# Linear Algebra/Comparing Set Descriptions

 Linear Algebra ← General = Particular + Homogeneous Comparing Set Descriptions Automation →

This subsection is optional. Later material will not require the work here.

## Comparing Set Descriptions

A set can be described in many different ways. Here are two different descriptions of a single set:

${\displaystyle \{{\begin{pmatrix}1\\2\\3\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}2\\4\\6\end{pmatrix}}w\,{\big |}\,w\in \mathbb {R} \}.}$

For instance, this set contains

${\displaystyle {\begin{pmatrix}5\\10\\15\end{pmatrix}}}$

(take ${\displaystyle z=5}$ and ${\displaystyle w=5/2}$) but does not contain

${\displaystyle {\begin{pmatrix}4\\8\\11\end{pmatrix}}}$

(the first component gives ${\displaystyle z=4}$ but that clashes with the third component, similarly the first component gives ${\displaystyle w=4/5}$ but the third component gives something different). Here is a third description of the same set:

${\displaystyle \{{\begin{pmatrix}3\\6\\9\end{pmatrix}}+{\begin{pmatrix}-1\\-2\\-3\end{pmatrix}}y\,{\big |}\,y\in \mathbb {R} \}.}$

We need to decide when two descriptions are describing the same set. More pragmatically stated, how can a person tell when an answer to a homework question describes the same set as the one described in the back of the book?

## Set Equality

Sets are equal if and only if they have the same members. A common way to show that two sets, ${\displaystyle S_{1}}$ and ${\displaystyle S_{2}}$, are equal is to show mutual inclusion: any member of ${\displaystyle S_{1}}$ is also in ${\displaystyle S_{2}}$, and any member of ${\displaystyle S_{2}}$ is also in ${\displaystyle S_{1}}$.[1]

Example 4.1

To show that

${\displaystyle S_{1}=\{{\begin{pmatrix}1\\-1\\0\end{pmatrix}}c+{\begin{pmatrix}1\\1\\0\end{pmatrix}}d\,{\big |}\,c,d\in \mathbb {R} \}}$

equals

${\displaystyle S_{2}=\{{\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}$

we show first that ${\displaystyle S_{1}\subseteq S_{2}}$ and then that ${\displaystyle S_{2}\subseteq S_{1}}$.

For the first half we must check that any vector from ${\displaystyle S_{1}}$ is also in ${\displaystyle S_{2}}$. We first consider two examples to use them as models for the general argument. If we make up a member of ${\displaystyle S_{1}}$ by trying ${\displaystyle c=1}$ and ${\displaystyle d=1}$, then to show that it is in ${\displaystyle S_{2}}$ we need ${\displaystyle m}$ and ${\displaystyle n}$ such that

${\displaystyle {\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n={\begin{pmatrix}2\\0\\0\end{pmatrix}}}$

that is, this relation holds between ${\displaystyle m}$ and ${\displaystyle n}$.

${\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&2\\1m&-&3n&=&0\\&&0&=&0\end{array}}}$

Similarly, if we try ${\displaystyle c=2}$ and ${\displaystyle d=-1}$, then to show that the resulting member of ${\displaystyle S_{1}}$ is in ${\displaystyle S_{2}}$ we need ${\displaystyle m}$ and ${\displaystyle n}$ such that

${\displaystyle {\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n={\begin{pmatrix}3\\-3\\0\end{pmatrix}}}$

that is, this holds.

${\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&3\\1m&-&3n&=&-3\\&&0&=&0\end{array}}}$

In the general case, to show that any vector from ${\displaystyle S_{1}}$ is a member of ${\displaystyle S_{2}}$ we must show that for any ${\displaystyle c}$ and ${\displaystyle d}$ there are appropriate ${\displaystyle m}$ and ${\displaystyle n}$. We follow the pattern of the examples; fix

${\displaystyle {\begin{pmatrix}c+d\\-c+d\\0\end{pmatrix}}\in S_{1}}$

and look for ${\displaystyle m}$ and ${\displaystyle n}$ such that

${\displaystyle {\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n={\begin{pmatrix}c+d\\-c+d\\0\end{pmatrix}}}$

that is, this is true.

