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數學證明代寫-108A
時間:2021-01-28
November 3, 2020
Math 108A (section 300)
Solutions to Midterm #1
1.) or 2.) Let V be a vector space. Prove that (v) = v for all v 2 V .
Solution: We need to show that the additive inverse of v is v (since x
denotes the additive inverse of x). To prove this, we need to show that
v + v = 0. But that is true because the additive inverse of v is v (by
definition of v).
1.) or 2.) Give a careful definition of a basis of a vector space.
Solution: A basis of a vector space V iover F is a list of vectors v1, . . . , vn
which is both linearly independent (i.e., the only solution to
c1v1 + · · · cnvn = 0
is c1 = · · · = cn = 0) and spans V (i.e. every v 2 V can be written in the
form
v = a1v1 + · · ·+ anvn
for some a1, . . . an in F.
Note that a careful definiton would need to include definitions of “linearly
independent” and “span,” as I did above. An alternate acceptable answer
is: a basis of a vector space V iover F is a list of vectors v1, . . . , vn such that
every v 2 V can be written uniquely in the form
v = a1v1 + · · ·+ anvn
with a1, . . . an in F.
3.) Suppose v1, v2, v3, v4 is linearly independent in V . Prove that the list
v1 v2, v2 v3, v3 v4, v4
is also linearly independent.
Same question for the list
v1, v2 v1, v3 v2, v4 v3
Solution: These are very similar, so I will only do the first one. Consider
a general linear combination
c1(v1 v2) + c2(v2 v3) + c3(v3 v4) + c4v4 = 0.
We may rewrite this (c0lecting terms according to vj) as
c1v1 + (c2 c1)v2 + (c3 c2)v3 + (c4 c3)v4 = 0.

O
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Since v1, v2, v3, v4 is a linearly independent in V , we may conclude that
c1 = c2 c1 = c3 c2 = c4 c3 = 0.
We now solve this: the first equation tells us c1 = 0; substituting into the
second equation, we see that c2 = 0; substituting into the third equation,
we see that c3 = 0; and finally, substituting into the fourth equation, we see
that c4 = 0. Thus, the list
v1 v2, v2 v3, v3 v4, v4
is linearly independent.
4.) Recall that P2(R) denotes the space of polynomials of degree at most 2
with real coecients.
(a) Let U = {p 2 P2(R) : p(a) = 0}. (Note: on your exam, a was
either 5 or 6.) Find a basis for U .
(b) Extend the basis in part (a) to a basis of P2(R).
(c) Find a subspace W of P2(R) such that P2(R) = U W .
Solution: For part (a), we take a general polynomial of degree 2, namely
c0x2+c1x+c2 and substitute x = a to find c0a2+c1a+c2 = 0. We can solve
the latter as c2 = c0a2 c1a so a basis is given by the two polynomials
x2 a2 and x a.
There are many possible solutions to part (b); one choice is to use the
constant polynomial 1 as the third basis member. Whatever choice v you
make in part (b), you can take W to be the space spanned by your choice
v.
5.) Prove or give a counterexample: if U1, U2, and W are subspaces of a
vector space V such that
V = U1 W and V = U2 W,
then U1 = U2.
Solution: There are many possible counter examples. Let V = R2, U1 be
the x-axis, W be the y-axis, and U2 be the line {y = x}. A complete solution
would now verify that the claimed properties hold, which involves the two
facts (x, y) = (x, 0) + (0, y) and (x, y) = (x, x) + (0, y x).
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