Mathematical Immaturity

1.10 Exercises

• In this exercise, $L(S)$ denotes the subspace spanned by a subset $S$ of a linear space $V.$ Prove each of the statements (a) through (f).
  $\quad\text{(a)}\quad$ $S \subseteq L(S).$
  $\quad\text{(b)}\quad$ If $S \subseteq T \subseteq V$ and if $T$ is a subspace of $V,$ then $L(S) \subseteq T.$ This property is described by saying that $L(S)$ is the smallest subspace of $V$ which contains $S.$
  $\quad\text{(c)}\quad$ A subset $S$ of $V$ is a subspace of $V$ if and only if $L(S) = S.$
  $\quad\text{(d)}\quad$ If $S \subseteq T \subseteq V,$ then $L(S) \subseteq L(T).$
  $\quad\text{(e)}\quad$ If $S$ and $T$ are subspaces of $V,$ then so is $S \cap T.$
  $\quad\text{(f)}\quad$ If $S$ and $T$ are subsets of $V,$ then $L(S \cap T) \subseteq L(S) \cap L(T).$
  $\quad\text{(g)}\quad$ Give an example in which $L(S \cap T) \neq L(S) \cap L(T).$
• Recall from Section 1.6, Theorem 1.4, that a nonempty subset $T$ of a linear space $V$ is a subspace if and only if it satisfies the closure axioms.
• 
  $\text{(a)}\quad$ Let $S = \{s_1, \dots, s_n\}$ be a subset of $n$ elements of $V,$ and for $i = 1, \dots, n,$ let let $a_i$ be a coefficient of the $i^{th}$ element in the sum \begin{align*} \sum_{i = 1}^n a_is_i \end{align*} Then, $L(S)$ is the set of all such sums for real coefficients $a_i.$ Now, suppose that for some $k$ in $\{1, \dots, n\},$ we set $a_k = 1$ and the remaining coefficients to zero. Then, we know that the sum \begin{align*} \sum_{i = 1}^n a_is_i &= s_k \end{align*} is an element of $L(S).$ But if we do this for each $k$ in $\{1, ..., n\},$ then we prove that every element of $S$ is an element of $L(S)$ and thus $S \subseteq L(S).\quad\blacksquare$
  
  $\text{(b)}\quad$ By definition, $S$ is a subset of $T,$ so every element of $S$ is an element of $T.$ But, $T$ is a subspace of $V.$ Therefore, as we proved in Theorem 1.4, $T$ satisfies closure under multiplication by real numbers. So if $a_1$ is some real number and $s_1$ is an element of $S.$ This means that $a_1s_1$ is an element of $T.$ And likewise for real number $a_2$ and element $s_2$ in $S.$ But because $T$ is a subspace of $V,$ it satisfies closure under addition. In other words, the sum $$a_1s_1 + a_2s_2$$ is also a member of $T.$ Carrying out this process for $i = 1, \dots, n$ (where $n$ is the number of elements in $S$), we find that the sum \begin{align*} a_1s_1 + \dots + a_ns_n &= \sum_{i=1}^n a_is_i \end{align*} is an element of $T.$ But if each $a_i$ is an arbitrary real coefficient, and each $s_i$ is an element of $S,$ then the above sum represents the set of all linear combinations of the elements of $S.$ And if every such linear combination is an element of $T,$ then $L(S) \subseteq T.\quad\blacksquare$
  
  $\text{(c)}\quad$ If $S$ is a subspace of $V,$ then it satisfies closure under addition and closure under multiplication by real numbers. And as shown in part (b), this implies that for $i = 1, \dots, n$ (where $n$ is the number of elements in $S$), if each $a_i$ is a real coefficient and each $s_i$ is an element of $S,$ the sum \begin{align*} \sum_{i=1}^na_is_i \end{align*} is also an element of $S.$ But since the choice of each $a_i$ is arbitrary, it means that $S$ contains the set of all linear combinations of the elements of $S.$ In other words, $L(S) \subseteq S.$ And as we proved in part (a), $S \subseteq L(S).$ Hence, $L(S) = S.$ To prove that $L(S) = S$ implies that $S$ is a subspace of $V,$ note that if $L(S) \subseteq S,$ then it implies that $S$ satisfies the closure axioms. And as proven in Theorem 1.4, this means that $S$ is a subspace of $V.\quad\blacksquare$
  
