Linear Algebra (ASP!) Need this in less then an hour show that (u,w2} is an orth

Question

Linear Algebra (ASP!) Need this in less then an hour

show that (u,w2} is an orthogonal basis for R. Then express x as a linear combination of the us. 14 Which of the following criteria are necessary for a set of vectors to be an orthogonal basis for Rlal that apply A. The vectors must span R2 B. The distance between any pair of distinct vectors must be constant. C. The vectors must form an orthogonal set. D. The vectors must all have a length of 1 Which theorem could help prove one of these criteria from another? O A. ,up} is a basis in RP, then the members of S span Rp and hence form an orthogonal set. up} is an orthogonal set of nonzero vectors in Rn, then s is linearly independent and hence is a basis for the subspace spanned by S If s-ful , lfs-ful , If s·ful,. .Mp} and the distance between any pair of distinct vectors is constant, then the vectors are evenly spaced and hence form an orthogonal set. If s-ful, ,up} and each ui has length 1, then s is an orthogonal set and hence is a basis for the subspace spanned by S B. C. D. What calculation shows that (u, u2) is an orthogonal basis for R2? Since the | is | | , the vectors | From the theorem above, this proves that the vectors are also because they are two Express x as a linear combination of the u's.

Explanation / Answer

We have u1.u2 = (3,-7)T.(14,6)T=3*14-7*6 = 42-42 = 0. Thus, (u1,u2) is an orthogonal set of vectors. Now, let A = [u1,u2,e1,e2,x] =

3

14

1

0

5

-7

6

0

1

-2

We can reduce A to its RREF as under:

Multiply the 1st row by 1/3

Add 7 times the 1st row to the 2nd row

Multiply the 2nd row by 3/116

Add -14/3 times the 2nd row to the 1st row

Then the RREF of A is

1

0

3/58

-7/58

1/2

0

1

7/116

3/116

1/4

It implies that u and u are linearly independent, e1= (3/58)u1+(7/116)u2, e2 = -(7/58)u1+(3/116)u2 and x = (1/2)u1+(1/4)u2. Since u1,u2 are linear combinations of e1,e2 and are linearly independent, and since (u1,u2) is an orthogonal set of vectors, hence (u1,u2) is an orthogonal basis for R2. Also, x = (1/2)u1+(1/4)u2.

For a set of vectors   to be an orthogonal basis for R2, the following criteria is necessary:

C.The vectors must form an orthogonal set.

If S = {u1,u2,…,up} isan orthogonal set of non-zero vectors in Rn, then S is linearly independent and hence a basis for the subspace spanned by S. Option B is the correct answer.

Since the dot product of u1,u2 is 0, the vectors u1,u2 are orthogonal.From the theorem above, this proves that the vectors u1,u2are also linearly independent because, these are two orthogonal vectors.

x = (1/2)u1+(1/4)u2.

3

14

1

0

5

-7

6

0

1

-2

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