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# 9.2: Building Subspaces

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Consider the set

$U= \left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix} \right\} \subset \Re^{3}.$

Because $$U$$ consists of only two vectors, it clear that $$U$$ is $$\textit{not}$$ a vector space, since any constant multiple of these vectors should also be in $$U$$. For example, the $$0$$-vector is not in $$U$$, nor is $$U$$ closed under vector addition.

But we know that any two vectors define a plane: In this case, the vectors in $$U$$ define the $$xy$$-plane in $$\Re^{3}$$. We can view the $$xy$$-plane as the set of all vectors that arise as a linear combination of the two vectors in $$U$$. We call this set of all linear combinations the $$\textit{span}$$ of $$U$$:

$span(U)=\left\{ x \begin{pmatrix}1\\0\\0\end{pmatrix}+y \begin{pmatrix}0\\1\\0\end{pmatrix} \middle| x,y\in \Re \right\}.$

Notice that any vector in the $$xy$$-plane is of the form

$\begin{pmatrix}x\\y\\0\end{pmatrix} = x \begin{pmatrix}1\\0\\0\end{pmatrix}+y \begin{pmatrix}0\\1\\0\end{pmatrix} \in span(U).$

Definition: Span

Let $$V$$ be a vector space and $$S=\{ s_{1}, s_{2}, \ldots \} \subset V$$ a subset of $$V$$. Then the $$\textit{span of S}$$ is the set:

$span(S):=\{ r^{1}s_{1}+r^{2}s_{2}+\cdots + r^{N}s_{N} | r^{i}\in \Re, N\in \mathbb{N} \}. \label{span}$

That is, the span of $$S$$ is the set of all finite linear combinations (usually our vector spaces are defined over $$\mathbb{R}$$, but in general we can have vector spaces defined over different base fields such as $$\mathbb{C}$$ or $$\mathbb{Z}_{2}$$. The coefficients $$r^{i}$$ should come from whatever our base field is (usually $$\mathbb{R}$$).} of elements of $$S$$. Any $$\textit{finite}$$ sum of the form "a constant times $$s_{1}$$ plus a constant times $$s_{2}$$ plus a constant times $$s_{3}$$ and so on'' is in the span of $$S$$.

It is important that we only allow finite linear combinations. In Equation \ref{span}, $$N$$ must be a finite number. It can be any finite number, but it must be finite.

Example 101

Let $$V=\Re^{3}$$ and $$X\subset V$$ be the $$x$$-axis. Let $$P=\begin{pmatrix}0\\1\\0\end{pmatrix}$$, and set

$S=X \cup \{P\}\, .$

The vector $$\begin{pmatrix}2 \\ 3 \\ 0\end{pmatrix}$$ is in $$span(S),$$ because $$\begin{pmatrix}2\\3\\0\end{pmatrix}=\begin{pmatrix}2\\0\\0\end{pmatrix}+3\begin{pmatrix}0\\1\\0\end{pmatrix}.$$ Similarly, the vector $$\begin{pmatrix}-12 \\ 17.5 \\ 0\end{pmatrix}$$ is in $$span(S),$$ because $$\begin{pmatrix}-12\\17.5\\0\end{pmatrix}=\begin{pmatrix}-12\\0\\0\end{pmatrix}+17.5\begin{pmatrix}0\\1\\0\end{pmatrix}.$$
Similarly, any vector of the form

$\begin{pmatrix}x\\0\\0\end{pmatrix}+y \begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}x\\y\\0\end{pmatrix}$

is in $$span(S)$$. On the other hand, any vector in $$span(S)$$ must have a zero in the $$z$$-coordinate. (Why?)

So $$span(S)$$ is the $$xy$$-plane, which is a vector space. (Try drawing a picture to verify this!)

Lemma

For any subset $$S\subset V$$, $$span(S)$$ is a subspace of $$V$$.

Proof

We need to show that $$span(S)$$ is a vector space.

It suffices to show that $$span(S)$$ is closed under linear combinations. Let $$u,v\in span(S)$$ and $$\lambda, \mu$$ be constants. By the definition of $$span(S)$$, there are constants $$c^{i}$$ and $$d^{i}$$ (some of which could be zero) such that:

\begin{eqnarray*}
u & = & c^{1}s_{1}+c^{2}s_{2}+\cdots \\
v & = & d^{1}s_{1}+d^{2}s_{2}+\cdots \\
\Rightarrow \lambda u + \mu v & = & \lambda (c^{1}s_{1}+c^{2}s_{2}+\cdots ) + \mu (d^{1}s_{1}+d^{2}s_{2}+\cdots ) \\
& = & (\lambda c^{1}+\mu d^{1})s_{1} + (\lambda c^{2}+\mu d^{2})s_{2} + \cdots
\end{eqnarray*}

This last sum is a linear combination of elements of $$S$$, and is thus in $$span(S)$$. Then $$span(S)$$ is closed under linear combinations, and is thus a subspace of $$V$$.

