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数学证明代写-108A

时间：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

t

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).

0

pH

C ck atta k a cox'tax

ta

s

学霸联盟

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

t

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).

0

pH

C ck atta k a cox'tax

ta

s

学霸联盟