What Values of m and n Make These Sets a Subspace of R^4?

[Johnathon1]

D is the set and the set contains the solutions to

x + (1 - m)y^-1 + 2z + n²w = 0

I'm trying to find m, n values which means the set is a subspace of R (four dimensions).

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Similarly, trying to find the m, n values that makes the following two expressions two separate subspaces, too.

mx + y - 3z + (m - m²)|w| = m³ - m

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x + (m - n)y + z + 2m²w = n

I've been reviewing the three properties of subspaces over and over and don't know how to apply them in these scenarios.

 

[jvicens]

Hi there,

I am happy to help you understand the concept of subspaces and how they apply to the given scenarios.

First, let's review the three properties of subspaces:

1. Closure under addition: This means that if two vectors are in the subspace, their sum must also be in the subspace.

2. Closure under scalar multiplication: This means that if a vector is in the subspace, multiplying it by any scalar will result in a vector that is still in the subspace.

3. Contains the zero vector: This means that the subspace must contain the zero vector (a vector where all components are equal to zero).

Now, let's apply these properties to the given sets:

1. For the first set, we have x + (1 - m)y-1 + 2z + n2w = 0. To determine if this set is a subspace, we need to see if it satisfies the three properties mentioned above.

- Closure under addition: Let's take two vectors from the set, say (x1, y1, z1, w1) and (x2, y2, z2, w2). If we add them together, we get (x1 + x2, y1 + y2, z1 + z2, w1 + w2). Plugging this into the original equation, we get:

(x1 + x2) + (1 - m)(y1 + y2)-1 + 2(z1 + z2) + n2(w1 + w2) = 0

Since the original equation is true for both (x1, y1, z1, w1) and (x2, y2, z2, w2), it must also be true for their sum. Therefore, the set is closed under addition.

- Closure under scalar multiplication: Let's take a vector from the set, say (x, y, z, w). If we multiply it by a scalar, say k, we get (kx, ky, kz, kw). Plugging this into the original equation, we get:

kx + (1 - m)(ky)-1 + 2kz + n2(kw) = 0

Since the original equation is true for (x, y, z, w), it must also be true for its scalar multiple. Therefore, the set is closed under scalar multiplication