Learning Outcomes
-
Determine if a subset of \(\IR^n\) is a subspace or not.
Section 2.3 Subspaces (EV3)
Subsection 2.3.1 Warm Up
Activity 2.3.1.
Consider the linear equation
\begin{equation*}
x+2y+z=0.
\end{equation*}
(a)
Verify that both \(\vec{v}=\left[\begin{array}{c}1\\-1\\1\end{array}\right]\) and \(\vec{w}=\left[\begin{array}{c}1\\0\\-1\end{array}\right]\) are solutions.
(b)
Is the vector \(2\vec{v}-3\vec{w}\) also a solution?
Subsection 2.3.2 Class Activities
Observation 2.3.2.
Recall that if \(S=\left\{\vec{v}_1,\dots, \vec{v}_n\right\}\) is subset of vectors in \(\IR^n\text{,}\) then \(\vspan(S)\) is the set of all linear combinations of vectors in \(S\text{.}\) In EV2 (Section 2.2), we learned how to decide whether \(\vspan(S)\) was equal to all of \(\IR^n\) or something strictly smaller.
Activity 2.3.3.
Let \(S\) denote a set of vectors in \(\IR^n\) and suppose that \(\vec{u},\vec{v}\in\vspan(S)\text{,}\) \(c\in\IR\) and that \(\vec{w}\in\IR^n\text{.}\) Which of the following vectors might not belong to \(\vspan(S)\text{?}\)
-
\(\displaystyle \vec{0}\)
-
\(\displaystyle \vec{u}+\vec{w}\)
-
\(\displaystyle \vec{u}+\vec{v}\)
-
\(\displaystyle c\vec{u}\)
Definition 2.3.4.
A homogeneous system of linear equations is one of the form:
\begin{alignat*}{5}
a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& 0 \\
a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& 0 \\
\vdots& &\vdots& && &\vdots&&\vdots\\
a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& 0
\end{alignat*}
This system is equivalent to the vector equation:
\begin{equation*}
x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}
\end{equation*}
and the augmented matrix:
\begin{equation*}
\left[\begin{array}{cccc|c}
a_{11} & a_{12} & \cdots & a_{1n} & 0\\
a_{21} & a_{22} & \cdots & a_{2n} & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn} & 0
\end{array}\right]
\end{equation*}
Activity 2.3.5.
Consider the homogeneous vector equation \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\text{.}\)
(a)
Is this equation consistent?
-
no.
-
yes.
-
more information is required.
(b)
Note that if \(\left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] \) and \(\left[\begin{array}{c} b_1 \\ \vdots \\ b_n \end{array}\right] \) are both solutions, we know that
\begin{equation*}
a_1 \vec{v}_1+\cdots+a_n \vec{v}_n = \vec{0}
\text{ and }
b_1 \vec{v}_1+\cdots+b_n \vec{v}_n = \vec{0} \text{.}
\end{equation*}
Therefore by adding these equations:
\begin{equation*}
(a_1 + b_1) \vec{v}_1+\cdots+(a_n+b_n) \vec{v}_n = \vec{0},
\end{equation*}
we may conclude that the vector \(\left[\begin{array}{c} a_1+ b_1 \\ \vdots \\ a_n+b_n \end{array}\right] \) is...
-
another solution.
-
not a solution.
-
is equal to \(\vec{0}\text{.}\)
(c)
Similarly, if \(c \in \IR\text{,}\) then since multiplying by \(c\) yields:
\begin{equation*}
(ca_1)\vec{v}_1+\cdots+(ca_n)\vec{v}_n=\vec{0},
\end{equation*}
we may conclude that the vector \(\left[\begin{array}{c} ca_1 \\ \vdots \\ ca_n \end{array}\right] \) is...
-
another solution.
-
not a solution.
-
is equal to \(\vec{0}\text{.}\)
-
The empty set.
Observation 2.3.6.
If \(S\) is any set of vectors in \(\IR^n\text{,}\) then the set \(\vspan(S)\) has the following properties:
-
the set \(\vspan(S)\) is non-empty.
-
the set \(\vspan(S)\) is closed under addition: for any \(\vec{u},\vec{v}\in \vspan(S)\text{,}\) the sum \(\vec{u}+\vec{v}\) is also in \(\vspan(S)\text{.}\)
-
the set \(\vspan(S)\) is closed under scalar multiplication: for any \(\vec{u}\in\vspan(S)\) and scalar \(c\in\IR\text{,}\) the product \(c\vec{u}\) is also in \(\vspan(S)\text{.}\)
Likewise, if \(W\) is the solution set to a homogenous vector equation, it too satisfies:
-
the set \(W\) is non-empty.
-
the set \(W\) is closed under addition: for any \(\vec{u},\vec{v}\in W\text{,}\) the sum \(\vec{u}+\vec{v}\) is also in \(W\text{.}\)
-
the set \(\vspan(S)\) is closed under scalar multiplication: for any \(\vec{u}\in W\) and scalar \(c\in\IR\text{,}\) the product \(c\vec{u}\) is also in \(W\text{.}\)
Definition 2.3.7.
A subset \(W\) of a vector space is called a subspace provided that it satisfies the following properties:
-
the subset is non-empty.
-
the subset is closed under addition: for any \(\vec{u},\vec{v} \in W\text{,}\) the sum \(\vec{u}+\vec{v}\) is also in \(W\text{.}\)
-
the subset is closed under scalar multiplication: for any \(\vec{u} \in W\) and scalar \(c \in \IR\text{,}\) the product \(c\vec{u}\) is also in \(W\text{.}\)
Observation 2.3.8.
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.
