Showing that H is Not a Subspace of R^3

In linear algebra, a subspace is a subset of a vector space that satisfies certain properties. Specifically, a subset H of a vector space V is a subspace if it is closed under vector addition and scalar multiplication. In other words, for any vectors u and v in H, and any scalar c, the following conditions must hold:

1. u + v is in H.
2. c * u is in H.

Let's consider the set H = ((x, y, z): x + 2y - z + 1 = 0) and see if it satisfies the properties of a subspace of the vector space R^3.

Step 1: Check Closure Under Vector Addition

Suppose we have two vectors (x1, y1, z1) and (x2, y2, z2) in H. This means that x1 + 2y1 - z1 + 1 = 0 and x2 + 2y2 - z2 + 1 = 0.

Now, let's consider the sum of these two vectors: (x1 + x2, y1 + y2, z1 + z2)

Substituting the given equations, we get: (x1 + x2) + 2(y1 + y2) - (z1 + z2) + 1 = 0 + 0 = 0

This means that the sum of the two vectors is also in H, and H is closed under vector addition.

Step 2: Check Closure Under Scalar Multiplication

Now, let's consider the scalar multiplication of a vector (x, y, z) in H by a scalar c. c * (x, y, z) = (c*x, c*y, c*z)

Substituting the equation for H, we get: (c*x) + 2(c*y) - (c*z) + 1 = c(x + 2y - z + 1)

However, since x + 2y - z + 1 = 0 for any vector in H, we have: c(x + 2y - z + 1) = c * 0 = 0

This means that the scalar multiple of a vector in H is not necessarily in H, and H is not closed under scalar multiplication.

Conclusion

Since H is not closed under scalar multiplication, it does not satisfy the properties of a subspace. Therefore, H = ((x, y, z): x + 2y - z + 1 = 0) is not a subspace of the vector space R^3.