Linear Algebra Exercises-n-Answers.pdf

84 Linear Algebra, by Hefferon
(c) Yes, they have the same dimension. One isomorphism is this.
⎛ ⎞
( ) a
a b c ⎜
↦→
d e f
⎝
⎟
. ⎠
f
(d) Yes, they have the same dimension. This is an isomorphism.
( )
a + bx + · · · + fx 5 a b c
↦→
d e f
(e) Yes, both have dimension 2k.
(
5
Three.I.2.9 (a) Rep B (3 − 2x) =
−2)
Three.I.2.10
Three.I.2.11
Three.I.2.12
Three.I.2.13
They have different dimensions.
Yes, both are mn-dimensional.
(b)
(
0
2)
(c)
( )
−1
1
Yes, any two (nondegenerate) planes are both two-dimensional vector spaces.
There are many answers, one is the set of P k (taking P −1 to be the trivial vector space).
Three.I.2.14 False (except when n = 0). For instance, if f : V → R n is an isomorphism then multiplying
by any nonzero scalar, gives another, different, isomorphism. (Between trivial spaces the
isomorphisms are unique; the only map possible is ⃗0 V ↦→ 0 W .)
Three.I.2.15 No. A proper subspace has a strictly lower dimension than it’s superspace; if U is
a proper subspace of V then any linearly independent subset of U must have fewer than dim(V )
members or else that set would be a basis for V , and U wouldn’t be proper.
Three.I.2.16
Three.I.2.17
Where B = 〈 β ⃗ 1 , . . . , β ⃗ n 〉, the inverse is this.
⎛ ⎞
1..
⎜ ⎟
⎝c
⎠ ↦→ c 1β1 ⃗ + · · · + c nβn
⃗
c n
All three spaces have dimension equal to the rank of the matrix.
Three.I.2.18 We must show that if ⃗a = ⃗ b then f(⃗a) = f( ⃗ b). So suppose that a 1β1 ⃗ + · · · + a nβn ⃗ =
b 1β1 ⃗ + · · · + b nβn ⃗ . Each vector in a vector space (here, the domain space) has a unique representation
as a linear combination of basis vectors, so we can conclude that a 1 = b 1 , . . . , a n = b n . Thus,
⎛ ⎛
and so the function is well-defined.
Three.I.2.19
f(⃗a) =
⎜
⎝
⎞
a 1
.
a n
⎟
⎠ =
⎞
1
⎜
⎝b
. b n
⎟
⎠ = f( ⃗ b)
Yes, because a zero-dimensional space is a trivial space.
Three.I.2.20 (a) No, this collection has no spaces of odd dimension.
(b) Yes, because P k
∼ = R k+1 .
(c) No, for instance, M 2×3
∼ = M3×2 .
Three.I.2.21 One direction is easy: if the two are isomorphic via f then for any basis B ⊆ V , the set
D = f(B) is also a basis (this is shown in Lemma 2.3). The check that corresponding vectors have the
same coordinates: f(c 1β1 ⃗ + · · · + c nβn ⃗ ) = c 1 f( β ⃗ 1 ) + · · · + c n f( β ⃗ n ) = c 1
⃗ δ1 + · · · + c n
⃗ δn is routine.
For the other half, assume that there are bases such that corresponding vectors have the same
coordinates with respect to those bases. Because f is a correspondence, to show that it is an
isomorphism, we need only show that it preserves structure. Because Rep B (⃗v ) = Rep D (f(⃗v )),
the map f preserves structure if and only if representations preserve addition: Rep B (⃗v 1 + ⃗v 2 ) =
Rep B (⃗v 1 ) + Rep B (⃗v 2 ) and scalar multiplication: Rep B (r ·⃗v ) = r · Rep B (⃗v ) The addition calculation is
this: (c 1 +d 1 ) β ⃗ 1 +· · ·+(c n +d n ) β ⃗ n = c 1β1 ⃗ +· · ·+c nβn ⃗ +d 1β1 ⃗ +· · ·+d nβn ⃗ , and the scalar multiplication
calculation is similar.
Three.I.2.22 (a) Pulling the definition back from R 4 to P 3 gives that a 0 + a 1 x + a 2 x 2 + a 3 x 3 is
orthogonal to b 0 + b 1 x + b 2 x 2 + b 3 x 3 if and only if a 0 b 0 + a 1 b 1 + a 2 b 2 + a 3 b 3 = 0.