Show that W = {(x1, x2, x3, x4)|x4 − x3 = x2 − x1} is a subspace of R4, spanned by (1, 0, 0, −1), (0, 1, 0, 1) and (0, 0, 1, 1)

To show that W is a subspace of R4, we need to prove that it satisfies three properties:

1. The zero vector of R4 is in W.

2. W is closed under vector addition.

3. W is closed under scalar multiplication.

4. The zero vector in R4 is (0,0,0,0). If we substitute these values into the equation x4 - x3 = x2 - x1, we get 0 - 0 = 0 - 0, which is true. Therefore, the zero vector is in W.

5. To show that W is closed under vector addition, we need to show that for any two vectors u and v in W, the sum u + v is also in W. Let u = (u1, u2, u3, u4) and v = (v1, v2, v3, v4) be any two vectors in W. Then, by the definition of W, we have u4 - u3 = u2 - u1 and v4 - v3 = v2 - v1. If we add these two equations, we get (u4 + v4) - (u3 + v3) = (u2 + v2) - (u1 + v1), which means that the vector (u1 + v1, u2 + v2, u3 + v3, u4 + v4) is in W. Therefore, W is closed under vector addition.

6. To show that W is closed under scalar multiplication, we need to show that for any vector u in W and any scalar c, the product cu is also in W. Let u = (u1, u2, u3, u4) be any vector in W and let c be any scalar. Then, by the definition of W, we have u4 - u3 = u2 - u1. If we multiply this equation by c, we get cu4 - cu3 = cu2 - cu1, which means that the vector (cu1, cu2, cu3, cu4) is in W. Therefore, W is closed under scalar multiplication.

Now, to show that W is spanned by the vectors (1, 0, 0, -1), (0, 1, 0, 1) and (0, 0, 1, 1), we need to show that any vector in W can be written as a linear combination of these three vectors. Let (x1, x2, x3, x4) be any vector in W. Then, by the definition of W, we have x4 - x3 = x2 - x1. This equation can be rewritten as x1 - x2 + x3 - x4 = 0, which means that the vector (x1, x2, x3, x4) can be written as a linear combination of the vectors (1, 0, 0, -1), (0, 1, 0, 1) and (0, 0, 1, 1). Therefore, W is spanned by these three vectors.