A planar subset of \(\IR^3\) compared with the plane \(\IR^2\text{.}\)
Activity 2.3.9.
Let \(W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)
(a)
Is \(W\) the empty set?
(b)
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 \(W\text{.}\) What are we allowed to assume?
-
\(x+2y+z=0\text{.}\)
-
\(a+2b+c=0\text{.}\)
-
Both of these.
-
Neither of these.
(c)
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 \(W\text{?}\)
-
\((x+a)+2(y+b)+(z+c)=0\text{.}\)
-
\(x+a+2y+b+z+c=0\text{.}\)
-
\(x+2y+z=a+2b+c\text{.}\)
(d)
Use the assumptions from (a) to verify the equation from (b).
(e)
Is \(W\) is a subspace of \(\IR^3\text{?}\)
-
Yes
-
No
-
Not enough information
(f)
Show that \(k\vec v=\left[\begin{array}{c}kx\\ky\\kz\end{array}\right]\) also belongs to \(W\) for any \(k\in\IR\) by verifying \((kx)+2(ky)+(kz)=0\) under these assumptions.
(g)
Is \(W\) is a subspace of \(\IR^3\text{?}\)
-
Yes
-
No
-
Not enough information
Activity 2.3.10.
Let \(W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\)
(a)
Is \(W\) the empty set?
(b)
Which of these statements is valid?
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in W\text{,}\) so \(W\) is a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in W\text{,}\) so \(W\) is not a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in W\text{,}\) so \(W\) is a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in W\text{,}\) so \(W\) is not a subspace.
(c)
Which of these statements is valid?
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in W\text{,}\) so \(W\) is a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in W\text{,}\) so \(W\) is not a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in W\text{,}\) so \(W\) is a subspace.
-
\(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in W\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in W\text{,}\) so \(W\) is not a subspace.
Remark 2.3.11.
In summary, any one of the following is enough to prove that a nonempty subset \(W\) is not a subspace:
-
Find specific values for \(\vec u,\vec v\in W\) such that \(\vec u+\vec v\not\in W\text{.}\)
-
Find specific values for \(c\in\IR,\vec v\in W\) such that \(c\vec v\not\in W\text{.}\)
-
Show that \(\vec 0\not\in W\text{.}\)
If you cannot do any of these, then \(W\) can be proven to be a subspace by doing all of the following:
-
Show that \(W\) is non-empty.
-
For all \(\vec v,\vec w\in W\) (not just specific values), \(\vec u+\vec v\in W\text{.}\)
-
For all \(\vec v\in W\) and \(c\in \IR\) (not just specific values), \(c\vec v\in W\text{.}\)
Activity 2.3.12.
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 2.3.13.
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 2.3.14.
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 2.3.15.
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 2.3.16.
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 2.3.3 Cool Down
Remark 2.3.17.
Activity 2.2.1vectorlinear combinationspanrecipeActivity 2.3.18.
(a)
Given the set of ingredients \(S=\{\textrm{flour}, \textrm{yeast}, \textrm{salt}, \textrm{water}, \textrm{sugar}, \textrm{milk}\}\text{,}\) how should we think of the subspace \(\vspan(S)\text{?}\)
(b)
What is one meal that lives in the subspace \(\vspan(S)\text{?}\)
(c)
What is one meal that does not live in the subspace \(\vspan(S)\text{?}\)
Activity 2.3.19.
Let
\begin{equation*}
W=\left\{\left[\begin{array}{c}x\\y\\z\\w\end{array}\right]\middle|x+y=3z+2w\right\}.
\end{equation*}
The set \(W\) is a subspace. Below are two attempted proofs of the fact that \(W\) is closed under vector addition. Both of them are invalid; explain why.
(a)
Let \(\vec{u}=\left[\begin{array}{c}1\\4\\1\\1\end{array}\right],\vec{v}=\left[\begin{array}{c}2\\-1\\1\\-1\end{array}\right].\) Then both \(\vec{u},\vec{w}\) are elements of \(W\text{.}\) Their sum is
\begin{equation*}
\vec{w}=\left[\begin{array}{c}3\\3\\2\\0\end{array}\right]
\end{equation*}
and since
\begin{equation*}
3+3=3\cdot (2)+2\cdot (0),
\end{equation*}
it follows that \(\vec{w}\) is also in \(W\) and so \(W\) is closed under vector addition.
(b)
If \(\left[\begin{array}{c}x\\y\\z\\w\end{array}\right],\left[\begin{array}{c}a\\b\\c\\d\end{array}\right]\) are in \(W\text{,}\) we need to show that \(\left[\begin{array}{c}x+a\\y+b\\z+c\\w+d\end{array}\right]\) is also in W. To be in \(W\text{,}\) we need
\begin{equation*}
(x+a)+(y+b)=3(z+c)+2(w+d).
\end{equation*}
Well, if
\begin{equation*}
(x+a)+(y+b)=3(z+c)+2(w+d),
\end{equation*}
then we know that
\begin{equation*}
x+y-3z-2w+a+b-3c-2d=0
\end{equation*}
by moving everything over to the left hand side. Since we are assumming that \(x+y-3z-2w=0\) and \(a+b-3c-2d=0\text{,}\) it follows that \(0=0\text{,}\) which is true, which proves that vector addition is closed.
Subsection 2.3.4 Videos
Exercises 2.3.5 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024/exercises/#/bank/EV3/.
Subsection 2.3.6 Mathematical Writing Explorations
Exploration 2.3.20.
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 2.3.21.
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 2.3.22.
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 2.3.23.
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 2.3.7 Sample Problem and Solution
Sample problem Example B.1.7.
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