Linear Algebra/Comparing Set Descriptions

From Wikibooks, open books for an open world
< Linear Algebra
Jump to: navigation, search
Linear Algebra
 ← General = Particular + Homogeneous Comparing Set Descriptions Automation → 

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

Comparing Set Descriptions[edit]

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[edit]

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.[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[edit]

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

Footnotes[edit]

  1. More information on set equality is in the appendix.
Linear Algebra
 ← General = Particular + Homogeneous Comparing Set Descriptions Automation →