Mathematical Immaturity

1.10 Exercises

• Let $V$ be a finite-dimensional linear space, and let $S$ be a subspace of $V.$ Prove each of the following statements.
  $\quad\text{(a)}\quad$ $S$ is finite dimensional and $\dim S \leq \dim V.$
  $\quad\text{(b)}\quad$ $\dim S = \dim V$ if and only if $S = V.$
  $\quad\text{(c)}\quad$ Every basis for $S$ is part of a basis for $V.$
  $\quad\text{(d)}\quad$ A basis for $V$ need not contain a basis for $S.$
• 
  $\text{(a)}\quad$ If $S$ is a subset of $V$ and $\dim V = n,$ then each element $x$ in $S$ can be represented as a finite linear combination of $n$ independent elements $v_1, \dots, v_n$ in $V.$ But since this is the case for all $x$ in $S,$ then $S$ is spanned by a set of no more than $n$ independent elements. In other words, $S$ is finite-dimensional with $\dim S \leq \dim V.\quad \blacksquare$
  
  $\text{(b)}\quad$ We proved in Theorem 1.7 (b) that any set of $n$ independent elements in $V$ is a basis for $V.$ Thus, if $\dim S = \dim V = n,$ then there exists a set of $n$ independent elements $s_1, \dots, s_n$ in $S$ that span $S.$ But, $S$ is a subset of $V,$ so this means that the same set of $n$ elements also spans $V.$ But if each element in $V$ can be represented as a finite linear combination of $s_1, \dots, s_n,$ then it must be that $V \subseteq S.$ And since $S \subseteq V,$ we conclude that $S = V.$
  
  If $S = V,$ then for each element $x$ in $V$ we have \begin{align*} x &= \sum_{i = 1}^{n}a_iv_i \end{align*} where $n$ is the dimension of $V$ and $v_1, \dots, v_n$ are $n$ independent elements of $V.$ But since $S = V,$ each $x$ is an element of $S,$ which means $S$ is spanned by the same basis $\{v_1, \dots, v_n\}.$ Hence, $\dim S = \dim V,$ and we have proven that $\dim S = \dim V$ if and only if $S = V.\quad \blacksquare$
  
  $\text{(c)}\quad$ Since $S$ is a subset of $V,$ then every basis for $S$ is set of independent elements in $V.$ But as we proved in Theorem 1.7 (a), any set of independent elements in $V$ is a subset of some basis for $V.$ Thus, every basis for $S$ is part of a basis for $V.\quad \blacksquare$
  
  $\text{(d)}\quad$ Suppose $V$ has basis $\{x, y\}$ and $S$ has basis $\{x + y\}.$ As we can see, $S$ is a subspace of $V,$ but the basis for $V$ does not contain the basis for $S.\quad \blacksquare$