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, S₁ and S₂, are equal is to show mutual inclusion: any member of S₁ is also in S₂, and any member of S₂ is also in S₁.

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 S₁ is also in S₂. We first consider two examples to use them as models for the general argument. If we make up a member of S₁ by trying c = 1 and d = 1, then to show that it is in S₂ 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 S₁ is in S₂ 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 S₁ is a member of S₂ 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 S₁ is a member of S₂ 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 S₂

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

is also of the right form for S₁

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

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