Learning Outcomes
- Determine if a subset of \(\IR^n\) is a subspace or not.
Section 2.3 Subspaces (EV3)
Subsection 2.3.1 Class Activities
Definition 2.3.1.
A subset \(S\) of a vector space is called a subspace provided it is equal to the span of a set of vectors from that space.
Activity 2.3.2.
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 planar subspace within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)
(a)
For any unspecified \(\vec{u}, \vec{v} \in S\text{,}\) is it the case that \(\vec{u} + \vec{v} \in S\text{?}\)
- Yes.
- No.
(b)
For any unspecified \(\vec{u} \in S\) and \(c\in\IR\text{,}\) is it the case that \(\vec{u}+\left[\begin{array}{c} c \\ c \\ c \end{array}\right] \in S\text{?}\)
- Yes.
- No.
(c)
For any unspecified \(\vec{u} \in S\) and \(c\in\IR\text{,}\) is it the case that \(c\vec{u} \in S\text{?}\)
- Yes.
- No.
Fact 2.3.3.
A subset \(S\) of a vector space is a subspace provided:
- the subset 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 subset 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 \(S\text{.}\)
Observation 2.3.4.
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.
Activity 2.3.5.
Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)
(a)
Let’s assume that \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) and \(\vec{w} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] \) are in \(S\text{.}\) What are we allowed to assume?
- \(x+2y+z=0\text{.}\)
- \(a+2b+c=0\text{.}\)
- Both of these.
- Neither of these.
(b)
Which equation must be verified to show that \(\vec v+\vec w = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\) also belongs to \(S\text{?}\)
- \((x+a)+2(y+b)+(z+c)=0\text{.}\)
- \(x+a+2y+b+z+c=0\text{.}\)
- \(x+2y+z=a+2b+c\text{.}\)
(c)
Use the assumptions from (a) to verify the equation from (b).
(d)
Is \(S\) is a subspace of \(\IR^3\text{?}\)
- Yes
- No
- Not enough information
(e)
Show that \(k\vec v=\left[\begin{array}{c}kx\\ky\\kz\end{array}\right]\) also belongs to \(S\) for any \(k\in\IR\) by verifying \((kx)+2(ky)+(kz)=0\) under these assumptions.
(f)
Is \(S\) is a subspace of \(\IR^3\text{?}\)
- Yes
- No
- Not enough information
Activity 2.3.6.
Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\)
(a)
Which of these statements is valid?
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in S\text{,}\) so \(S\) is a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in S\text{,}\) so \(S\) is not a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in S\text{,}\) so \(S\) is a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in S\text{,}\) so \(S\) is not a subspace.
(b)
Which of these statements is valid?
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in S\text{,}\) so \(S\) is a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in S\text{,}\) so \(S\) is not a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in S\text{,}\) so \(S\) is a subspace.
- \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in S\text{,}\) so \(S\) is not a subspace.
Remark 2.3.7.
In summary, you can check any of the following to show that a nonempty subset \(W\) isn’t a subspace:
- 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 \(c\vec v\not\in W\text{.}\)
- Show that \(\vec 0\not\in W\) (same as the last step, with \(c=0\)).
If you cannot do any of these, then \(W\) can be proven to be a subspace by doing both of 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 2.3.8.
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 2.3.9.
Consider the following two sets of Euclidean vectors:
\begin{equation*}
U=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle|\,7 \, x + 4 \, y = 0\right\} \hspace{2em} W=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle|\,3 \, x y^{2} = 0\right\}
\end{equation*}
Explain why one of these sets is a subspace of \(\mathbb{R}^2\) and one is not.
Activity 2.3.10.
Consider the following attempted proof that
\begin{equation*}
U=\left\{ \left[\begin{array}{c} x \\ y \end{array}\right] \middle| x+y=xy\right\}
\end{equation*}
is closed under scalar multiplication.
Let \(\left[\begin{array}{c} x \\ y \end{array}\right]\in U\text{,}\) so we know that \(x+y=xy\text{.}\) We want to show \(k\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} kx \\ ky \end{array}\right]\in U\text{,}\) that is, \((kx)+(ky)=(kx)(ky)\text{.}\) This is verified by the following calculation:\begin{align*} (kx)+(ky)&=(kx)(ky)\\ k(x+y)&=k^2xy\\ 0[k(x+y)]&=0[k^2xy]\\ 0&=0 \end{align*}
Is this reasoning valid?
- Yes
- No
Remark 2.3.11.
Proofs of an equality \(\mathrm{LEFT}=\mathrm{RIGHT}\) should generally be of one of these forms:
- Using a chain of equalities:\begin{align*} \mathrm{LEFT} &= \cdots\\ &= \cdots\\ &= \cdots\\ &= \mathrm{RIGHT} \end{align*}Alternatively:\begin{align*} \mathrm{LEFT} &= \cdots & \mathrm{RIGHT} &=\cdots\\ &= \cdots & &= \cdots\\ &= \cdots & &= \cdots\\ &= \mathrm{SAME}& &= \mathrm{SAME} \end{align*}
- When the assumption \(\mathrm{THIS}=\mathrm{THAT}\) is already known or assumed to be true :\begin{align*} && \mathrm{THIS} &= \mathrm{THAT}\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \mathrm{LEFT} &= \mathrm{RIGHT} \end{align*}
Warning 2.3.12.
The following proof is invalid.
\begin{align*}
&& \mathrm{LEFT} &= \mathrm{RIGHT}\\
& \Rightarrow& \cdots &= \cdots\\
& \Rightarrow& \cdots &= \cdots\\
& \Rightarrow& 0 &= 0\\
& \Rightarrow& \mathrm{ANYTHING} &= \mathrm{ANYTHING}
\end{align*}
Basically, you cannot prove something is true by assuming it’s true, and it’s not helpful to prove to someone that zero equals itself (they probably already know that).
Subsection 2.3.2 Videos
Subsection 2.3.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/EV3.slides.html.Exercises 2.3.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/exercises/#/bank/EV3/.Subsection 2.3.5 Mathematical Writing Explorations
Exploration 2.3.13.
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 2.3.14.
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 2.3.15.
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 2.3.16.
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 2.3.6 Sample Problem and Solution
Sample problem Example B.1.7.
