Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 77
(b) Yes, this is an isomorphism.
It is one-to-one:
( ) ( )
a1 b
if f( 1 a2 b
) = f( 2
) then
c 1 d 1 c 2 d 2
⎛
⎞ ⎛
⎞
a 1 + b 1 + c 1 + d 1 a 2 + b 2 + c 2 + d 2
⎜ a 1 + b 1 + c 1
⎟
⎝ a 1 + b 1
⎠ = ⎜
⎟
⎝
⎠
a 1
a 2 + b 2 + c 2
a 2 + b 2
a 2
gives that a 1 = a 2 , and that b 1 = b 2 , and that c 1 = c 2 , and that d 1 = d 2 .
It is onto, since this shows ⎛ ⎞
x
( )
⎜y
⎟ w z − w
⎝z
⎠ = f( )
y − z x − y
w
that any four-tall vector is the image of a 2×2 matrix.
Finally, it preserves combinations
( ) ( ) ( )
a1 b
f( r 1 · 1 a2 b
+ r
c 1 d 2 · 2 r1 a
) = f( 1 + r 2 a 2 r 1 b 1 + r 2 b 2
)
1 c 2 d 2 r 1 c 1 + r 2 c 2 r 1 d 1 + r 2 d 2
⎛
⎞
r 1 a 1 + · · · + r 2 d 2
= ⎜r 1 a 1 + · · · + r 2 c 2
⎟
⎝r 1 a 1 + · · · + r 2 b 2
⎠
r 1 a 1 + r 2 a 2
⎛
⎞ ⎛
⎞
a 1 + · · · + d 1 a 2 + · · · + d 2
= r 1 · ⎜a 1 + · · · + c 1
⎟
⎝ a 1 + b 1
⎠ + r 2 · ⎜a 2 + · · · + c 2
⎟
⎝ a 2 + b 2
⎠
a 1
a 2
( )
a1 b
= r 1 · f( 1
) + r
c 1 d 2 · f(
1
( )
a2 b 2
)
c 2 d 2
and so item (2) of Lemma 1.9 shows that it preserves structure.
(c) Yes, it is an isomorphism.
To show that it is one-to-one, we suppose that two members of the domain have the same image
under f.
( ) ( )
a1 b
f( 1 a2 b
) = f( 2
)
c 1 d 1 c 2 d 2
This gives, by the definition of f, that c 1 + (d 1 + c 1 )x + (b 1 + a 1 )x 2 + a 1 x 3 = c 2 + (d 2 + c 2 )x + (b 2 +
a 2 )x 2 + a 2 x 3 and then the fact that polynomials are equal only when their coefficients are equal
gives a set of linear equations
c 1 = c 2
d 1 + c 1 = d 2 + c 2
b 1 + a 1 = b 2 + a 2
a 1 = a 2
that has only the solution a 1 = a 2 , b 1 = b 2 , c 1 = c 2 , and d 1 = d 2 .
To show that f is onto, we note that p + qx + rx 2 + sx 3 is the image under f of this matrix.
( ) s r − s
p
q − p
We can check that f preserves structure by using item (2) of Lemma 1.9.
( ) ( ) ( )
a1 b
f(r 1 · 1 a2 b
+ r
c 1 d 2 · 2 r1 a
) = f( 1 + r 2 a 2 r 1 b 1 + r 2 b 2
)
1 c 2 d 2 r 1 c 1 + r 2 c 2 r 1 d 1 + r 2 d 2
= (r 1 c 1 + r 2 c 2 ) + (r 1 d 1 + r 2 d 2 + r 1 c 1 + r 2 c 2 )x
+ (r 1 b 1 + r 2 b 2 + r 1 a 1 + r 2 a 2 )x 2 + (r 1 a 1 + r 2 a 2 )x 3
= r 1 · (c
1 + (d 1 + c 1 )x + (b 1 + a 1 )x 2 + a 1 x 3)
+ r 2 · (c
2 + (d 2 + c 2 )x + (b 2 + a 2 )x 2 + a 2 x 3)
( ) ( )
a1 b
= r 1 · f( 1
a2 b
) + r
c 1 d 2 · f( 2
)
1 c 2 d 2