${\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&c+d\\m&-&3n&=&-c+d\\&&0&=&0\end{array}}}$

Applying Gauss' method

${\displaystyle {\begin{array}{rcl}{\begin{array}{*{2}{rc}r}4m&-&n&=&c+d\\m&-&3n&=&-c+d\end{array}}&{\xrightarrow[{}]{-(1/4)\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}4m&-&n&=&c+d\\&&-(11/4)n&=&-(5/4)c+(3/4)d\end{array}}\end{array}}}$

gives ${\displaystyle n=(5/11)c-(3/11)d}$ and ${\displaystyle m=(4/11)c+(2/11)d}$. This shows that for any choice of ${\displaystyle c}$ and ${\displaystyle d}$ there are appropriate ${\displaystyle m}$ and ${\displaystyle n}$. We conclude any member of ${\displaystyle S_{1}}$ is a member of ${\displaystyle S_{2}}$ because it can be rewritten in this way:

${\displaystyle {\begin{pmatrix}c+d\\-c+d\\0\end{pmatrix}}={\begin{pmatrix}4\\1\\0\end{pmatrix}}((4/11)c+(2/11)d)+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}((5/11)c-(3/11)d).}$

For the other inclusion, ${\displaystyle S_{2}\subseteq S_{1}}$, we want to do the opposite. We want to show that for any choice of ${\displaystyle m}$ and ${\displaystyle n}$ there are appropriate ${\displaystyle c}$ and ${\displaystyle d}$. So fix ${\displaystyle m}$ and ${\displaystyle n}$ and solve for ${\displaystyle c}$ and ${\displaystyle d}$:

${\displaystyle {\begin{array}{rcl}{\begin{array}{*{2}{rc}r}c&+&d&=&4m-n\\-c&+&d&=&m-3n\end{array}}&{\xrightarrow[{}]{\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}c&+&d&=&4m-n\\&&2d&=&5m-4n\end{array}}\end{array}}}$

shows that ${\displaystyle d=(5/2)m-2n}$ and ${\displaystyle c=(3/2)m+n}$. Thus any vector from ${\displaystyle S_{2}}$

${\displaystyle {\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n}$

is also of the right form for ${\displaystyle S_{1}}$

${\displaystyle {\begin{pmatrix}1\\-1\\0\end{pmatrix}}((3/2)m+n)+{\begin{pmatrix}1\\1\\0\end{pmatrix}}((5/2)m-2n).}$
Example 4.2

Of course, sometimes sets are not equal. The method of the prior example will help us see the relationship between the two sets. These

${\displaystyle P=\{{\begin{pmatrix}x+y\\2x\\y\end{pmatrix}}\,{\big |}\,x,y\in \mathbb {R} \}\quad {\text{and}}\quad R=\{{\begin{pmatrix}m+p\\n\\p\end{pmatrix}}\,{\big |}\,m,n,p\in \mathbb {R} \}}$

are not equal sets. While ${\displaystyle P}$ is a subset of ${\displaystyle R}$, it is a proper subset of ${\displaystyle R}$ because ${\displaystyle R}$ is not a subset of ${\displaystyle P}$.

To see that, observe first that given a vector from ${\displaystyle P}$ we can express it in the form for ${\displaystyle R}$— if we fix ${\displaystyle x}$ and ${\displaystyle y}$, we can solve for appropriate ${\displaystyle m}$, ${\displaystyle n}$, and ${\displaystyle p}$:

${\displaystyle {\begin{array}{*{3}{rc}r}m&&&+&p&=&x+y\\&&n&&&=&2x\\&&&&p&=&y\end{array}}}$

shows that that any

${\displaystyle {\vec {v}}={\begin{pmatrix}1\\2\\0\end{pmatrix}}x+{\begin{pmatrix}1\\0\\1\end{pmatrix}}y}$

can be expressed as a member of ${\displaystyle R}$ with ${\displaystyle m=x}$, ${\displaystyle n=2x}$, and ${\displaystyle p=y}$:

${\displaystyle {\vec {v}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}x+{\begin{pmatrix}0\\1\\0\end{pmatrix}}2x+{\begin{pmatrix}1\\0\\1\end{pmatrix}}y.}$

Thus ${\displaystyle P\subseteq R}$.

