# Vector Space of Functions from a Set to a Vector Space

## Problem 705

For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$

\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]

For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by

\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]

**(a)** Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

**(b)** Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

**(c)** Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.

**(d)** Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as

\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]

**(e)** Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

**(f)** Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

Contents

- Problem 705
- Proof.
- (a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
- (b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
- (c) Find a basis of $\Fun(S_1 , V)$.
- (d) Construct a linear isomorphism between $\Fun(S, V)$ and the vector space $V^m$
- (e) Constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
- (f) Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

## Proof.

### (a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

We will prove that each vector space axioms holds for $\Fun(S, V)$.

**Closed under addition and scalar multiplication**: For $f , g \in \Fun(S, V)$ and $c, d \in \K$, $cf+dg$ is another function with domain $S$ defined by

\[ (cf+dg)(s) = cf(s) + dg(s) \in V \, \mbox{ for all } s \in S . \]
Thus $cf+dg \in \Fun(S, V)$. This proves that $\Fun(S, V)$ is closed under addition and scalar multiplication.

**Addition is associative**: For $f, g, h \in \Fun(S, V)$ and $s \in S$ we have

\begin{align*}

((f+g)+h)(s) &= (f+g)(s) + h(s) \\

&= ( f(s) + g(s) ) + h(s) \\

&= f(s) + ( g(s) + h(s) ) \\

&= f(s) + (g+h)(s) \\

&= (f+(g+h))(s) . \end{align*}

Because $((f+g)+h)(s) = (f+(g+h))(s)$ holds for each $s \in S$, we can say that $(f+g)+h = f+(g+h)$.

**Addition is commutative**: Notice that for $s \in S$ we have

\[ (f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s) . \]
Because this holds for every $s \in S$, we can say that $f+g = g+f$.

**Zero vector**: The zero vector is the function $\mathbf{0}$, defined by $\mathbf{0}(s) = 0 \in V$ for each $s \in S$. To check that $\mathbf{0}$ is the additive identity, we check for any $f \in \Fun(S, V)$ and $s \in S$,

\[ (f+ \mathbf{0})(s) = f(s) + \mathbf{0}(s) = f(s) + 0 = f(s) . \]
Because this holds for each $s \in S$, we can say that $f + \mathbf{0} = f$.

**Inverse vectors**: For $f \in \Fun(S, V)$, its additive inverse is the function $(-f)$ defined by

\[ (-f)(s) = – f(s) \, \mbox{ for all } s \in S . \]
To check that $-f$ is the inverse of $f$, we check for all $s \in S$

\[ (f + (-f))(s) = f(s) + (-f)(s) = f(s) – f(s) = 0 = \mathbf{0}(s) . \]

**Scalar multiplication is associative**: For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$ we have

\[ ((cd)(f))(s) = (cd) f(s) = c ( d f(s) ) = c ( df)(s) = (c (df) )(s) . \]
Because this holds for each $s \in S$, we conclude that $(cd)(f) = c (df)$.

**Scalar identity element**: Let $1 \in \K$ be the multiplicative identity. Then for $f \in \Fun(S, V)$ and $s \in S$ we have

\[ (1f)(s) = 1 f(s) = f(s) . \]
Because this holds for all $s \in S$, we conclude that $1f = f$.

**Distributivity**: Next we check distributivity. For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$, we have

\[ ( (c+d)(f) )(s) = (c+d) f(s) = c f(s) + d f(s) = (cf + df)(s) . \]
Because this holds for all $s \in S$, we can conclude that

\[ (c+d)f = cf + df . \]

For $c \in \K$, $f, g \in \Fun(S, V)$ and $s \in S$ we have

\begin{align*}

( c(f+g))(s) &= c ( (f+g)(s) ) \\

&= c ( f(s) + g(s) ) \\

&= cf(s) + cg(s) \\

&= (cf + cg)(s) . \end{align*}

Because this holds for all $s \in S$, we conclude that

\[ c(f+g) = cf + cg . \]

We have shown that $\Fun(S, V)$ satisfies all of the vector space axioms, so it is a vector space.

### (b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

Define the map $T : \Fun(S_1 , V) \rightarrow V$ by

\[ T(f) = f(s) \, \mbox{ for all } f \in \Fun(S_1 , V) . \]

We check that $T$ is a linear map. For $f, g \in \Fun(S_1, V)$ we have

\[ T(f+g) = (f+g)(s) = f(s) + g(s) = T(f) + T(g) . \]
Thus $T$ is additive. Next, for $c \in \K$ and $f \in \Fun(S_1, V)$ we have

\[ T(cf) = (cf)(s) = c f(s) = c T(f) . \]
This shows that $T$ respects scalar multiplication, and so $T$ is a linear map.

Next we prove that $T$ is an isomorphism. To see that $T$ is surjective, pick any $v \in V$. Define the map $f_v \in \Fun(S_1, V)$ by $f_v(s) = v$. Then $T(f_v) = f_v(s) = v$, and so $v$ lies in the range of $T$.

To see that $T$ is injective, it is enough to show that if $T(f) = 0 \in V$ then $f = \mathbf{0} \in \Fun(S_1, V)$. So suppose that $T(f) = 0$. Because $T(f) = f(s)$, this implies that $f(s) = 0$. But $\mathbf{0}(s) = 0$ as well. Because $f(s) = \mathbf{0}(s)$ for each $s \in S_1$ (which, remember, contains only the one element), we can conclude that $f = \mathbf{0}$.

