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{matrix}1 \\ 2 \\ 3\end{matrix} ) z \ | \ z ∈ ℝ \} \ \text{and} \ \{ ( \begin{matrix}2 \\ 4 \\ 6\end{matrix} ) w \ | \ w ∈ ℝ \} .

For instance, this set contains

( \begin{matrix}5 \\ 10 \\ 15\end{matrix} )

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

( \begin{matrix}4 \\ 8 \\ 11\end{matrix} )

(the first component gives z = 4 but that clashes with the third component,
similarly the first component gives w = 2 but the third component
gives something different).
Here is a third description of the same set:

\{ ( \begin{matrix}3 \\ 6 \\ 9\end{matrix} ) + ( \begin{matrix}− 1 \\ − 2 \\ − 3\end{matrix} ) y \ | \ y ∈ 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}.[1]

Example 4.1

To show that

{S}_{1} = \{ ( \begin{matrix}1 \\ − 1 \\ 0\end{matrix} ) c + ( \begin{matrix}1 \\ 1 \\ 0\end{matrix} ) d \ | \ c , d ∈ ℝ \}

equals

{S}_{2} = \{ ( \begin{matrix}4 \\ 1 \\ 0\end{matrix} ) m + ( \begin{matrix}− 1 \\ − 3 \\ 0\end{matrix} ) n \ | \ m , n ∈ ℝ \}

we show first that {S}_{1} ⊆ {S}_{2} and then that {S}_{2} ⊆ {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{matrix}4 \\ 1 \\ 0\end{matrix} ) m + ( \begin{matrix}− 1 \\ − 3 \\ 0\end{matrix} ) n = ( \begin{matrix}2 \\ 0 \\ 0\end{matrix} )

that is, this relation holds between m and n.

\begin{matrix}4 m & − & n & = & 2 \\ 1 m & − & 3 n & = & 0 \\ & & 0 & = & 0\end{matrix}

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{matrix}4 \\ 1 \\ 0\end{matrix} ) m + ( \begin{matrix}− 1 \\ − 3 \\ 0\end{matrix} ) n = ( \begin{matrix}3 \\ − 3 \\ 0\end{matrix} )

that is, this holds.

\begin{matrix}4 m & − & n & = & 3 \\ 1 m & − & 3 n & = & − 3 \\ & & 0 & = & 0\end{matrix}

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{matrix}c + d \\ − c + d \\ 0\end{matrix} ) ∈ {S}_{1}

and look for m and n such that

( \begin{matrix}4 \\ 1 \\ 0\end{matrix} ) m + ( \begin{matrix}− 1 \\ − 3 \\ 0\end{matrix} ) n = ( \begin{matrix}c + d \\ − c + d \\ 0\end{matrix} )

that is, this is true.

\begin{matrix}4 m & − & n & = & c + d \\ m & − & 3 n & = & − c + d \\ & & 0 & = & 0\end{matrix}

Applying Gauss' method

\begin{matrix}\begin{matrix}4 m & − & n & = & c + d \\ m & − & 3 n & = & − c + d\end{matrix} & {→}_{ \ }^{− ( 1 / 4 ) {ρ}_{1} + {ρ}_{2}} & \begin{matrix}4 m & − & n & = & c + d \\ & & − ( 11 / 4 ) n & = & − ( 5 / 4 ) c + ( 3 / 4 ) d\end{matrix}\end{matrix}

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

For the other inclusion, {S}_{2} ⊆ {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{matrix}\begin{matrix}c & + & d & = & 4 m − n \\ − c & + & d & = & m − 3 n\end{matrix} & {→}_{ \ }^{{ρ}_{1} + {ρ}_{2}} & \begin{matrix}c & + & d & = & 4 m − n \\ & & 2 d & = & 5 m − 4 n\end{matrix}\end{matrix}

shows that d = ( 5 / 2 ) m − 2 n and c = ( 3 / 2 ) m + n.
Thus any vector from {S}_{2}

( \begin{matrix}4 \\ 1 \\ 0\end{matrix} ) m + ( \begin{matrix}− 1 \\ − 3 \\ 0\end{matrix} ) n

is also of the right form for {S}_{1}

( \begin{matrix}1 \\ − 1 \\ 0\end{matrix} ) ( ( 3 / 2 ) m + n ) + ( \begin{matrix}1 \\ 1 \\ 0\end{matrix} ) ( ( 5 / 2 ) m − 2 n ) .

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{matrix}x + y \\ 2 x \\ y\end{matrix} ) \ | \ x , y ∈ ℝ \} \ \text{and} \ R = \{ ( \begin{matrix}m + p \\ n \\ p\end{matrix} ) \ | \ m , n , p ∈ 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{matrix}m & & & + & p & = & x + y \\ & & n & & & = & 2 x \\ & & & & p & = & y\end{matrix}

shows that that any

\overset{→}{v} = ( \begin{matrix}1 \\ 2 \\ 0\end{matrix} ) x + ( \begin{matrix}1 \\ 0 \\ 1\end{matrix} ) y

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

\overset{→}{v} = ( \begin{matrix}1 \\ 0 \\ 0\end{matrix} ) x + ( \begin{matrix}0 \\ 1 \\ 0\end{matrix} ) 2 x + ( \begin{matrix}1 \\ 0 \\ 1\end{matrix} ) y .

Thus P ⊆ R.

