Linear Algebra/Comparing Set Descriptions

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:

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

For instance, this set contains

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

(take $z=5$ and $w=5/2$) but does not contain

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

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

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

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, $S_{1}$ and $S_{2}$, are equal is to show mutual inclusion: any member of $S_{1}$ is also in $S_{2}$, and any member of $S_{2}$ is also in $S_{1}$.

Example 4.1

To show that

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

equals

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

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

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

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

that is, this relation holds between $m$ and $n$.

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

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

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

that is, this holds.

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

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

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

and look for $m$ and $n$ such that

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

that is, this is true.

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

Applying Gauss' method

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

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

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

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

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

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

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

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

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

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

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

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

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

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

shows that that any

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

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

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

Thus $P\subseteq R$.

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

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

shows that the only vectors

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

representable in the form

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

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

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

is in $R$ but not in $P$.

Exercises

Problem 1

Decide if the vector is a member of the set.

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

Problem 2

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

$\{{\begin{pmatrix}2\\-5\end{pmatrix}}k\,{\big |}\,k\in \mathbb {R} \}$

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

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

and that both describe the solution set of this system.

${\begin{array}{*{4}{rc}r}x&-&y&+&z&+&w&=&-1\\&&y&&&-&w&=&3\\x&&&+&z&+&2w&=&4\end{array}}$

This exercise is recommended for all readers.

Problem 5

Decide if the sets are equal.

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

Solutions