Linear Algebra Subspaces

Question

Let w = (1 1 2 -1), W = { u € R^4 | u • w = 2 } , and

V = { u - (1 0 1 1) | for all u € W }.

(a) Is W a subspace ? Explain your answer

(b) Show that V is a subspace and that T = { (-1 1 0 0) , (-2 0 1 0) , (1 0 0 1) } is a basis for V.

(c) Let T be the basis of V defined in (b)

Find [(4 -1 1 5)]^T (Transpose). Show your workings, however, you do not need to show your elementary row operations.

Answer 1

a) W is not a subspace since it doesn't contain the zero vector.

b) To show V is a subspace, we first show that it is closed under scalar multiplication. Let c be a scalar and u - (1 0 1 1) and element of V (where u is an element of W). Scalar multiplication yields cu-(c 0 c c). To show that this is an element of V, we must show that adding (1 0 1 1) to it gives an element of W, i.e. that cu-(c 0 c c) + (1 0 1 1) belongs to W, or equivalently, that the dot product of cu - (c 0 c c) + (1 0 1 1) and w is equal to 2. Now, the dot product of cu and w is 2c (since u belongs to W), the dot product of (c 0 c c) and w is c + 2c - c = 2c, and the dot product of (1 0 1 1) and w is 2. Thus the dot product of u - (c 0 c c) + (1 0 1 1) and w is equal to 2c - 2c + 2 = 2. This shows that V is closed under scalar multiplication.

Next, we show that V is closed under addition. Let u_1 - (1 0 1 1) and u_2 - (1 0 1 1) be elements of V. To show that their sum belongs to V, we must show that their sum minus (1 0 1 1) has dot product with w equal to 2. This is easily verified using the fact that the dot product of u_1 and w is 2, the dot product of u_2 and w is 2, and the dot product of (1 0 1 1) and w is 2.

We conclude that V is a vector space. The elements in T all belong to V and are linearly independent, so form a basis (to verify linear independence, look at the 2nd, 3rd, and 4th coordinates).

c) Not sure I understand what this part is asking for ...