Learning Outcomes
- Determine if a subset of \(\IR^n\) is a subspace or not.
Section 2.4 Subspaces (VS4)
Subsection 2.4.1 Class Activities
Activity 2.4.1.
Consider two non-colinear vectors in \(\IR^3\text{.}\) If we look at all linear combinations of those two vectors (that is, their span), we end up with a plane within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)
(a)
Are all of the vectors in \(S\) also in \(\IR^3\text{?}\)
(b)
Let \(\vec{z}\) be the additive identity in \(\IR^3\text{.}\) Is \(\vec{z} \in S\text{?}\)
(c)
For any unspecified \(\vec{u}, \vec{v} \in S\text{,}\) is it the case that \(\vec{u} + \vec{v} \in S\text{?}\)
(d)
For any unspecified \(\vec{u} \in S\) and \(c\in\IR\text{,}\) is it the case that \(c\vec{u} \in S\text{?}\)
Definition 2.4.2.
A subset of a vector space is called a subspace if it is a vector space on its own. The operations of addition and scalar from the parent vector space are inherited by the subspace.
Observation 2.4.3.
Note the similarities between a planar subspace spanned by two non-colinear vectors in \(\IR^3\text{,}\) and the Euclidean plane \(\IR^2\text{.}\) While they are not the same thing (and shouldn't be referred to interchangably), algebraists call such similar spaces isomorphic; we'll learn what this means more carefully in a later chapter.
Fact 2.4.4.
In order for a nonempty subset \(S\) of a vector space \(V\) to be a subspace, we must first check that the operations \(S\) inherits from \(V\) are defined on \(S\text{.}\) In other words, we need to check two things:
- The set is closed under addition: for any \(\vec{u},\vec{v} \in S\text{,}\) the sum \(\vec{u}+\vec{v}\) is also in \(S\text{.}\)
- The set is closed under scalar multiplication: for any \(\vec{u} \in S\) and scalar \(c \in \IR\text{,}\) the product \(c\vec{u}\) is also in \(S\text{.}\)
Once we have verified the operations are closed and thus well-defined on \(S\text{,}\) we get the eight vector space properties for free.
- Six of the eight (e.g. commutativity) propertieshold directly since they hold in \(V\text{.}\)
- Since \(0 \in \IR\) and \(S\) is closed under scalar multiplication, we see \(\vec{0}=0\odot \vec{v}\) is in \(S\) for any \(\vec{v} \in S\text{.}\)
- Since \(-1 \in \IR\) and \(S\) is closed under scalar multiplication, we see \(-\vec{v}=-1\odot \vec{v}\) is in \(S\) for any \(\vec{v} \in S\text{.}\)
Activity 2.4.5.
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 2.4.6.
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 2.4.7.
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 2.4.8.
Consider these subsets of \(\IR^3\text{:}\)
\begin{equation*}
R=
\setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=z+1}
\hspace{2em}
S=
\setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{y=|z|}
\hspace{2em}
T=
\setBuilder{ \left[\begin{array}{c}x\\y\\z\end{array}\right]}{z=xy}\text{.}
\end{equation*}
(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 2.4.9.
Consider these subsets of \(M_{2 \times 2}\text{,}\) the vector space of all \(2 \times 2\) matrices with real entries. Show that each of these sets is or is not a subspace of \(M_{2 \times 2}\text{.}\)
(a)
\begin{equation*}
\setBuilder{ \left[\begin{array}{cc}a&0\\0&b\end{array}\right]}{a,b \in \IR}\text{.}
\end{equation*}
(b)
\begin{equation*}
\setBuilder{ \left[\begin{array}{cc}a&0\\0&b\end{array}\right]}{a + b = 0}\text{.}
\end{equation*}
(c)
\begin{equation*}
\setBuilder{ \left[\begin{array}{cc}a&0\\0&b\end{array}\right]}{a + b = 5}\text{.}
\end{equation*}
(d)
\begin{equation*}
\setBuilder{ \left[\begin{array}{cc}a&c\\0&b\end{array}\right]}{a + b = 0, c \in \IR}\text{.}
\end{equation*}
Activity 2.4.10.
Let \(W\) be a subspace of a vector space \(V\text{.}\) How are \(\vspan W\) and \(W\) related?
- \(\vspan W\) may include vectors that aren't in \(W\)
- \(W\) may include vectors that aren't in \(\vspan W\)
- \(W\) and \(\vspan W\) always contain the same vectors
Fact 2.4.11.
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 2.4.2 Videos
Subsection 2.4.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/VS4.slides.html.Exercises 2.4.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/exercises/#/bank/VS4/.Subsection 2.4.5 Mathematical Writing Explorations
Exploration 2.4.12.
A square matrix \(M\) is symmetric if, for each index \(i,j\text{,}\) the entries \(m_{ij} = m_{ji}\text{.}\) That is, the matrix is itself when reflected over the diagonal from upper left to lower right. Prove that the set of \(n \times n\) symmetric matrices is a subspace of \(M_{n \times n}\text{.}\)Exploration 2.4.13.
The space of all real-valued function of one real variable is a vector space. First, define \(\oplus\) and \(\odot\) for this vector space. Check that you have closure (both kinds!) and show what the zero vector is under your chosen addition. Decide if each of the following is a subspace. If so, prove it. If not, provide the counterexample.- The set of even functions, \(\{f:\mathbb{R} \rightarrow \mathbb{R}: f(-x) = f(x) \mbox{ for all } x\}\text{.}\)
- The set of odd functions, \(\{f:\mathbb{R} \rightarrow \mathbb{R}: f(-x) = -f(x) \mbox{ for all } x\}\text{.}\)
Exploration 2.4.14.
Give an example of each of these, or explain why it's not possible that such a thing would exist.- A nonempty subset of \(M_{2 \times 2}\) that is not a subspace.
- A set of two vectors in \(\mathbb{R}^2\) that is not a spanning set.
Exploration 2.4.15.
Let \(V\) be a vector space and \(S = \{\vec{v}_1,\vec{v}_2,\ldots,\vec{v}_n\}\) a subset of \(V\text{.}\) Show that the span of \(S\) is a subspace. Is it possible that there is a subset of \(V\) containing fewer vectors than \(S\text{,}\) but whose span contains all of the vectors in the span of \(S\text{?}\)Subsection 2.4.6 Sample Problem and Solution
Sample problem Example B.1.8.
