# Variables used to define functions in finite dimensional vector spcaes

Let's say I have a finite one-dimensional vector space $$V$$ with basis $$\{\alpha\}$$. Then the product vector space $$V^k$$ has basis $$\{e_1, \dots, e_k\}$$ where $$e_i = (0, \dots, \alpha, \dots, 0)$$ where $$\alpha$$ is in the $$i$$-th position in the $$k$$-tuple.

Let's say I have a function $$f : V^k \to V$$ defined in some way $$f(x_1, \dots, x_k) = \dots$$

Now the $$x_1, \dots x_k$$ are variables above, and are just placeholders for values really. Also technically $$f(x_1, \dots, x_k)$$ is shorthand for $$f((x_1, \dots, x_k))$$. That being said is it correct to view $$(x_1, \dots, x_k)$$ as $$x_1\cdot e_1 + x_2\cdot e_2 + \dots + x_k\cdot e_k$$ or as $$(x_1 \alpha, \dots, x_k\alpha)$$?

Is it thus correct to think of $$f(x_1, \dots, x_k)$$ as $$f(x_1\cdot e_1 + x_2\cdot e_2 + \dots + x_k\cdot e_k)$$ or as $$f(x_1\alpha, \dots, x_k\alpha)$$?

• I cannot make sense out of the equation $e_i=(0,0,...,\alpha,...,0)$ where $\{\alpha\}$ is a basis. – Kavi Rama Murthy Jan 15 at 5:48
• @KaviRamaMurthy I think the OP means that $\{\alpha\}$ is a basis of $V$, so a basis $\{e_i\}_{i=1}^{k}$ of $V^k$ can be constructed by placing $\alpha$ in the $i^\text{th}$ entry, and having the zero element of $V$ as the others. – user445909 Jan 15 at 9:16

## 1 Answer

Suppose $$V$$ is a vector space over a field $$\mathbb{F}$$.

Since the domain of $$f$$ is $$V^k$$, each entry of the $$k$$-tuple $$(x_1,\dots,x_k)$$ seen in $$f(x_1,\dots,x_k)$$ is a vector in $$V$$. Otherwise, $$f(x_1,\dots, x_k)$$ is not defined. Thus, for every $$i \in \{1,\dots , k \}$$, the product $$x_i \alpha$$ is between two members of $$V$$, which is not a standard operation on a vector space. Since this operation is undefined, we cannot make sense of $$f(x_1 \alpha, \dots, x_k \alpha)$$ as the situation stands.

However, if we define the coordinate map $$C: \mathbb{F}^k \to V^k$$, where for every $$(x_1,\dots,x_n) \in \mathbb{F}^k,$$ we define $$C(x_1,\dots,x_n) := (x_1 \alpha, \dots, x_n \alpha)$$, then we could write \begin{align} (f\circ C)(x_1,\dots,x_n) &= f(x_1 \alpha,\dots,x_n \alpha) \\ &= f(x_1 e_1 + \dots + x_n e_n), \end{align} under the usual definition of entry-wise vector addition and scalar multiplication in $$V^k$$.

In short, in order to make sense of $$f(x_1,\dots,x_k)$$ as you expressed it, each $$x_i$$ should be thought of as a scalar, not a vector. But the domain of $$f$$ is $$V^k$$, so we cannot do this unless we define a coordinate map.