Section 2.4 Subspaces (V4)
Definition 2.4.1.
A subset of a vector space is called a subspace if it is a vector space on its own.
For example, the span of these two vectors forms a planar subspace inside of the larger vector space \(\IR^3\text{.}\)
Fact 2.4.3.
Any subset \(S\) of a vector space \(V\) that contains the additive identity \(\vec 0\) satisfies the eight vector space properties automatically, since it is a collection of known vectors.
However, to verify that it's a subspace, we need to check that addition and multiplication still make sense using only vectors from \(S\text{.}\) So we need to check two things:
The set is closed under addition: for any \(\vec{x},\vec{y} \in S\text{,}\) the sum \(\vec{x}+\vec{y}\) is also in \(S\text{.}\)
The set is closed under scalar multiplication: for any \(\vec{x} \in S\) and scalar \(c \in \IR\text{,}\) the product \(c\vec{x}\) is also in \(S\text{.}\)
Activity 2.4.4.
Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)
(a)
Let \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) and \(\vec{w} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] \) be vectors in \(S\text{,}\) so \(x+2y+z=0\) and \(a+2b+c=0\text{.}\) Show that \(\vec v+\vec w = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\) also belongs to \(S\) by verifying that \((x+a)+2(y+b)+(z+c)=0\text{.}\)
(b)
Let \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\in S\text{,}\) so \(x+2y+z=0\text{.}\) Show that \(c\vec v=\left[\begin{array}{c}cx\\cy\\cz\end{array}\right]\) also belongs to \(S\) for any \(c\in\IR\) by verifying an appropriate equation.
(c)
Is \(S\) is a subspace of \(\IR^3\text{?}\)
Activity 2.4.5.
Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\) Choose a vector \(\vec v=\left[\begin{array}{c} \unknown\\\unknown\\\unknown \end{array}\right]\) in \(S\) and a real number \(c=\unknown\text{,}\) and show that \(c\vec v\) isn't in \(S\text{.}\) Is \(S\) a subspace of \(\IR^3\text{?}\)
Remark 2.4.6.
Since \(0\) is a scalar and \(0\vec{v}=\vec{z}\) for any vector \(\vec{v}\text{,}\) a nonempty set that is closed under scalar multiplication must contain the zero vector \(\vec{z}\) for that vector space.
Put another way, you can check any of the following to show that a nonempty subset \(W\) isn't a subspace:
Show that \(\vec 0\not\in W\text{.}\)
Find \(\vec u,\vec v\in W\) such that \(\vec u+\vec v\not\in W\text{.}\)
Find \(c\in\IR,\vec v\in W\) such that \(c\vec v\not\in W\text{.}\)
If you cannot do any of these, then \(W\) can be proven to be a subspace by doing the following:
Prove that \(\vec u+\vec v\in W\) whenever \(\vec u,\vec v\in W\text{.}\)
Prove that \(c\vec v\in W\) whenever \(c\in\IR,\vec v\in W\text{.}\)
Activity 2.4.7.
Consider these subsets of \(\IR^3\text{:}\)
(a)
Show \(R\) isn't a subspace by showing that \(\vec 0\not\in R\text{.}\)
(b)
Show \(S\) isn't a subspace by finding two vectors \(\vec u,\vec v\in S\) such that \(\vec u+\vec v\not\in S\text{.}\)
(c)
Show \(T\) isn't a subspace by finding a vector \(\vec v\in T\) such that \(2\vec v\not\in T\text{.}\)
Activity 2.4.8.
Let \(W\) be a subspace of a vector space \(V\text{.}\) How are \(\vspan W\) and \(W\) related?
\(\vspan W\) is bigger than \(W\)
\(\vspan W\) is the same as \(W\)
\(\vspan W\) is smaller than \(W\)
Fact 2.4.9.
If \(S\) is any subset of a vector space \(V\text{,}\) then since \(\vspan S\) collects all possible linear combinations, \(\vspan S\) is automatically a subspace of \(V\text{.}\)
In fact, \(\vspan S\) is always the smallest subspace of \(V\) that contains all the vectors in \(S\text{.}\)
Subsection 2.4.1 Videos
Exercises 2.4.2 Exercises
Exercises available at checkit.clontz.org 1 .
https://checkit.clontz.org/#/banks/tbil-la/V4/