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

\begin{matrix}\begin{matrix}x & + & y & = & m + p \\ 2 x & & & = & n \\ & & y & = & p\end{matrix} & {→}_{ \ }^{− 2 {ρ}_{1} + {ρ}_{2}} & \begin{matrix}x & + & y & = & m + p \\ & & − 2 y & = & − 2 m + n − 2 p \\ & & y & = & p\end{matrix} \\ & {→}_{ \ }^{( 1 / 2 ) {ρ}_{2} + {ρ}_{3}} & \begin{matrix}x & + & y & = & m + p \\ & & − 2 y & = & − 2 m + n − 2 p \\ & & 0 & = & m + ( 1 / 2 ) n\end{matrix}\end{matrix}

shows that the only vectors

( \begin{matrix}m + p \\ n \\ p\end{matrix} ) ∈ R

representable in the form

( \begin{matrix}x + y \\ 2 x \\ y\end{matrix} )

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

( \begin{matrix}0 \\ 1 \\ 0\end{matrix} )

is in R but not in P.

Exercises

Problem 1

Decide if the vector is a member of the set.

• ( \begin{matrix}2 \\ 3\end{matrix} ),
  \{ ( \begin{matrix}1 \\ 2\end{matrix} ) k \ | \ k ∈ ℝ \}
• ( \begin{matrix}− 3 \\ 3\end{matrix} ),

\{ ( \begin{matrix}1 \\ − 1\end{matrix} ) k \ | \ k ∈ ℝ \}

• ( \begin{matrix}− 3 \\ 3 \\ 4\end{matrix} ),
  \{ ( \begin{matrix}1 \\ − 1 \\ 2\end{matrix} ) k \ | \ k ∈ ℝ \}
• ( \begin{matrix}− 3 \\ 3 \\ 4\end{matrix} ),

\{ ( \begin{matrix}1 \\ − 1 \\ 2\end{matrix} ) k + ( \begin{matrix}0 \\ 0 \\ 2\end{matrix} ) m \ | \ k , m ∈ ℝ \}

• ( \begin{matrix}1 \\ 4 \\ 14\end{matrix} ),
  \{ ( \begin{matrix}2 \\ 2 \\ 5\end{matrix} ) k + ( \begin{matrix}− 1 \\ 0 \\ 2\end{matrix} ) m \ | \ k , m ∈ ℝ \}
• ( \begin{matrix}1 \\ 4 \\ 6\end{matrix} ),

\{ ( \begin{matrix}2 \\ 2 \\ 5\end{matrix} ) k + ( \begin{matrix}− 1 \\ 0 \\ 2\end{matrix} ) m \ | \ k , m ∈ ℝ \}

Problem 2

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

\{ ( \begin{matrix}2 \\ − 5\end{matrix} ) k \ | \ k ∈ 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{matrix}1 \\ 4 \\ 1 \\ 1\end{matrix} ) + ( \begin{matrix}− 1 \\ 0 \\ 1 \\ 0\end{matrix} ) z \ | \ z ∈ ℝ \} \ \text{and} \ \{ ( \begin{matrix}0 \\ 4 \\ 2 \\ 1\end{matrix} ) + ( \begin{matrix}− 1 \\ 0 \\ 1 \\ 0\end{matrix} ) k \ | \ k ∈ ℝ \} ,

and that both describe the solution set of this system.

\begin{matrix}x & − & y & + & z & + & w & = & − 1 \\ & & y & & & − & w & = & 3 \\ x & & & + & z & + & 2 w & = & 4\end{matrix}

This exercise is recommended for all readers.

Problem 5

Decide if the sets are equal.

• \{ ( \begin{matrix}1 \\ 2\end{matrix} ) + ( \begin{matrix}0 \\ 3\end{matrix} ) t \ | \ t ∈ ℝ \}

and
\{ ( \begin{matrix}1 \\ 8\end{matrix} ) + ( \begin{matrix}0 \\ − 1\end{matrix} ) s \ | \ s ∈ ℝ \}

• \{ ( \begin{matrix}1 \\ 3 \\ 1\end{matrix} ) t + ( \begin{matrix}2 \\ 1 \\ 5\end{matrix} ) s \ | \ t , s ∈ ℝ \}

and
\{ ( \begin{matrix}4 \\ 7 \\ 7\end{matrix} ) m + ( \begin{matrix}− 4 \\ − 2 \\ − 10\end{matrix} ) n \ | \ m , n ∈ ℝ \}

• \{ ( \begin{matrix}1 \\ 2\end{matrix} ) t \ | \ t ∈ ℝ \}

and
\{ ( \begin{matrix}2 \\ 4\end{matrix} ) m + ( \begin{matrix}4 \\ 8\end{matrix} ) n \ | \ m , n ∈ ℝ \}

• \{ ( \begin{matrix}1 \\ 0 \\ 2\end{matrix} ) s + ( \begin{matrix}− 1 \\ 1 \\ 0\end{matrix} ) t \ | \ s , t ∈ ℝ \}

and
\{ ( \begin{matrix}− 1 \\ 1 \\ 1\end{matrix} ) m + ( \begin{matrix}0 \\ 1 \\ 3\end{matrix} ) n \ | \ m , n ∈ ℝ \}

• \{ ( \begin{matrix}1 \\ 3 \\ 1\end{matrix} ) t + ( \begin{matrix}2 \\ 4 \\ 6\end{matrix} ) s \ | \ t , s ∈ ℝ \}

and
\{ ( \begin{matrix}3 \\ 7 \\ 7\end{matrix} ) t + ( \begin{matrix}1 \\ 3 \\ 1\end{matrix} ) s \ | \ t , s ∈ ℝ \}

Solutions

Footnotes

• ↑ More information on set equality is in the appendix.

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