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Linear Algebra/Comparing Set Descriptions/Solutions

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Solutions

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Problem 1

Decide if the vector is a member of the set.

  1. (23), {(12)k|k∈ℝ}
  2. (−33), {(1−1)k|k∈ℝ}
  3. (−334), {(1−12)k|k∈ℝ}
  4. (−334), {(1−12)k+(002)m|k,m∈ℝ}
  5. (1414), {(225)k+(−102)m|k,m∈ℝ}
  6. (146), {(225)k+(−102)m|k,m∈ℝ}
Answer
  1. No.
  2. Yes.
  3. No.
  4. Yes.
  5. Yes; use Gauss' method to get k=4 and m=−3.
  6. No; use Gauss' method to conclude that there is no solution.
Problem 2

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

{(2−5)k|k∈ℝ}
Answer

One easy thing to do is to double and triple the vector:

{(4−10)k|k∈ℝ}and{(6−15)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.

Answer

Instead of showing all three equalities, we can show that the first equals the second, and that the second equals the third. Both equalities are easy, using the methods of this subsection.

This exercise is recommended for all readers.
Problem 4

Show that these sets are equal

{(1411)+(−1010)z|z∈ℝ}and{(0421)+(−1010)k|k∈ℝ},

and that both describe the solution set of this system.

x−y+z+w=−1y−w=3x+z+2w=4
Answer

That system reduces like this:

→−ρ1+ρ2x−y+z+w=−1y−w=3y+w=5→−ρ2+ρ3x−y+z+w=−1y−w=32w=2

showing that w=1, y=4 and x=2−z.

This exercise is recommended for all readers.
Problem 5

Decide if the sets are equal.

  1. {(12)+(03)t|t∈ℝ} and {(18)+(0−1)s|s∈ℝ}
  2. {(131)t+(215)s|t,s∈ℝ} and {(477)m+(−4−2−10)n|m,n∈ℝ}
  3. {(12)t|t∈ℝ} and {(24)m+(48)n|m,n∈ℝ}
  4. {(102)s+(−110)t|s,t∈ℝ} and {(−111)m+(013)n|m,n∈ℝ}
  5. {(131)t+(246)s|t,s∈ℝ} and {(377)t+(131)s|t,s∈ℝ}
Answer

For each item, we call the first set S1 and the other S2.

  1. They are equal. To see that S1⊆S2, we must show that any element of the first set is in the second, that is, for any vector of the form
    v→=(12)+(03)t
    there is an appropriate s such that
    v→=(18)+(0−1)s.
    Restated, given t we must find s so that this holds.
    1+0s=1+0t8−1s=2+3t
    That system reduces to
    1=1s=6−3t
    That is,
    (12)+(03)t=(18)+(0−1)(6−3t)
    and so any vector in the form for S1 can be stated in the form needed for inclusion in S2. For S2⊆S1, we look for t so that these equations hold.
    1+0t=1+0s2+3t=8−1s
    Rewrite that as
    1=1t=2−(1/3)s
    and so
    (18)+(0−1)s=(12)+(03)(2−(1/3)s).
  2. These two are equal. To show that S1⊆S2, we check that for any t,s we can find an appropriate m,n so that these hold.
    4m−4n=1t+2s7m−2n=3t+1s7m−10n=1t+5s
    Use Gauss' method
    (4−41t+2s7−23t+1s7−101t+5s)→(−7/4)ρ1+ρ3(−7/4)ρ1+ρ2(4−41t+2s05(5/4)t−(10/4)s0−3−(3/4)t+(6/4)s)→(3/5)ρ2+ρ3(4−41t+2s05(5/4)t−(10/4)s000)
    to conclude that
    (131)t+(215)s=(477)((1/2)t)+(−4−2−10)((1/4)t−(1/2)s)
    and so S1⊆S2. For S2⊆S1, solve
    1t+2s=4m−4n3t+1s=7m−2n1t+5s=7m−10n
    with Gaussian reduction
    (124m−4n317m−2n157m−10n)→−ρ1+ρ3−3ρ1+ρ2(124m−4n0−5−5m+10n033m−6n)→(3/5)ρ2+ρ3(124m−4n0−5−5m+10n000)
    to get
    (477)m+(−4−2−10)n=(131)(2m)+(215)(m−2n)
    and so any member of S2 can be expressed in the form needed for S1.
  3. These sets are equal. To prove that S1⊆S2, we must be able to solve
    2m+4n=1t4m+8n=2t
    for m and n in terms of t. Apply Gaussian reduction
    (241t482t)→−2ρ1+ρ2(241t000)
    to conclude that any pair m,n where 2m+4n=t will do. For instance,
    (12)t=(24)((1/2)t)+(48)(0)
    or
    (12)t=(24)((−3/2)t)+(48)(t).
    Thus S1⊆S2. For S2⊆S1, we solve
    1t=2m+4n2t=4m+8n
    with Gauss' method
    (12m+4n24m+8n)→−2ρ1+ρ2(12m+4n00)
    to deduce that any vector in S2 is also in S1.
    (24)m+(48)n=(12)(2m+4n)∈S1.
  4. Neither set is a subset of the other. For S1⊆S2 to hold we must be able to solve
    −1m+0n=1s−1t1m+1n=0s+1t1m+3n=2s+0t
    for m and n in terms of t and s. Gauss' method
    (−101s−1t110s+1t132s+0t)→ρ1+ρ3ρ1+ρ2(−101s−1t011s+0t033s−1t)→−3ρ2+ρ3(−101s−1t011s+0t030s−1t)
    shows that we can only find an appropriate pair m,n when t=0. That is,
    (−110)
    has no expression of the form
    (−111)m+(013)n.
    Having shown that S1 is not a subset of S2, we know S1≠S2 so, strictly speaking, we need not go further. But we shall also show that S2 is not a subset of S1. For S2⊆S1 to hold, we must be able to solve
    1s−1t=−1m+0n0s+1t=1m+1n2s+0t=1m+3n
    for s and t. Apply row reduction
    (1−1−1m+0n011m+1n201m+3n)→−2ρ1+ρ3(1−1−1m+0n011m+1n023m+3n)→−2ρ2+ρ3(1−1−1m+0n011m+1n001m+1n)
    to deduce that the only vectors from S2 that are also in S1 are of the form
    (−111)m+(013)(−m).
    For instance,
    (−111)
    is in S2 but not in S1.
  5. These sets are equal. First we change the parameters:
    S2={(377)m+(131)n|m,n∈ℝ}.
    Now, to show that S1⊆S2, we solve
    3m+1n=1t+2s7m+3n=3t+4s7m+1n=1t+6s
    with Gauss' method
    (311t+2s733t+4s711t+6s)→(−7/3)ρ1+ρ3(−7/3)ρ1+ρ2(311t+2s02/3(2/3)t−(2/3)s0−4/3(−4/3)t+(4/3)s)→2ρ2+ρ3(311t+2s02/3(2/3)t−(2/3)s000)
    to get that
    (131)t+(246)s=(377)(s)+(131)(t−s)
    and so S1⊆S2. The proof that S2⊆S1 involves solving
    1t+2s=3m+1n3t+4s=7m+3n1t+6s=7m+1n
    with Gaussian reduction
    (123m+1n347m+3n167m+1n)→−ρ1+ρ3−3ρ1+ρ2(123m+1n0−2−2m044m)→2ρ2+ρ3(123m+1n0−2−2m000)
    to conclude
    (377)m+(131)n=(131)(m+n)+(246)(m)
    and so any vector in S2 is also in S1.
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