This subsection is optional. Later material will not require the work here.
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} \}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6007eb4e7853d6a20927f63b9ea17b652d7d5484)
For instance, this set contains
![{\displaystyle {\begin{pmatrix}5\\10\\15\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/508a3e20c814c8626b3510b21af09e1daac183b2)
(take
and
) but does not contain
![{\displaystyle {\begin{pmatrix}4\\8\\11\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8592d6e9c51230d87b555e9efc64ebd402f1d95f)
(the first component gives
but that clashes with the third component,
similarly the first component gives
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} \}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c350a8ffc7a70bfd908659a88427e4c90ead081)
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?
Sets are equal if and only if they have the same members.
A common way to show that two sets,
and
, are equal is to show mutual inclusion:
any member of
is also in
, and
any member of
is also in
.[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} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86213b955ab53b04cc4429b58c72dc9e0d6cfd32)
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} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca528a6f81064b9a853444509d663e9dbd33aa5f)
we show first that
and then that
.
For the first half we must check that any vector
from
is also in
.
We first consider two examples to use them as models for the general argument.
If we make up a member of
by trying
and
,
then to show that it is in
we need
and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/734e448b2b907d6da45baf5d610500312c2fad32)
that is, this relation holds between
and
.
![{\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&2\\1m&-&3n&=&0\\&&0&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba6c13f98dc77fa9651d80e704fd0a39dd046ce4)
Similarly,
if we try
and
, then to show that the resulting
member of
is in
we need
and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bc758b71d266d99ebbe5d80d7d0d156d05b766f)
that is, this holds.
![{\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&3\\1m&-&3n&=&-3\\&&0&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96948edcd8df888ba4a95a272367a64a797d1465)
In the general case,
to show that any vector from
is a member of
we must show
that for any
and
there are appropriate
and
.
We follow the pattern of the examples; fix
![{\displaystyle {\begin{pmatrix}c+d\\-c+d\\0\end{pmatrix}}\in S_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/591ea9b25dd4bd22f8db5b96b600460e7aba3d59)
and look for
and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf3832aa9a3abf4cff1f435f43c72e6e62cb291a)
that is, this is true.
![{\displaystyle {\begin{array}{*{2}{rc}r}4m&-&n&=&c+d\\m&-&3n&=&-c+d\\&&0&=&0\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82e455b248e5582a234766a30aeba609fb0eaa7f)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a77e02189911ac183e08399164ce4904545b0618)
gives
and
.
This shows that for any choice of
and
there are appropriate
and
.
We conclude any member of
is a member of
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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5b1db1cde71a11f45defdc7c7b0ca53d6f1ff3e)
For the other inclusion,
, we want to do the opposite.
We want to show that for any choice of
and
there are appropriate
and
.
So fix
and
and solve for
and
:
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/426f6fd6b84615075a38081be952831d9c1858ba)
shows that
and
.
Thus any vector from
![{\displaystyle {\begin{pmatrix}4\\1\\0\end{pmatrix}}m+{\begin{pmatrix}-1\\-3\\0\end{pmatrix}}n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c81c4dcde5d585bc5272563b2f334f0ae5bbf35d)
is also of the right form for
![{\displaystyle {\begin{pmatrix}1\\-1\\0\end{pmatrix}}((3/2)m+n)+{\begin{pmatrix}1\\1\\0\end{pmatrix}}((5/2)m-2n).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/432e71718bf470ceaac21caa3c2ae1f1e0bb3a5e)
- 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} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d73657126d2915b6cd82a365ce8fbc977a5e92)
are not equal sets.
While
is a subset of
, it is a proper subset of
because
is not a subset of
.
To see that, observe first that given a vector from
we can express it in the form for
— if
we fix
and
, we can solve for appropriate
,
, and
:
![{\displaystyle {\begin{array}{*{3}{rc}r}m&&&+&p&=&x+y\\&&n&&&=&2x\\&&&&p&=&y\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a06492d9bd24e5a8a5727bdcd577b991efd5c6c)
shows that that any
![{\displaystyle {\vec {v}}={\begin{pmatrix}1\\2\\0\end{pmatrix}}x+{\begin{pmatrix}1\\0\\1\end{pmatrix}}y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdc93ea222b46a4e3c142a9db319d3ae745201d8)
can be expressed as a member of
with
,
, and
:
![{\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/903665d694a6eb7fb9db20a7cf91e9452912715b)
Thus
.
But, for the other direction, the reduction
resulting from fixing
,
, and
and looking for
and
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f31f6502a17d50bba5d6db09993eef416f025f30)
shows that the only vectors
![{\displaystyle {\begin{pmatrix}m+p\\n\\p\end{pmatrix}}\in R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8424346c3a642d5a2d83c4d5f45d6c974dcbd640)
representable in the form
![{\displaystyle {\begin{pmatrix}x+y\\2x\\y\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ca3a8d3a17ccd261d54fdd54980ad0c4a4008b6)
are those where
.
For instance,
![{\displaystyle {\begin{pmatrix}0\\1\\0\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7d57e492d0987e3eee9ec524db5606ea9cc3c31)
is in
but not in
.
- Problem 1
Decide if the vector is a member of the set.
-
,
-
,
-
,
-
,
-
,
-
,
- 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} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e06af85e2accf02b0ea16ea73385e8b5ab4602d7)
- 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} \},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07565c1847b3fc6ec96930b4e99fe75d65dd0dd0)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1b11ad37320e7b8dfb3a6861e2efb121429a17e)
- This exercise is recommended for all readers.
- Problem 5
Decide if the sets are equal.
-
and
-
and
-
and
-
and
-
and
Solutions
- ↑ More information on set equality is in the appendix.