Linear Algebra Exercises-n-Answers.pdf

76 Linear Algebra, by Hefferon
and this shows
that it preserves scalar multiplication.
f( r(a 0 + a 1 x + a 2 x 2 ) ) = f( (ra 0 ) + (ra 1 )x + (ra 2 )x 2 )
⎛
= ⎝ ra ⎞
0
ra 1
⎠
ra 2
⎛
= r · ⎝ a ⎞
0
a 1
⎠
a 2
= r f(a 0 + a 1 x + a 2 x 2 )
Three.I.1.11 ( These are ( the images. ( )
5 0 −1
(a) (b) (c)
−2)
2)
1
To prove that f is one-to-one, assume that it maps two linear polynomials to the same image
f(a 1 + b 1 x) = f(a 2 + b 2 x). Then
( ) ( )
a1 − b 1 a2 − b
=
2
b 1 b 2
and so, since column vectors are equal only when their components are equal, b 1 = b 2 and a 1 = a 2 .
That shows that the two linear polynomials are equal, and so f is one-to-one.
To show that f is onto, note that
( s
t)
is the image of (s − t) + tx.
To check that f preserves structure, we can use item (2) of Lemma 1.9.
f (c 1 · (a 1 + b 1 x) + c 2 · (a 2 + b 2 x)) = f ((c 1 a 1 + c 2 a 2 ) + (c 1 b 1 + c 2 b 2 )x)
( )
(c1 a
= 1 + c 2 a 2 ) − (c 1 b 1 + c 2 b 2 )
c 1 b 1 + c 2 b 2
( ) ( )
a1 − b
= c 1 ·
1 a2 − b
+ c
b 2 ·
2
1 b 2
= c 1 · f(a 1 + b 1 x) + c 2 · f(a 2 + b 2 x)
Three.I.1.12 To verify it is one-to-one, assume that f 1 (c 1 x + c 2 y + c 3 z) = f 1 (d 1 x + d 2 y + d 3 z). Then
c 1 + c 2 x + c 3 x 2 = d 1 + d 2 x + d 3 x 2 by the definition of f 1 . Members of P 2 are equal only when
they have the same coefficients, so this implies that c 1 = d 1 and c 2 = d 2 and c 3 = d 3 . Therefore
f 1 (c 1 x + c 2 y + c 3 z) = f 1 (d 1 x + d 2 y + d 3 z) implies that c 1 x + c 2 y + c 3 z = d 1 x + d 2 y + d 3 z, and so f 1 is
one-to-one.
To verify that it is onto, consider an arbitrary member of the codomain a 1 +a 2 x+a 3 x 2 and observe
that it is indeed the image of a member of the domain, namely, it is f 1 (a 1 x+a 2 y +a 3 z). (For instance,
0 + 3x + 6x 2 = f 1 (0x + 3y + 6z).)
The computation checking that f 1 preserves addition is this.
f 1 ( (c 1 x + c 2 y + c 3 z) + (d 1 x + d 2 y + d 3 z) ) = f 1 ( (c 1 + d 1 )x + (c 2 + d 2 )y + (c 3 + d 3 )z )
= (c 1 + d 1 ) + (c 2 + d 2 )x + (c 3 + d 3 )x 2
The check that f 1 preserves scalar multiplication is this.
= (c 1 + c 2 x + c 3 x 2 ) + (d 1 + d 2 x + d 3 x 2 )
= f 1 (c 1 x + c 2 y + c 3 z) + f 1 (d 1 x + d 2 y + d 3 z)
f 1 ( r · (c 1 x + c 2 y + c 3 z) ) = f 1 ( (rc 1 )x + (rc 2 )y + (rc 3 )z )
= (rc 1 ) + (rc 2 )x + (rc 3 )x 2
= r · (c 1 + c 2 x + c 3 x 2 )
= r · f 1 (c 1 x + c 2 y + c 3 z)
Three.I.1.13 (a) No; this map is not one-to-one. In particular, the matrix of all zeroes is mapped
to the same image as the matrix of all ones.