linear algebra, a mapping between vector spaces is surjective if every vector in the codomain has a corresponding vector in the domain that maps to it under the given transformation.