  $\text{(d)}\quad$ Let $n$ be the number of elements in $T.$ We can rearrange the elements of $T$ such that for some $k \leq n,$ we have $T = \{t_1, \dots, t_n\}$ and $S = \{t_1, \dots, t_k\}.$ Then, $L(T)$ is the set of all $x$ in $V$ of the form \begin{align*} x &= \sum_{i=1}^n a_i t_i \end{align*} where $a_1, \dots, a_n$ are scalars. Now, suppose we take the subset of elements in $T$ whose coefficients $a_{k+1},\dots,a_n$ are zero. In other words, the subset of $x'$ in $L(T)$ of the form \begin{align*} x' &= \sum_{i=1}^k a_i t_i \end{align*} As we can see, each $x'$ is a finite linear combination of the elements of $S,$ and the set of all $x'$ is $L(S).$ But we know that each $x'$ is an element of $L(T),$ hence $L(S) \subseteq L(T).\quad\blacksquare$
  
  $\text{(e)}\quad$ Let $x$ and $y$ be elements of $S \cap T$ and let $c$ be a scalar. Because $S$ and $T$ are both subspaces of $V,$ they satisfy the closure axioms. As such, $x + y$ and $cx$ are elements of both $S$ and $T.$ But since this is the case for all elements of $S \cap T,$ it implies that $S \cap T$ satisfies the closure axioms and is thus a subspace of $V. \quad \blacksquare$
  
  $\text{(f)}\quad$ If $S \cap T$ has no elements, then $L(S \cap T) = \{O\}$ which is trivially a subset of $L(S) \cap L(T).$ Suppose then, that $X = \{x_1, \dots, x_k\}$ is the intersection of $S$ and $T$ containing $k \geq 1$ elements. Each element in its linear span is of the form: \begin{align*} x' &= \sum_{i = 1}^k a_ix_i \end{align*} where $a_1, \dots, a_k$ are scalars. If there are no elements in $S$ but not in $S \cap T,$ then the set of all $x'$ is $L(S).$ Otherwise, we can add the following sum to each $x'$ \begin{align*} s' &= \sum_{i = 1}^m b_i s_i \end{align*} where $b_1, \dots, b_m$ are scalars and $s_1, \dots, s_m$ are the $m \geq 1$ elements of $S - T.$ The the resulting sum is a finite linear combination of the elements of $S,$ the set of which is $L(S).$ If we set $b_1, \dots, b_m$ to zero, we can see that $x'$ is an element of $L(S),$ thus $L(S \cap T) \subseteq L(S).$ Similarly, if $T$ has no elements that are not in $S \cap T,$ then $L(T)$ is the set of all finite combinations $x'$ as defined above. Otherwise, we add the following sum to each $x'$ \begin{align*} t' &= \sum_{i = 1}^n c_i t_i \end{align*} where $c_1, \dots, c_n$ are scalars and $t_1, \dots, t_n$ are the $n \geq 1$ elements of $T - S.$ The resulting sum is a finite linear combination of the elements of $T,$ the set of which is $L(T),$ If we set $c_1, \dots, c_n$ to zero, we can see that $x'$ is an element of $L(T),$ thus $L(S \cap T) \subseteq L(T).$ Combining these results, we see that $L(S \cap T) \subseteq L(S) \cap L(T).\quad\blacksquare$
  
  $\text{(g)}\quad$ Let $S = \{1 +x\}$ and $T = \{1, x\}.$ Then, $L(S \cap T) = \{O\}$ but the element $1 + x$ is in $L(S) \cap L(T),$ thus $L(S \cap T) \neq L(S) \cap L(T).\quad \blacksquare$