Show that H = ((x y z): x + 2y - z + 1 = 0) is not a subspace of R^3.

Show that H = ((x y z): x + 2y - z + 1 = 0) is not a subspace of R^3.

Proving that H is Not a Subspace of R^3

In linear algebra, a subspace is a subset of a vector space that satisfies the properties of a vector space. To be considered a subspace, a set must be closed under vector addition and scalar multiplication. In other words, the sum of any two vectors in the set and the scalar multiple of any vector in the set must also belong to the set.

Let's consider the set H = {(x, y, z): x + 2y - z + 1 = 0} and determine whether it is a subspace of the vector space R^3.

To show that H is not a subspace of R^3, we need to find a counterexample that violates at least one of the subspace properties.

Step 1: Consider the zero vector (0, 0, 0). The zero vector satisfies the equation x + 2y - z + 1 = 0, as 0 + 2(0) - 0 + 1 = 1. Therefore, the zero vector belongs to the set H.

Step 2: Check the closure under vector addition. Let's consider two vectors in H, (x1, y1, z1) and (x2, y2, z2), where x1 + 2y1 - z1 + 1 = 0 and x2 + 2y2 - z2 + 1 = 0.

The sum of these two vectors is (x1 + x2, y1 + y2, z1 + z2). We need to check if this sum also satisfies the equation x + 2y - z + 1 = 0.

(x1 + x2) + 2(y1 + y2) - (z1 + z2) + 1 = (x1 + 2y1 - z1 + 1) + (x2 + 2y2 - z2 + 1) = 0 + 0 = 0

However, this is not always the case. Consider the vectors (1, 0, 1) and (-1, 1, 0), both of which belong to H. Their sum is (0, 1, 1), which does not satisfy the equation x + 2y - z + 1 = 0.

Therefore, H is not closed under vector addition and is not a subspace of R^3.

Step 3: Check the closure under scalar multiplication. Let's consider a vector (x, y, z) in H and a scalar k.

k(x, y, z) = (kx, ky, kz)

We need to check if the scalar multiple also satisfies the equation x + 2y - z + 1 = 0.

k(x + 2y - z + 1) = kx + 2ky - kz + k = x + 2y - z + k

Since k can be any real number, the scalar multiple will not always satisfy the equation x + 2y - z + 1 = 0 (unless k = 1).

Therefore, H is not closed under scalar multiplication and is not a subspace of R^3.

In conclusion, the set H = {(x, y, z): x + 2y - z + 1 = 0} is not a subspace of the vector space R^3 because it fails to satisfy the properties of a subspace, specifically the closure under vector addition and scalar multiplication.

Copyright © 2024 Multiplication Chart  All rights reserved.