But, for the other direction, the reduction resulting from fixing ${\displaystyle m}$, ${\displaystyle n}$, and ${\displaystyle p}$ and looking for ${\displaystyle x}$ and ${\displaystyle y}$

${\displaystyle {\begin{array}{rcl}{\begin{array}{*{2}{rc}r}x&+&y&=&m+p\\2x&&&=&n\\&&y&=&p\end{array}}&{\xrightarrow[{}]{-2\rho _{1}+\rho _{2}}}&{\begin{array}{*{2}{rc}r}x&+&y&=&m+p\\&&-2y&=&-2m+n-2p\\&&y&=&p\end{array}}\\&{\xrightarrow[{}]{(1/2)\rho _{2}+\rho _{3}}}&{\begin{array}{*{2}{rc}r}x&+&y&=&m+p\\&&-2y&=&-2m+n-2p\\&&0&=&m+(1/2)n\end{array}}\end{array}}}$

shows that the only vectors

${\displaystyle {\begin{pmatrix}m+p\\n\\p\end{pmatrix}}\in R}$

representable in the form

${\displaystyle {\begin{pmatrix}x+y\\2x\\y\end{pmatrix}}}$

are those where ${\displaystyle 0=m+(1/2)n}$. For instance,

${\displaystyle {\begin{pmatrix}0\\1\\0\end{pmatrix}}}$

is in ${\displaystyle R}$ but not in ${\displaystyle P}$.

## Exercises

Problem 1

Decide if the vector is a member of the set.

1. ${\displaystyle {\begin{pmatrix}2\\3\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
2. ${\displaystyle {\begin{pmatrix}-3\\3\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
3. ${\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
4. ${\displaystyle {\begin{pmatrix}-3\\3\\4\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}1\\-1\\2\end{pmatrix}}k+{\begin{pmatrix}0\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
5. ${\displaystyle {\begin{pmatrix}1\\4\\14\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
6. ${\displaystyle {\begin{pmatrix}1\\4\\6\end{pmatrix}}}$, ${\displaystyle \{{\begin{pmatrix}2\\2\\5\end{pmatrix}}k+{\begin{pmatrix}-1\\0\\2\end{pmatrix}}m\,{\big |}\,k,m\in \mathbb {R} \}}$
Problem 2

Produce two descriptions of this set that are different than this one.

${\displaystyle \{{\begin{pmatrix}2\\-5\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}}$
This exercise is recommended for all readers.
Problem 3

Show that the three descriptions given at the start of this subsection all describe the same set.

This exercise is recommended for all readers.
Problem 4

Show that these sets are equal

${\displaystyle \{{\begin{pmatrix}1\\4\\1\\1\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}z\,{\big |}\,z\in \mathbb {R} \}\quad {\text{and}}\quad \{{\begin{pmatrix}0\\4\\2\\1\end{pmatrix}}+{\begin{pmatrix}-1\\0\\1\\0\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \},}$

and that both describe the solution set of this system.

${\displaystyle {\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\x&&&+&z&+&2w&=&4\end{array}}}$
This exercise is recommended for all readers.
Problem 5

Decide if the sets are equal.

1. ${\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}+{\begin{pmatrix}0\\3\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}}$ and ${\displaystyle \{{\begin{pmatrix}1\\8\end{pmatrix}}+{\begin{pmatrix}0\\-1\end{pmatrix}}s\,{\big |}\,s\in \mathbb {R} \}}$
2. ${\displaystyle \{{\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\1\\5\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}$ and ${\displaystyle \{{\begin{pmatrix}4\\7\\7\end{pmatrix}}m+{\begin{pmatrix}-4\\-2\\-10\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}$
3. ${\displaystyle \{{\begin{pmatrix}1\\2\end{pmatrix}}t\,{\big |}\,t\in \mathbb {R} \}}$ and ${\displaystyle \{{\begin{pmatrix}2\\4\end{pmatrix}}m+{\begin{pmatrix}4\\8\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}$
4. ${\displaystyle \{{\begin{pmatrix}1\\0\\2\end{pmatrix}}s+{\begin{pmatrix}-1\\1\\0\end{pmatrix}}t\,{\big |}\,s,t\in \mathbb {R} \}}$ and ${\displaystyle \{{\begin{pmatrix}-1\\1\\1\end{pmatrix}}m+{\begin{pmatrix}0\\1\\3\end{pmatrix}}n\,{\big |}\,m,n\in \mathbb {R} \}}$
5. ${\displaystyle \{{\begin{pmatrix}1\\3\\1\end{pmatrix}}t+{\begin{pmatrix}2\\4\\6\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}$ and ${\displaystyle \{{\begin{pmatrix}3\\7\\7\end{pmatrix}}t+{\begin{pmatrix}1\\3\\1\end{pmatrix}}s\,{\big |}\,t,s\in \mathbb {R} \}}$