Learning Outcomes
- Determine if a subset of \(\IR^n\) is a subspace or not.
Section 19.3 Subspaces (EV3)
Subsection 19.3.1 Warm Up
Activity 19.29.
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 19.3.2 Class Activities
Observation 19.30.
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 19.2), we learned how to decide whether \(\vspan(S)\) was equal to all of \(\IR^n\) or something strictly smaller.
Activity 19.31.
Let’s consider the relationship between vectors within a spanning set.
(a)
Let \(S\) denote a set of vectors in \(\IR^3\) and suppose that \(\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}4\\5\\6\end{array}\right]\in\vspan(S)\text{.}\) Which of the following vectors might not belong to \(\vspan(S)\text{?}\)
- \(\displaystyle \left[\begin{array}{c}0\\0\\0\end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c}1\\2\\3\end{array}\right]+ \left[\begin{array}{c}4\\5\\6\end{array}\right]\)
- \(\displaystyle \left[\begin{array}{c}1\\2\\3\end{array}\right]+ \left[\begin{array}{c}1\\1\\1\end{array}\right]\)
- \(\displaystyle -2\left[\begin{array}{c}1\\2\\3\end{array}\right]\)
Answer.
C
(b)
More generally, let \(S\) denote a set of vectors in \(\IR^n\) and suppose that \(\vec v,\vec w\in\vspan(S)\) and \(c\in\mathbb R\text{.}\) Which of the following vectors must belong to \(\vspan(S)\text{?}\)
- \(\displaystyle \vec 0\)
- \(\displaystyle \vec v+\vec w\)
- \(\displaystyle c\vec v\)
- All of these
Answer.
D
Definition 19.32.
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 19.33.
Consider an arbitrary homogeneous vector equation \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\text{.}\)
(a)
Is this equation consistent?
- No, it has no solutions.
- Yes, it is guaranteed to have at least one solution.
- More information is required.
(b)
Suppose that \(\left[\begin{array}{c} 1 \\ 2 \\ 3 \end{array}\right] \) and \(\left[\begin{array}{c} 4 \\ 5 \\ 6 \end{array}\right] \) are both solutions to the homogeneous vector equation \(x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3=\vec{0}\text{.}\) This means that
\begin{equation*}
1 \vec{v}_1+2 \vec{v}_2+3\vec{v}_3 = \vec{0}
\text{ and }
4 \vec{v}_1+5 \vec{v}_2+6\vec{v}_3 = \vec{0} \text{.}
\end{equation*}
Therefore by adding these equations:
\begin{equation*}
(1+4) \vec{v}_1+(2+5) \vec{v}_2+(3+6)\vec{v}_3= \vec{0},
\end{equation*}
we may conclude that the vector \(\left[\begin{array}{c} 1+4 \\ 2+5 \\ 3+6 \end{array}\right] \) is...
- another solution.
- not a solution.
- is equal to \(\vec{0}\text{.}\)
(c)
More generally, 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 to \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\text{,}\) 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 \(\vec{0}\text{.}\)
(d)
Similarly, if \(\left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] \) is a solution and \(c\in\mathbb R\text{,}\) we know first that
\begin{equation*}
a_1\vec{v}_1+\cdots+a_n\vec{v}_n=\vec{0}
\end{equation*}
and by multiplying both sides by \(c\) we also know
\begin{equation*}
(ca_1)\vec{v}_1+\cdots+(ca_n)\vec{v}_n=\vec{0}.
\end{equation*}
Thus 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 \(\vec{0}\text{.}\)
Observation 19.34.
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 \(W\) 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 19.35.
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 19.36.
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 interchangeably), algebraists call such similar spaces isomorphic; we’ll learn what this means more carefully in a later chapter.
A planar subset of \(\IR^3\text{.}\)
The plane \(\IR^2\) itself.
Activity 19.37.
To show that sets of Euclidean vectors form subspaces, we will need to prove that certain equalities hold.
(a)
Consider the following argument that \(5=7\text{:}\)
\begin{align*}
&& 5&=7\\
& \Rightarrow& 5-6&=7-6\\
& \Rightarrow& -1&=1\\
& \Rightarrow& (-1)^2&=(1)^2\\
& \Rightarrow& 1&=1
\end{align*}
Is this reasoning valid?
- Yes
- No
Answer.
B. No, this proves \(1=1\) from \(5=7\text{,}\) not the other way around.
(b)
Consider the following argument that \(5=7\text{:}\)
\begin{align*}
& & 1&=1\\
& \Rightarrow& (-1)^2&=(1)^2\\
& \Rightarrow& -1&=1\\
& \Rightarrow& 5-6&=7-6\\
& \Rightarrow& 5&=7
\end{align*}
Is this reasoning valid?
- Yes
- No
Answer.
B. No, it’s not true that \((-1)^2=(1)^2\) implies \(-1=1\text{.}\)
Remark 19.38.