We have shown that $T$ is linear, injective, and surjective. This finishes the proof that $T$ is a linear isomorphism.

### (c) Find a basis of $\Fun(S_1 , V)$.

We can use the basis $B$ of $V$ and the linear isomorphism $T$ found in the previous part to define a basis of $\Fun(S_1, V)$. More specifically, the image of $B$ under the inverse isomorphism

\[ T^{-1} : V \rightarrow \Fun(S_1 , V) \]
will define a basis of $\Fun(S_1 , V)$.

For $v \in V$ define the function $f_v$ by $f_v(s) = v$. It is clear that $T(f_v) = v$, and so $T^{-1}(v) = f_v$ for all $v \in V$. In particular, $T^{-1}(e_i) = f_{e_i}$ for each $1 \leq i \leq n$, and so the set $\{ f_{e_1} , f_{e_2} , \cdots , f_{e_n} \}$ forms a basis of $\Fun(S, V)$.

### (d) Construct a linear isomorphism between $\Fun(S, V)$ and the vector space $V^m$

Define the map $T : \Fun(S, V) \rightarrow V^m$ by

\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) \in V^m \, \mbox{ for all } f \in \Fun(S, V) . \]

First we show that $T$ is linear. For $f, g \in \Fun(S, V)$ and $c, d \in \K$ we have

\begin{align*}

T(cf+dg) &= \left( (cf + dg)(s_1) , (cf + dg)(s_2) , \cdots , (cf + dg)(s_m) \right) \\

&= \left( (cf)(s_1) + (dg)(s_1) , (cf)(s_2) + (dg)(s_2) , \cdots , (cf)(s_m) + (dg)(s_m) \right) \\

&= \left( c f(s_1) , c f(s_2) , \cdots , c f(s_m) \right) + \left( d g(s_1) , d g(s_2) , \cdots , d g(s_m) \right) \\

&= c \left( f(s_1) , f(s_2) , \cdots , f(s_m) \right) + d \left( g(s_1) , g(s_2) , \cdots , g(s_m) \right) \\

&= c T(f) + d T(g) . \end{align*}

This proves that $T$ is a linear map.

Next we show that it is an isomorphism.

First we show surjectivity. For $(v_1 , v_2 , \cdots , v_m) \in V^m$ define the function $f$ by

\[ f(s_i) = v_i \, \mbox{ for } 1 \leq i \leq m . \]

Then

\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) = ( v_1 , v_2 , \cdots , v_m ) . \]
Thus every element of $V^m$ lies in the range of $T$.

Next we check injectivity. Suppose that $T(f) = (0 , 0 , \cdots , 0) \in V^m$. This means that $f(s_i) = 0$ for each $1 \leq i \leq m$. Thus $f(s) = \mathbf{0}(s)$ for all $s \in S$, and so $f = \mathbf{0}$. This proves that $T$ is injective. Together with surjectivity and linearity, we have proven that $T$ is an isomorphism.

### (e) Constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

We will use the basis $B = \{ e_1 , \cdots , e_n \}$ of $V$ to create a basis for $V^m$. We will translate this basis into a basis of $\Fun(S, V)$ using the isomorphism $T$, or actually its inverse $T^{-1}$.

For integers $i, j$ with $1 \leq i \leq n$ and $1 \leq j \leq m$, define the vector

\[ d_i^j = ( 0 , \cdots , e_i , \cdots , 0 ) . \]
Notice that every component is $0$ except for the $j$-th component, which is the vector $e_i$. The set $\{ d_i^j \}_{ \substack{1 \leq i \leq n \\ 1 \leq j \leq m} }$ is the standard basis of $V^m$ defined by $B$. The set $\{ T^{-1} ( d_i^j ) \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m}}$ will then be a basis of $\Fun(S, V)$.

To find $T^{-1}$, suppose that $T(f) = d_i^j$. Because the $j$-th component of $d_i^j$ is $e_i$, we must have that $f(s_j) = e_i$. Because all of the other components are $0$, we must have $f(s_k) = 0$ for all $k \neq j$. This information completely determines the function $f$. Define the function $\delta_i^j$ by

\[ \delta_i^j ( s_k ) = \left\{ \begin{array}{cl} e_i & \mbox{ if } k = j \\ 0 & \mbox{ if } k \neq j \end{array} \right. . \]

Then it is clear that $T(\delta_i^j) = d_i^j$; that is, $T^{-1}(d_i^j) = \delta_i^j$. Thus a basis of $\Fun(S, V)$ is formed by the set $\{ \delta_i^j \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m }}$. There are exactly $nm$ elements in this basis, thus $\Fun(S, V)$ has dimension $nm$.

### (f) Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

We will show in one step that $\Fun(S, W)$ is closed under addition and scalar multiplication. For $f, g \in \Fun(S, W)$, $c, d \in \K$ and $s \in S$ we have

\[ (cf + dg)(s) = c f(s) + d g(s) \in W . \]
This proves that $(cf + dg) : S \rightarrow W$, and so $(cf + dg) \in \Fun(S, W)$.

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