Consider the real vector space v = r 4 . for each of the following five statements, provide either a proof or a counterexample. (a) dim v = 4. (b) span((1, 1, 0, , 1, 1, , 0, 1, 1)) = v . (c) the list ((1, −1, 0, , 1, −1, , 0, 1, −−1, 0, 0, 1)) is linearly independent. (d) every list of four vectors v1, . . , v4 ∈ v , such that span(v1, . . , v4) = v , is linearly independent. (页码61).

Answer with Step-by-step explanation:

We are given that a real vector space

V=

a.We have to prove that dim V =4

Let (a,b,c,d) is an element of V

Suppose that four elements

(1,0,0,0),(0,1,0,0),(0,0,1,0) and (0,0,0,1)

a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)

=(a,b,c,d)

All four elements are independent and can span each element of V.

Hence, dimension of V=4

b.We are given that (1,1,0,0),(0,1,1,0,(0,0,1,1)

We have to prove that these elements can span V

If number of elements are  less than dimension of then the elements can not span V.

Rank of matrix =3

Hence, rank is less than the dimension of V.Therefore, given elements can not span V.

c.We are given that (1,-1,0,0),(0,1,-1,0),(0,0,1,-1),(-1,0,0,1)

We have to show that these elements are linearly independent.

Every row or column is not a linear combination of other rows or columns.

Therefore, these elements are linearly independent.

d.We are given that and span

We have to prove that given vectors are linear independent.

Let

(a,b,c,d)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1)

All four vectors are linearly independent because any element is not a linear combination other elements .

Hence, every four vectors are linearly independent.

if the probability is 1/12, 75 people will receive a $25 gift card

step-by-step explanation:

step-by-step explanation: