Page 144 - Advanced Linear Algebra
P. 144
128 Advanced Linear Algebra
ZZ
?
" b 4 ~ " b 4
?
But this follows from the fact that
Z
ZZ
Z
Z
" c " ~ ²" c " ³ b ² c ³" 4
?
Hence, scalar multiplication is well-defined. We leave it to the reader to show
that 4° 4 is a vector space over 9° ? .
?
Consider now a set 8 ~¸ 0¹ 4 and the corresponding set
4
8 ? b4 ~ ¸ b4 0¹
?
?4
If spans 4 over , then b 8 ? 4 spans 4 ° ? 4 over ° 9 ? . To see this, note
8
9
that any #4 has the form #~ for 9 and so
'
?
?
#b 4 ~ 8 9 b 4
~ ² b 4 ³?
~ ² b ³² b 4³
?
?
?
?
8
which shows that b4 spans 4° 4 .
Now suppose that 8 ~¸ 0¹ is a basis for 4 over . We show that
9
8 ? b4 is a basis for ? 4° 4 over ?9° . We have seen that 8 ? b4 spans
4° 4. Also, if
?
² b ³² b 4³ ~ 4? ? ?
then 4? and so
~
where ? . From the linear independence of we deduce that ? 8 for all
and so b ? ~ ? . Hence b ? 4 8 is linearly independent and therefore a
basis, as desired.
?
To see that (( ~ ( b4 ( , note that if b4 ~ b4 , then
8
8
?
?
c ~
where ? . If £ , then the coefficient of on the right must be equal to