Note that this proof, like many proofs, consisted of little more than just writing out the definitions.

Example 102

For which values of $$a$$ does

$span \left\{ \begin{pmatrix}1\\0\\a\end{pmatrix} , \begin{pmatrix}1\\2\\-3\end{pmatrix} , \begin{pmatrix}a\\1\\0\end{pmatrix} \right\} = \Re^{3}?$

Given an arbitrary vector $$\begin{pmatrix}x\\y\\z\end{pmatrix}$$ in $$\Re^{3}$$, we need to find constants $$r^{1}, r^{2}, r^{3}$$ such that

$r^{1} \begin{pmatrix}1\\0\\a\end{pmatrix} + r^{2}\begin{pmatrix}1\\2\\-3\end{pmatrix} +r^{3} \begin{pmatrix}a\\1\\0\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}.$

We can write this as a linear system in the unknowns $$r^{1}, r^{2}, r^{3}$$ as follows:

$\begin{pmatrix} 1 & 1 & a \\ 0 & 2 & 1 \\ a & -3 & 0 \end{pmatrix} \begin{pmatrix}r^{1}\\r^{2}\\r^{3}\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}.$

If the matrix

$$M=\begin{pmatrix} 1 & 1 & a \\ 0 & 2 & 1 \\ a & -3 & 0 \end{pmatrix}$$ is invertible, then we can find a solution

$M^{-1}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}r^{1}\\r^{2}\\r^{3}\end{pmatrix}$ for $$\textit{any}$$ vector $$\begin{pmatrix}x\\y\\z\end{pmatrix} \in \Re^{3}$$.

Therefore we should choose $$a$$ so that $$M$$ is invertible:

$i.e.,\; 0 \neq \det M = -2a^{2} + 3 + a = -(2a-3)(a+1).$

Then the span is $$\Re^{3}$$ if and only if $$a \neq -1, \frac{3}{2}$$.

Some other very important ways of building subspaces are given in the following examples.

Example 103: The kernel of a linear map

Suppose $$L:U\to V$$ is a linear map between vector spaces. Then if

$L(u)=0=L(u')\, ,$

linearity tells us that

$L(\alpha u + \beta u') = \alpha L(u) + \beta L(u') =\alpha 0 + \beta 0 = 0\, .$

Hence, thanks to the subspace theorem, the set of all vectors in $$U$$ that are mapped to the zero vector is a subspace of $$V$$.

It is called the kernel of $$L$$:

${\rm ker} L:=\{u\in U| L(u) = 0\}\subset U.$

Note that finding kernels is a homogeneous linear systems problem.

Example 104: The image of a linear map

Suppose $$L:U\to V$$ is a linear map between vector spaces. Then if

$v=L(u) \mbox{ and } v'=L(u')\, ,$

linearity tells us that

$\alpha v + \beta v' = \alpha L(u) + \beta L(u') =L(\alpha u +\beta u')\, .$

Hence, calling once again on the subspace theorem, the set of all vectors in $$V$$ that are obtained as outputs of the map $$L$$ is a subspace.

It is called the image of $$L$$:

${\rm im} L:=\{L(u)\ u\in U \}\subset V.$

Example 105: An eigenspace of a linear map

Suppose $$L:V\to V$$ is a linear map and $$V$$ is a vector space. Then if

$L(u)=\lambda u \mbox{ and } L(v)=\lambda v\, ,$

linearity tells us that

$L(\alpha u + \beta v) = \alpha L(u) + \beta L(v) =\alpha L(u) + \beta L(v) =\alpha \lambda u + \beta \lambda v = \lambda (\alpha u + \beta v)\, .$

Hence, again by subspace theorem, the set of all vectors in $$V$$ that obey the $$\textit{eigenvector equation}$$ $$L(v)=\lambda v$$ is a subspace of $$V$$. It is called an eigenspace

$V_{\lambda}:=\{v\in V| L(v) = \lambda v\}.$

For most scalars $$\lambda$$, the only solution to $$L(v) = \lambda v$$ will be $$v=0$$, which yields the trivial subspace $$\{0\}$$.

When there are nontrivial solutions to $$L(v)=\lambda v$$, the number $$\lambda$$ is called an eigenvalue, and carries essential information about the map $$L$$.

Kernels, images and eigenspaces are discussed in great depth in chapters 16 and 12.