Section 2.2 Constructing Subspaces
Given a vector space
\(V\) over a field
\(F\text{,}\) recall what it means for a subset
\(W\) of
\(V\) to be a
subspace. While it is not hard to check whether or not a subset of a vector space is a subspace, it can be a bit subtle at first blush.
Example 2.2.1. Is every line or plane a subspace of \(\R^3?\)
The answer is no, and it is not hard to see, but it will take a while before we understand the significance. It is true that every line or plane containing the origin is a subspace of \(\R^3,\) and except for the addition of the zero subspace and all of \(\R^3,\) these are all the subspaces of \(\R^3.\)
So we can immediately exclude lines or planes that do not pass through the original simply because they fail to have the additive identity in the set. But this sounds awfully picky, doesn’t it? Actually is it not; without the origin in the set, everything goes wrong.
For example consider the plane
\begin{equation*}
W = \{(x,y,z)\in \R^3\mid
x+y+z=1\}.
\end{equation*}
Not only is \((0,0,0)\notin W\text{,}\) but it is also not closed under addition or scalar multiplication:
\begin{align*}
\amp(1,0,0), (0,1,0) \in W \text{ but } (1,1,0) \notin
W,\\
\amp (1,0,0) \in W \text{ but } \lambda (1,0,0) \notin W
\text{ unless } \lambda=1.
\end{align*}
The next example may bother you at first, but linear algebra may be the first course in which being mathematically precise is essential. We shall discuss the possible misconceptions, and this will lead us to a more sophisticated notion, that of an isomorphism.
Example 2.2.2. Is \(\R^2\) a subspace of \(\R^3?\)
The answer is no, and the reason is simple, but let’s start with some false reasoning, and then see how resolving our mistake leads to interesting ideas.
The reasoning starts with the correct statement that both \(\R^2\) and \(\R^3\) are vector spaces over \(\R.\) Where false reasoning intrudes is the claim that \(\R^2
\subseteq \R^3.\)
You may protest! The \(xy\)-plane is a subspace of \(\R^3!\) And I would agree, but \(\R^2\) is not. Why? Simply because \(\R^2\) consists of ordered pairs while \(\R^3\) consists of ordered triples; pairs are not triples.
But how does that help with the \(xy\)-plane? The \(xy\)-plane (in \(\R^3\)) is the set
\begin{equation*}
W = \{ (x,y,z)\in \R^3 \mid z=0\}.
\end{equation*}
At least \(W \subset \R^3\text{,}\) and we check the closure axioms easily.
Similarly, we see that the \(yz\) and \(xz\)-planes are subspaces of \(\R^3.\) Indeed each of these subspaces is an exact replica of \(\R^2.\) One might go so far as to define a map to justify this, for example: \(T: \R^2 \to W\subset \R^3\) by \(T((x,y)) = (x,y,0).\)
Do you think you could define a map from \(\R^2\) to any plane in \(\R^3\) (containing the origin)?
Having suggested we do need to be careful, let’s now recall some important, but familiar examples of a subspace of \(F^n\) for an integer \(n\ge 1.\) Let \(F\) be a field, say \(F=\Q, \R,\) or \(\C,\) and let \(A \in M_{m\times n}(F)\text{.}\) The rows of \(A\) are elements of \(F^n,\) while the columns are a subset of \(F^m.\) These sets are not themselves subspaces since they are not closed under vector addition and scalar multiplication.
We can make subspaces out of the rows and columns by creating the
row space (resp.
column space), the set of all linear combinations of the rows (resp. columns). Taking the
span of a set of vectors is one of the most common ways in which to construct a subspace of a vector space. The notion of span as well as of
linear independence are two fundamental notions in linear algebra that involve the construction of
linear combinations.
The following is an absolutely critical observation concerning spans.
Theorem 2.2.4.
Let \(V\) be a vector space and \(S,T\) two subsets of \(V.\)
\begin{align*}
\amp\text{If } S \subseteq \Span(T), \text{ then } \Span(S)
\subseteq \Span(T),\text{ and so,}\\
\amp\Span(S) = \Span(T) \iff
S\subseteq\Span(T)\text{ and }
T\subseteq\Span(S).
\end{align*}
Let’s consider some examples.
-
In \(V=\R^3\text{,}\) let \(S=\left\{\ba{r}1\\0\\0\ea,
\ba{r}0\\1\\0\ea,\ba{r}0\\0\\1\ea\right\}\text{,}\) and \(T=\left\{\ba{r}5\\0\\0\ea,
\ba{r}5\\6\\0\ea,\ba{r}5\\6\\7\ea\right\}\text{.}\) It is clear that \(V=\Span(S)\text{,}\) and so of course \(T \subset
\Span(S)\text{.}\) It is also easy to see that \(S \subset
\Span(T)\) by thinking algorithmicly.
Let’s observe that since \(\Span(T)\) is a subspace of \(V\text{,}\) it is closed under linear combinations, so that \(\ba{r}0\\6\\0\ea=\ba{r}5\\6\\0\ea - \ba{r}5\\0\\0\ea\) and \(\ba{r}0\\0\\7\ea=\ba{r}5\\6\\7\ea-\ba{r}5\\6\\0\ea\) are both in \(\Span(T)\text{.}\) Now it is easy to check that \(S\subset \Span(T)\text{,}\) so that \(V = \Span(S) =
\Span(T).\)
What if \(T=\left\{\ba{r}5\\0\\0\ea,
\ba{r}5\\6\\0\ea,\ba{r}5\\6\\7\ea,\ba{r}2\\3\\4\ea\right\}\text{.}\) Is is still true that \(S,T\) have the same spans?
Is \(T=\{2, 3+x, 4+5x+6x^2, 7+8x+9x^2+10x^3\}\) a spanning set for \(P_3(\Q)?\)
