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Please provide the process for solution. solution to each problem must show all

ID: 2967081 • Letter: P

Question

Please provide the process for solution.

solution to each problem must show all major steps, and be organized.

13. W is a subset of R2 defined as

W ={(x,y)|x?0, y?0}.

(a) Is W closed under vector addition? If your answer is no,then find two vectors u,v?W such that u+v??W (not included)

(b) Is W closed under scalar multiplication?
If your answer is no, then find a vector u ? W and a scalar c such that cu ?? W(not included).

(c) Is W a subspace of R2?

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14. Let

v1 = (1,1,1,1), v2 = (1,2,1,4), v3 = (1,0,1,0).

Determine whether the vectors v1, v2, v3 are linearly independent.

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15. Let

v1 =(1,2,3), v2 =(3,2,1), v3 =(1,1,1).

Determine whether the vectors v1, v2, v3 are linearly independent.

___________________________________________

16. Let

v1 =(1,2,3), v2 =(1,1,1), v3 =(1,0,1), v4 =(1,3,5).

  Determine whether the vectors v1, v2, v3, v4 span R3.

Explanation / Answer

For answers use these conditions....

If v1, v2, v3, v4 are linearly independent vectors in R^4, then {v1, v2, v3} must also be linearly independent.

And

If v1, v2, v3, v4 are in R^4 and v3 = 2v1 + v2, then {v1, v2, v3, v4} must be linearly dependent

and

If v1, v2, v3, v4 are in R^4 and v3 = 0, then {v1, v2, v3, v4} must be linearly dependent.

and

If v1, v2, v3, v4 are in R^4 and {v1, v2, v3}is linearly dependent, then {v1, v2, v3, v4} must also be linearly dependent

If {v1, v2, v3}were dependent, then there would be an equation if x1v1 + x2v2 + x3v3 = 0 with x1, x2, x3 not all zero. But then x1v1 + x2v2 + x3v3 + 0 v4 = 0 and still the coefficients are notALL zero so this contradicts the fact that {v1, v2, v3, v4} is linearly independent. Note: Questions 4 and 5 are logically equivalent. so  If v1, v2, v3, v4 are linearly independent vectors in R^4, then {v1, v2, v3} must also be linearly independent.

For span ::::

span of v1 = (1, 1) and v2 = (2, ?1) in R2?
Answer: R2

For R3 check Ax =v and if you can get the solution then it spans..

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