Linear Algebra Please show all work and clearly so I can understand. Thank you!

Question

Linear Algebra

Please show all work and clearly so I can understand. Thank you!
Sorry for long question. Help appreciated!

8-suppose v is a subspace of R4 consisting of all vectors of the form (b, b-c b+c, c) and a) Find a basis for V b) Give an example of a vector in W. c) Prove that W is also a subspace of R d) Find a basis for W. e) Explain that the zero vector in R4 is the only vector in both V and W f Show that any vector in R4 can be expressed as a linear combination of basis vectors in V and W.

Explanation / Answer

8. (a). An arbitrary vector v in V is of the form (b,b-c, b+c,c) = b(1,1,1,0)+c(0,-1,1,1). Hence {(1,1,1,0),(0,-1 ,1,1) } is a basis for V.

(b). Let w = (p,q,r,s) be an arbitrary vector in W . Then p+q+r= 0 or, and –q+r+s = 0. On solving these equations, we get p = -2r-s and q = r+s. Then w = (-2r-s,r+s,r,s) = r(-2,1,1,0)+s(-1,1,0,1). Hence any vector in W is a linear combination of the vectors (-2,1,1,0) and(-1,1,0,1). Both these vectors are in W.

(c ). Let (-2,1,1,0)= w1 and(-1,1,0,1)= w2. Then any 2 arbitrary vectors in W are of the form a w1+b w2 and c w1+d w2 where a,b,c,d are arbitrary real numbers. Then (a w1+b w2)+( c w1+d w2) = (a+c)w1 +(b+d)w2. It implies that (a w1+b w2)+( c w1+d w2) is also in W so that W is closed under vector addition. Further, if k is an arbitrary scalar, then k(a w1+b w2)= ka w1+kb w2 is also in W. Hence W is closed under scalar multiplication. Also, since (0,0,0,0).v = 0, hence the zero vector is also in W. Therefore, W is a subspace of R4.

(d). {w1,w2}= { (-2,1,1,0),(-1,1,0,1)}is a basis for W.

(e). Since every vector in W is orthogonal to every vector in V , hence the zero vector is the only vector common to both V and W.

(f). If A be the matrix with (1,1,1,0),(0,-1 ,1,1) , (-2,1,1,0),(-1,1,0,1) as columns, then the RREF of A is I4. It implies that any vector in R4 can be expressed as a linear combination of the vectors in V and W.

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