Linear Algebra Exercises-n-Answers.pdf

42 Linear Algebra, by Hefferon
(e) No. The equation
leads to this reduction.
⎛
⎝ 2 3 5 6 x
⎞
1 0 1 0 y
1 1 2 2 z
⎛
r 1
⎝ 2 ⎞ ⎛
1⎠ + r 2
⎝ 3 ⎞ ⎛
0⎠ + r 3
⎝ 5 ⎞ ⎛
1⎠ + r 4
⎝ 6 ⎞ ⎛
0⎠ = ⎝ x ⎞
y⎠
1 1 2 2 z
⎠ −(1/2)ρ1+ρ2 −(1/3)ρ 2+ρ 3
−→ −→
−(1/2)ρ 1 +ρ 3
⎛
⎝ 2 3 5 6 x
0 −3/2 −3/2 −3 −(1/2)x + y
0 0 0 0 −(1/3)x − (1/3)y + z
This shows that not every three-tall vector can be so expressed. Only the vectors satisfying the
restriction that −(1/3)x − (1/3)y + z = 0 are in the span. (To see that any such vector is indeed
expressible, take r 3 and r 4 to be zero and solve for r 1 and r 2 in terms of x, y, and z by backsubstitution.)
Two.I.2.25 (a) { ( c b c ) ∣ b, c ∈ R} = {b ( 0 1 0 ) + c ( 1 0 1 ) ∣ b, c ∈ R} The obvious choice
for the set that spans is { ( 0 1 0 ) , ( 1 0 1 ) ( ) ( ) (
}.
) ( )
−d b ∣∣ 0 1 0 0 −1 0 ∣∣
(b) {
b, c, d ∈ R} = {b + c + d
b, c, d ∈ R} One set that spans
c d
0 0 1 0 0 1
this space consists of those three matrices.
(c) The system
a + 3b = 0
2a −c − d = 0
gives b = −(c + d)/6 and a = (c + d)/2. So one description is this.
( ) ( )
1/2 −1/6 1/2 −1/6 ∣∣
{c
+ d
c, d ∈ R}
1 0 0 1
That shows that a set spanning this subspace consists of those two matrices.
(d) The a = 2b − c gives {(2b − c) + bx + cx 3 ∣ ∣ b, c ∈ R} = {b(2 + x) + c(−1 + x 3 ) ∣ ∣ b, c ∈ R}. So the
subspace is the span of the set {2 + x, −1 + x 3 }.
(e) The set {a + bx + cx 2 ∣ ∣ a + 7b + 49c = 0} paramatrized as {b(−7 + x) + c(−49 + x 2 ) ∣ ∣ b, c ∈ R}
has the spanning set {−7 + x, −49 + x 2 }.
Two.I.2.26 Each answer given is only one out of many possible.
(a) We can paramatrize in this way
⎛ ⎞
⎛ ⎞ ⎛ ⎞
giving this for a spanning set.
{ ⎝ x 0⎠ ∣ x, z ∈ R} = {x ⎝ 1 0
z
0
⎛
⎞
⎛
⎠ + z
⎞
⎝ 0 0⎠ ∣ x, z ∈ R}
1
{ ⎝ 1 0⎠ , ⎝ 0 0⎠}
0 1
⎛
(b) Paramatrize it with {y ⎝ −2/3
⎞ ⎛
1 ⎠ + z ⎝ −1/3
⎞
⎛
0 ⎠ ∣ y, z ∈ R} to get { ⎝ −2/3
⎞ ⎛
1 ⎠ , ⎝ −1/3
⎞
0 ⎠}.
⎛
0
1
0 1
⎞ ⎛ ⎞
1 −1/2
(c) { ⎜−2
⎟
⎝ 1 ⎠ , ⎜ 0
⎟
⎝ 0 ⎠ }
0 1
(d) Paramatrize the description as {−a 1 + a 1 x + a 3 x 2 + a 3 x ∣ 3 a1 , a 3 ∈ R} to get {−1 + x, x 2 + x 3 }.
(e) {1, ( x, x 2 ), x 3 (, x 4 } ) ( ) ( )
1 0 0 1 0 0 0 0
(f) { , , , }
0 0 0 0 1 0 0 1
Two.I.2.27 Technically, no. Subspaces of R 3 are sets of three-tall vectors, while R 2 is a set of two-tall
vectors. Clearly though, R 2 is “just like” this subspace of R 3 .
⎛ ⎞
{ ⎝ x y⎠ ∣ x, y ∈ R}
0
⎞
⎠