Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 79
Three.I.1.20 If n ≥ 1 then P n−1
∼ = R n . (If we take P −1 and R 0 to be trivial vector spaces, then the
relationship extends one dimension lower.) The natural isomorphism between them is this.
⎛ ⎞
a 0
a 0 + a 1 x + · · · + a n−1 x n−1 a 1
↦→ ⎜
⎝
⎟
. ⎠
a n−1
Checking that it is an isomorphism is straightforward.
Three.I.1.21 This is the map, expanded.
f(a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 ) = a 0 + a 1 (x − 1) + a 2 (x − 1) 2 + a 3 (x − 1) 3
+ a 4 (x − 1) 4 + a 5 (x − 1) 5
= a 0 + a 1 (x − 1) + a 2 (x 2 − 2x + 1)
+ a 3 (x 3 − 3x 2 + 3x − 1)
+ a 4 (x 4 − 4x 3 + 6x 2 − 4x + 1)
+ a 5 (x 5 − 5x 4 + 10x 3 − 10x 2 + 5x − 1)
= (a 0 − a 1 + a 2 − a 3 + a 4 − a 5 )
+ (a 1 − 2a 2 + 3a 3 − 4a 4 + 5a 5 )x
+ (a 2 − 3a 3 + 6a 4 − 10a 5 )x 2 + (a 3 − 4a 4 + 10a 5 )x 3
+ (a 4 − 5a 5 )x 4 + a 5 x 5
This map is a correspondence because it has an inverse, the map p(x) ↦→ p(x + 1).
To finish checking that it is an isomorphism, we apply item (2) of Lemma 1.9 and show that it
preserves linear combinations of two polynomials. Briefly, the check goes like this.
f(c · (a 0 + a 1 x + · · · + a 5 x 5 ) + d · (b 0 + b 1 x + · · · + b 5 x 5 ))
= · · · = (ca 0 − ca 1 + ca 2 − ca 3 + ca 4 − ca 5 + db 0 − db 1 + db 2 − db 3 + db 4 − db 5 ) + · · · + (ca 5 + db 5 )x 5
Three.I.1.22
= · · · = c · f(a 0 + a 1 x + · · · + a 5 x 5 ) + d · f(b 0 + b 1 x + · · · + b 5 x 5 )
No vector space has the empty set underlying it. We can take ⃗v to be the zero vector.
Three.I.1.23 Yes; where the two spaces are {⃗a} and { ⃗ b}, the map sending ⃗a to ⃗ b is clearly one-to-one
and onto, and also preserves what little structure there is.
Three.I.1.24 A linear combination of n = 0 vectors adds to the zero vector and so Lemma 1.8 shows
that the three statements are equivalent in this case.
Three.I.1.25 Consider the basis 〈1〉 for P 0 and let f(1) ∈ R be k. For any a ∈ P 0 we have that
f(a) = f(a · 1) = af(1) = ak and so f’s action is multiplication by k. Note that k ≠ 0 or else the map
is not one-to-one. (Incidentally, any such map a ↦→ ka is an isomorphism, as is easy to check.)
Three.I.1.26 In each item, following item (2) of Lemma 1.9, we show that the map preserves structure
by showing that the it preserves linear combinations of two members of the domain.
(a) The identity map is clearly one-to-one and onto. For linear combinations the check is easy.
id(c 1 · ⃗v 1 + c 2 · ⃗v 2 ) = c 1 ⃗v 1 + c 2 ⃗v 2 = c 1 · id(⃗v 1 ) + c 2 · id(⃗v 2 )
(b) The inverse of a correspondence is also a correspondence (as stated in the appendix), so we need
only check that the inverse preserves linear combinations. Assume that ⃗w 1 = f(⃗v 1 ) (so f −1 ( ⃗w 1 ) = ⃗v 1 )
and assume that ⃗w 2 = f(⃗v 2 ).
f −1 (c 1 · ⃗w 1 + c 2 · ⃗w 2 ) = f −1( c 1 · f(⃗v 1 ) + c 2 · f(⃗v 2 ) )
= f −1 ( f ( c 1 ⃗v 1 + c 2 ⃗v 2 ) )
= c 1 ⃗v 1 + c 2 ⃗v 2
= c 1 · f −1 ( ⃗w 1 ) + c 2 · f −1 ( ⃗w 2 )
(c) The composition of two correspondences is a correspondence (as stated in the appendix), so we
need only check that the composition map preserves linear combinations.
g ◦ f ( c 1 · ⃗v 1 + c 2 · ⃗v 2
)
= g
(
f(c1 ⃗v 1 + c 2 ⃗v 2 ) )
= g ( c 1 · f(⃗v 1 ) + c 2 · f(⃗v 2 ) )
= c 1 · g ( f(⃗v 1 )) + c 2 · g(f(⃗v 2 ) )
= c 1 · g ◦ f (⃗v 1 ) + c 2 · g ◦ f (⃗v 2 )