Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 61
for some a 3 , . . . , a n ∈ R. Therefore,
⎛ ⎞
−1
1 (
σ1 (⃗v) − σ 2 (⃗v) ) 1
=
0
y − x
⎜ ⎟
⎝ . ⎠
0
is in V . That is, ⃗e 2 − ⃗e 1 ∈ V , where ⃗e 1 , ⃗e 2 , . . . , ⃗e n is the standard basis for R n . Similarly, ⃗e 3 − ⃗e 2 ,
. . . , ⃗e n − ⃗e 1 are all in V . It is easy to see that the vectors ⃗e 2 − ⃗e 1 , ⃗e 3 − ⃗e 2 , . . . , ⃗e n − ⃗e 1 are linearly
independent (that is, form a linearly independent set), so dim V ≥ n − 1.
Finally, we can write
⃗v = x 1 ⃗e 1 + x 2 ⃗e 2 + · · · + x n ⃗e n
= (x 1 + x 2 + · · · + x n )⃗e 1 + x 2 (⃗e 2 − ⃗e 1 ) + · · · + x n (⃗e n − ⃗e 1 )
This shows that if x 1 + x 2 + · · · + x n = 0 then ⃗v is in the span of ⃗e 2 − ⃗e 1 , . . . , ⃗e n − ⃗e 1 (that is, is in
the span of the set of those vectors); similarly, each σ(⃗v) will be in this span, so V will equal this span
and dim V = n − 1. On the other hand, if x 1 + x 2 + · · · + x n ≠ 0 then the above equation shows that
⃗e 1 ∈ V and thus ⃗e 1 , . . . , ⃗e n ∈ V , so V = R n and dim V = n.
Subsection Two.III.3: Vector Spaces and Linear Systems
Two.III.3.16
(a)
( ) 2 3
1 1
(b)
( ) 2 1
1 3
(c)
⎛
⎝ 1 6
⎞
4 7⎠ (d) ( 0 0 0 ) (e)
3 8
( ) −1
−2
Two.III.3.17 (a) Yes. To see if there are c 1 and c 2 such that c 1 · (2
1 ) + c 2 · (3
1 ) = ( 1 0 ) we
solve
2c 1 + 3c 2 = 1
c 1 + c 2 = 0
and get c 1 = −1 and c 2 = ( 1. Thus) the vector ( is in ) the row( space. ) ( )
(b) No. The equation c 1 0 1 3 + c2 −1 0 1 + c3 −1 2 7 = 1 1 1 has no solution.
⎛
⎞
⎛
0 −1 −1 1
⎝1 0 2 1⎠ ρ 1↔ρ 2 −3ρ 1 +ρ 2 ρ 2 +ρ 3
−→ −→ −→ ⎝ 1 0 2 1 ⎞
0 −1 −1 1 ⎠
3 1 7 1
0 0 0 −1
Thus, the vector is not in the row space.
Two.III.3.18
(a) No. To see if there are c 1 , c 2 ∈ R such that
( ) ( ) (
1 1 1
c 1 + c
1 2 =
1 3)
we can use Gauss’ method on the resulting linear system.
c 1 + c 2 = 1 −ρ 1 +ρ 2 c −→ 1 + c 2 = 1
c 1 + c 2 = 3
0 = 2
There is no solution and so the vector is not in the column space.
(b) Yes. From this relationship
⎛
c 1
⎝ 1 ⎞ ⎛
2⎠ + c 2
⎝ 3 ⎞ ⎛
0 ⎠ + c 3
⎝ 1 ⎞ ⎛
4⎠ = ⎝ 1 ⎞
0⎠
1 −3 3 0
we get a linear system that, when Gauss’ method is applied,
⎛
⎝ 1 3 1 1
⎞
⎛
2 0 4 0⎠ −2ρ1+ρ2 −ρ 2+ρ 3
−→ −→ ⎝ 1 3 1 1
⎞
0 −6 2 −2⎠
−ρ
1 −3 −3 0 1+ρ 3
0 0 −6 1
yields a solution. Thus, the vector is in the column space.