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 $\Fun(S, V)$.

 
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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.

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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 $(f+g)+h = f+(g+h)$.

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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 $f+g = g+f$.

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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 $f + \mathbf{0} = f$.

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

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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 $(cd)(f) = c (df)$.

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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 $1f = f$.

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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.

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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.

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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 $T$.

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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 $\Fun(S, V)$.

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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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