Linear Algebra Exercises-n-Answers.pdf

104 Linear Algebra, by Hefferon
(c) Yes; we can simply observe that the vector is the first column minus the second. Or, failing that,
setting up the relationship among the columns
⎛
c 1
⎝ 1 ⎞ ⎛
1 ⎠ + c 2
⎝ −1
⎞ ⎛
1 ⎠ + c 3
⎝ 1
⎞ ⎛
−1⎠ = ⎝ 2 ⎞
0⎠
−1 −1 1 0
and considering the resulting linear system
c 1 − c 2 + c 3 = 2 c 1 − c 2 + c 3 = 2 c 1 − c 2 + c 3 = 2
−ρ 1+ρ 2
ρ 2+ρ 3
c 1 + c 2 − c 3 = 0 −→ 2c 2 − 2c 3 = −2 −→ 2c 2 − 2c 3 = −2
ρ
−c 1 − c 2 + c 3 = 0 1+ρ 3
−2c 2 + 2c 3 = 2
0 = 0
gives the additional information (beyond that there is at least one solution) that there are infinitely
many solutions. Paramatizing gives c 2 = −1 + c 3 and c 1 = 1, and so taking c 3 to be zero gives a
particular solution of c 1 = 1, c 2 = −1, and c 3 = 0 (which is, of course, the observation made at the
start).
Three.III.2.10 As described in the subsection, with respect to the standard bases, representations are
transparent, and so, for instance, the first matrix describes this map.
⎛
⎝ 1 ⎞ ⎛ ⎞
⎛ ⎞
⎛
1 ( ( 0 (
0⎠ = ⎝ 1 1
0 ↦→ = ⎝ 1
1⎠↦→
⎝
0)
0)
1) 0 ⎞
(
3
0⎠↦→
4)
0 0
E 2 0
1
⎠E 3
So, for this first one, we are asking whether thare are scalars such that
( ) ( ) ( ) (
1 1 3 1
c 1 + c
0 2 + c
1 3 =
4 3)
that is, whether the vector is in the column space of the matrix.
(a) Yes. We can get this conclusion by setting up the resulting linear system and applying Gauss’
method, as usual. Another way to get it is to note by inspection of the equation of columns that
taking c 3 = 3/4, and c 1 = −5/4, and c 2 = 0 will do. Still a third way to get this conclusion is to
note that the rank of the matrix is two, which equals the dimension of the codomain, and so the
map is onto — the range is all of R 2 and in particular includes the given vector.
(b) No; note that all of the columns in the matrix have a second component that is twice the first,
while the vector does not. Alternatively, the column space of the matrix is
( ) ( ) ( ( )
2 0 3 ∣∣ 1 ∣∣
{c 1 + c
4 2 + c
0 3 c
6)
1 , c 2 , c 3 ∈ R} = {c c ∈ R}
2
(which is the fact already noted, but was arrived at by calculation rather than inspiration), and the
given vector is not in this set.
Three.III.2.11 (a) The first member of the basis
( (
0 1
=
1)
0)
B
is mapped to ( )
1/2
−1/2
D
which is this member of the codomain.
( )
1 1
2 · − 1 ( (
1 0
1 2 −1)
· =
1)
(b) The second member of the basis is mapped
( (
1 0
=
0)
1)
B
↦→
( )
(1/2
1/2
D
to this member of the codomain.
( )
1 1
2 · + 1 ( (
1 1
1 2 −1)
· =
0)
(c) Because the map that the matrix represents is the identity map on the basis, it must be the
identity on all members of the domain. We can come to the same conclusion in another way by
considering ( (
x y
y)
x)
=
B