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*}
- Using two chains of equalities:\begin{align*} \mathrm{LEFT} &= \cdots & \mathrm{RIGHT} &=\cdots\\ &= \cdots & &= \cdots\\ &= \cdots & &= \cdots\\ &= \mathrm{SAME}& &= \mathrm{SAME} \end{align*}
- Manipulating a known fact \(\mathrm{THIS}=\mathrm{THAT}\) into the desired equation:\begin{align*} && \mathrm{THIS} &= \mathrm{THAT}\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \mathrm{LEFT} &= \mathrm{RIGHT} \end{align*}
Warning 19.39.
The following proof of \(\mathrm{LEFT} = \mathrm{RIGHT}\) is invalid.
\begin{align*}
&& \mathrm{LEFT} &= \mathrm{RIGHT}\\
& \Rightarrow& \cdots &= \cdots\\
& \Rightarrow& \cdots &= \cdots\\
& \Rightarrow& \mathrm{SAME} &= \mathrm{SAME}
\end{align*}
This is instead a proof that \(\mathrm{SAME}=\mathrm{SAME}\text{,}\) assuming that \(\mathrm{LEFT} = \mathrm{RIGHT}\) is true in the first place. But the fact that \(\mathrm{SAME}=\mathrm{SAME}\) doesn’t need a proof, and this work fails to guarantee that \(\mathrm{LEFT} = \mathrm{RIGHT}\) will be true, as we saw in Activity 19.37.
Activity 19.40.
Let \(W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\) Consider the following questions to prove that \(W\) is a subspace.
(a)
Is \(W\) non-empty?
- Yes.
- No.
(b)
Let’s assume that \(\vec{u}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) and \(\vec{v} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] \) are in \(W\text{.}\) What equations are we assuming to be true?
- \(x+2y+z=0\text{.}\)
- \(a+2b+c=0\text{.}\)
- Both of these.
- Neither of these.
(c)
Which equation must be verified to show that \(\vec u+\vec v = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\) 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 your assumptions to complete the following proof of \((x+a)+2(y+b)+(z+c)=0\text{.}\)
\begin{align*}
(x+a)+2(y+b)+(z+c) &= \unknown\\
&= (\unknown)+(\unknown)\\
&= 0 + 0\\
&= 0
\end{align*}
(e)
Have we proven \(W\) is a subspace of \(\IR^3\text{?}\)
- Yes
- Not yet
(f)
Assume that \(\vec u= \left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) belongs to \(W\text{,}\) and \(c\in\mathbb R\text{.}\) Which equation must be verified to show that \(c\vec u= \left[\begin{array}{c} cx \\ cy \\ cz \end{array}\right]\) also belongs to \(W\text{?}\)
- \((cx)+2(cy)+(cz)=0\text{.}\)
- \(x+2y+z=c\text{.}\)
- \(x+2y+z+c=0\text{.}\)
(g)
Complete the following proof of \((cx)+2(cy)+(cz)=0\) from the assumption \(x+2y+z=0\text{.}\)
\begin{align*}
&& x+2y+z &= 0\\
& \Rightarrow& \unknown[x+2y+z] &= \unknown[0]\\
& \Rightarrow& \unknown &= \unknown\\
& \Rightarrow& (cx)+2(cy)+(cz) &= 0
\end{align*}
(h)
Have we proven \(W\) is a subspace of \(\IR^3\text{?}\)
- Yes
- Not yet
Activity 19.41.
Let \(W=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\)
(a)
Is \(W\) non-empty?
- Yes.
- No.
(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 19.42.
In summary, any one of the following is enough to prove that a nonempty subset \(W\) is not a subspace:
- Show that \(W\) is empty. (Or even just show \(\vec 0\not\in W\text{.}\))
- 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 \(c\vec v\not\in W\text{.}\)
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. (Usually by showing \(\vec 0\in W\text{.}\))
- For all \(\vec u,\vec v\in W\) (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 19.43.
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 19.44.
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 19.45.
(a)
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)\\ &\Rightarrow&k(x+y)&=k^2xy\\ &\Rightarrow&0[k(x+y)]&=0[k^2xy]\\ &\Rightarrow&0&=0 \end{align*}
Is this reasoning valid?
- Yes
- No
(b)
Does this fix the proof?
\begin{align*} &&x+y&=xy\\ &\Rightarrow&k(x+y)&=k(xy)\\ &\Rightarrow&(kx)+(ky)&=(kx)(ky) \end{align*}
- Yes
- No
Subsection 19.3.3 Individual Practice
Remark 19.46.
Recall that in Activity 19.16 we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals. In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
Activity 19.47.
(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 19.48.
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 \(\vec{u},\vec{v}\) 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 assuming 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 19.3.4 Videos
Subsection 19.3.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/EV3/.Subsection 19.3.6 Mathematical Writing Explorations
Exploration 19.49.
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 19.50.
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 19.51.
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 19.52.
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 19.3.7 Sample Problem and Solution
Sample problem Example C.7.
