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Exercise Which of the following are subspaces of some euclidean spaces?

1) H = {(a + b, 3a - b, 2a + b): a, bR}

Answer H is a subspace.

u, vH
u = (a + b, 3a - b, 2a + b), v = (a' + b', 3a' - b', 2a' + b')
u + v = (a + b + a' + b', 3a - b + 3a' - b', 2a + b + 2a' + b') = (a'' + b'', 3a'' - b'', 2a'' + b'') ∈ H,

where a'' = a + a', b'' = b + b'.

uH, cR
u = (a + b, 3a - b, 2a + b)
cu = (ca + cb, 3ca - cb, 2ca + cb) = (a' + b', 3a' - b', 2a' + b') ∈ H

where a' = ca, b' = cb.

2) H = {(a + b, 3a - b, 2a + b + 1): a, bR}

Answer If H is a subspace, then it must contain the zero vector. In other words, we can find a and b, such that

a + b = 0, 3a - b = 0, 2a + b + 1 = 0.

However, the first two equations imply a = b = 0, which does not satisfy the third equation. We conclude that H is not a subspace.

3) H = {(a + b, 3a - b, 2a + b): a, bZ}

Answer By taking a = 1, b = 0, we see that (1, 2, 3) ∈ H. If H is a subspace, then we should also have 1/2(1, 2, 3) = (1/2, 1, 3/2) ∈ H. On the other hand, we cannot find two integers a and b such that a + b = 1/2. Therefore H is not a subspace.

4) H = {(x, y, z): x + y - z = 0, 2x - 3y + z = 0}

Answer H is a subspace.

u, vH
u = (x, y, z), x + y - z = 0, 2x - 3y + z = 0; v = (x', y', z'), x' + y' - z' = 0, 2x' - 3y' + z' = 0
u + v = (x + x', y + y', z + z'), (x + x') + (y + y') + (z + z') = x + y - z + x' + y' - z' = 0, 2(x + x') - 3(y + y') + (z + z') = 2x - 3y + z + 2x' - 3y' + z' = 0
u + vH.

The verification of uH, cRcuH is similar.

5) H = {(x, y, z): x + y - z = 0, 2x - 3y + z = 1}

Answer Since x = y = z = 0 does not satisfy the condition 2x - 3y + z = 1, H does not contain the zero vector. Therefore H is not a subspace.

6) H = {(x, y, z): x2 + y2 - z2 = 0}

Answer Since (1, 0, 1), (0, 1, 1) ∈ H and (1, 0, 1) + (0, 1, 1) = (1,1,2) ∉ H, we conclude that H is not a subspace.

7) H = {(x, y): xy ≥ 0}

Answer Since (2, 0), (-1, -1) ∈ H and (2, 0) + (-1, -1) = (1, -1) ∉ H, we conclude that H is not a subspace.

8) H = {(x, y, 0): x, yR}

Answer H is a subspace.

u, vH
u = (x, y, 0), v = (x', y', 0)
u + v = (x + x', y + y', 0 + 0) = (x + x', y + y', 0) ∈ H.

The verification of uH, cRcuH is similar.