How would I do this problem?

Question

Let W be the set of vectors of the form

4a + 3b
0
a + b + c
c - 2a


where a, b, and c represent arbitrary real numbers. Show that W is a subspace of R4 and
find a set S that spans W.

Answer 1

Let [a, b, c] denote the vector <4a+3b, 0, a+b+c, c-2a>. Vectors of this form are a subset of R^4. In order to show that they form a subspace, we must show that they are closed under linear combinations. It suffices to show that they are closed under vector addition and scalar multiplication. [a, b, c] + [a', b', c'] = <4a+3b, 0, a+b+c, c-2a> + <4a'+3b', 0, a'+b'+c', c'-2a'> = <4(a+a') + 3(b+b'), 0, (a+a') + (b+b') + (c+c'), (c+c') - 2(a+a')> = [a+a', b+b', c+c'], so the set is closed under addition. Similarly, k[a, b, c] = k<4a+3b, 0, a+b+c, c-2a> = <4ka+3kb, 0, ka+kb+kc, kc-2ka> = [ka, kb, kc], so the set is closed under scalar multiplication.

The spanning set is {[1, 0, 0], [0, 1, 0], [0, 0, 1]}. Writing this out in full: {<4, 0,1, -2>, <3, 0, 1, 0>, <0, 0, 1, 1>}

Answer 2

I couldn't send you an email before because your address has not been confirmed.

I assume that those are the 4 components of a vector in R4? Clearly the subspace is in R3 because of the zero component. Here again, chose arbitrary alpha, beta, gamma and show that it's closed. I'll have to think more on this one.

Answer 3

you need to:
-show that the 0 vector is an element of W.
this is the case since takin a=b=c=0 gives the 0 vector
-show that if two elements are in W, then their sum is also in W.

and a set that spans W is given by:
(4,0,1,-2), (3,0,1,0) and (0,0,1,1)
'

Answer 4

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