Subspaces Of R³ A Detailed Explanation And Examples

Hey guys! Today, we're diving deep into linear algebra, specifically focusing on subspaces within the vector space R³. This is a crucial concept, so let's break it down in a way that's easy to understand. We're going to analyze several sets, denoted as 'W', and determine whether they qualify as subspaces of R³. Buckle up, it's going to be an insightful ride!

What Makes a Subset a Subspace?

Before we jump into the examples, let's quickly recap what it means for a subset to be a subspace. A subset W of a vector space V is considered a subspace if it satisfies three key conditions:

1. The Zero Vector: The zero vector of V must be present in W.
2. Closure under Addition: If we take any two vectors in W and add them together, the resulting vector must also be in W. This means W is "closed" under addition.
3. Closure under Scalar Multiplication: If we take any vector in W and multiply it by a scalar (a real number in this case, since we're dealing with R³), the resulting vector must also be in W. This means W is "closed" under scalar multiplication.

If a subset fails to meet even one of these conditions, it's not a subspace. Simple as that! Now, let's put this knowledge to the test with some examples.

Analyzing the Subsets

We'll go through each set W provided, carefully checking if it satisfies the three subspace conditions. Let's get started!

(a) W = {(x, y, z) ∈ R³ | x = 0}

In this set, W consists of all vectors in R³ where the first component (x) is zero. Think of this as the yz-plane in 3D space. Does this qualify as a subspace? Let's check the conditions:

• The Zero Vector: The zero vector in R³ is (0, 0, 0). Since x = 0 in this vector, it is an element of W. So, condition 1 is satisfied.
• Closure under Addition: Let's take two arbitrary vectors in W, say u = (0, y₁, z₁) and v = (0, y₂, z₂). Adding them together, we get u + v = (0, y₁ + y₂, z₁ + z₂). The first component is still 0, so u + v is also in W. Condition 2 is satisfied.
• Closure under Scalar Multiplication: Now, let's take a vector u = (0, y, z) in W and multiply it by a scalar 'c'. We get cu = (0, cy, cz). Again, the first component is 0, so cu is in W. Condition 3 is satisfied.

Since W satisfies all three conditions, we can confidently say that W = {(x, y, z) ∈ R³ | x = 0} is a subspace of R³. This one checks out!

(b) W = {(x, y, z) ∈ R³ | x ∈ Z}

Here, W is the set of all vectors in R³ where the first component (x) is an integer. This one's a bit trickier. Let's go through the conditions:

• The Zero Vector: The zero vector (0, 0, 0) has an integer as its first component (0 is an integer), so it is in W. Condition 1 is satisfied.
• Closure under Addition: Let's take two vectors in W, u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂), where x₁ and x₂ are integers. Their sum is u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂). The sum of two integers (x₁ + x₂) is also an integer, so u + v is in W. Condition 2 is satisfied.
• Closure under Scalar Multiplication: Now, let's take a vector u = (x, y, z) in W (where x is an integer) and multiply it by a scalar 'c'. We get cu = (cx, cy, cz). This is where it gets interesting! If 'c' is not an integer, then 'cx' might not be an integer. For example, if x = 1 and c = 0.5, then cx = 0.5, which is not an integer. Therefore, cu is not necessarily in W. Condition 3 is not satisfied.

Because W fails the closure under scalar multiplication condition, W = {(x, y, z) ∈ R³ | x ∈ Z} is not a subspace of R³. This highlights the importance of checking all three conditions!

(c) W = {(x, y, z) ∈ R³ | y is irrational}

This set W contains all vectors in R³ where the second component (y) is an irrational number (like √2 or π). Let's see if it's a subspace:

• The Zero Vector: The zero vector (0, 0, 0) has a rational number (0) as its second component. Therefore, the zero vector is not in W. Condition 1 is not satisfied.

Since W fails the very first condition, we don't even need to check the others! W = {(x, y, z) ∈ R³ | y is irrational} is not a subspace of R³. This is a quick and easy way to disqualify a set if it doesn't even contain the zero vector.

(d) W = {(x, y, z) ∈ R³ | x - 3z = 0}

In this case, W consists of vectors where the first component (x) minus three times the third component (z) equals zero. This represents a plane in R³. Let's put it to the subspace test:

• The Zero Vector: For the zero vector (0, 0, 0), we have 0 - 3(0) = 0. So, the zero vector is in W. Condition 1 is satisfied.
• Closure under Addition: Let's take two vectors u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂) in W. This means x₁ - 3z₁ = 0 and x₂ - 3z₂ = 0. Their sum is u + v = (x₁ + x₂, y₁ + y₂, z₁ + z₂). Now, we need to check if (x₁ + x₂) - 3(z₁ + z₂) = 0. Expanding, we get x₁ + x₂ - 3z₁ - 3z₂ = (x₁ - 3z₁) + (x₂ - 3z₂) = 0 + 0 = 0. So, u + v is in W. Condition 2 is satisfied.
• Closure under Scalar Multiplication: Let's take a vector u = (x, y, z) in W (so x - 3z = 0) and multiply it by a scalar 'c'. We get cu = (cx, cy, cz). Now, we need to check if cx - 3(cz) = 0. This simplifies to c(x - 3z) = c(0) = 0. So, cu is in W. Condition 3 is satisfied.

Since W satisfies all three conditions, W = {(x, y, z) ∈ R³ | x - 3z = 0} is a subspace of R³. This is another one that makes the cut!

(e) W = {(x, y, z) ∈ R³ | ax + by + cz = d, where d ≠ 0}

This is a general form of a plane in R³. The key here is that 'd' is not equal to zero. Let's see how this affects the subspace conditions:

• The Zero Vector: For the zero vector (0, 0, 0), we have a(0) + b(0) + c(0) = 0. Since d ≠ 0, the zero vector does not satisfy the equation and is therefore not in W. Condition 1 is not satisfied.

Because W fails to contain the zero vector, W = {(x, y, z) ∈ R³ | ax + by + cz = d, where d ≠ 0} is not a subspace of R³. This is a classic example of a plane that doesn't pass through the origin, and therefore isn't a subspace.

Conclusion: Subspaces Unveiled

Okay, guys, we've dissected five different sets and determined which ones are subspaces of R³. To recap, only (a) W = {(x, y, z) ∈ R³ | x = 0} and (d) W = {(x, y, z) ∈ R³ | x - 3z = 0} satisfy all three conditions: containing the zero vector, closure under addition, and closure under scalar multiplication.

Understanding subspaces is fundamental to linear algebra. It allows us to work with subsets of vector spaces that behave nicely and maintain the vector space structure. Keep practicing, and you'll master this concept in no time!