Exploring Subspace W In R^3 Properties Basis And Linear Algebra Concepts

Introduction

In the realm of linear algebra, understanding vector spaces and their subspaces is fundamental. A vector space is a set of objects, called vectors, that can be added together and multiplied by scalars, adhering to certain axioms. Subspaces are subsets of vector spaces that themselves satisfy the axioms of a vector space. This article delves into the specifics of a subspace ${ W }$ within the three-dimensional real space ${ R^3 }$, defined by a linear equation. We will rigorously demonstrate that ${ W }$ is indeed closed under the operations of vector addition and scalar multiplication, thereby establishing it as a subspace. Furthermore, we will embark on a journey to pinpoint a basis for ${ W }$, which is a set of linearly independent vectors that span the subspace. This exploration will provide a comprehensive understanding of the structure and properties of ${ W }$.

Defining the Subspace W

Let's begin by precisely defining the subspace ${ W }$ under consideration. We are given that ${ W }$ is a subset of ${ R^3 }$, the set of all three-dimensional vectors with real-number components. Specifically, ${ W }$ consists of all vectors ${ (x, y, z) }$ in ${ R^3 }$ that satisfy the linear equation ${ x + 2y - z = 0 }$. This equation represents a plane passing through the origin in ${ R^3 }$. To solidify our understanding, consider a few examples. The vectors ${ (0, 0, 0) }$, ${ (1, 0, 1) }$, and ${ (0, 1, 2) }$ all belong to ${ W }$ because they satisfy the defining equation. Conversely, the vector ${ (1, 1, 1) }$ does not belong to ${ W }$ because ${ 1 + 2(1) - 1 = 2 \neq 0 }$. The key question we address in this article is: Does this set ${ W }$ form a subspace of ${ R^3 }$?

Demonstrating Closure under Addition and Scalar Multiplication

To prove that ${ W }$ is a subspace of ${ R^3 }$, we must demonstrate that it satisfies two crucial conditions: closure under addition and closure under scalar multiplication. These conditions are the cornerstones of subspace identification.

Closure under Addition

Closure under addition means that if we take any two vectors in ${ W }$ and add them together, the resulting vector must also be in ${ W }$. Mathematically, if ${ extbf{u} }$ and ${ extbf{v} }$ are in ${ W }$, then ${ extbf{u} + extbf{v} }$ must also be in ${ W }$. Let's rigorously demonstrate this for our specific ${ W }$. Suppose we have two vectors ${ extbf{u} = (x_1, y_1, z_1) }$ and ${ extbf{v} = (x_2, y_2, z_2) }$ that both belong to ${ W }$. This implies that they satisfy the equation defining ${ W }$:

${ x_1 + 2y_1 - z_1 = 0 }$ ${ x_2 + 2y_2 - z_2 = 0 }$

Now, let's add these two vectors:

${ extbf{u} + extbf{v} = (x_1 + x_2, y_1 + y_2, z_1 + z_2) }$

To check if ${ extbf{u} + extbf{v} }$ is in ${ W }$, we need to verify if its components satisfy the defining equation of ${ W }$. Let's substitute the components of ${ extbf{u} + extbf{v} }$ into the equation:

${ (x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) }$

We can rearrange this expression:

${ (x_1 + 2y_1 - z_1) + (x_2 + 2y_2 - z_2) }$

Since both ${ x_1 + 2y_1 - z_1 }$ and ${ x_2 + 2y_2 - z_2 }$ are equal to 0 (because ${ extbf{u} }$ and ${ extbf{v} }$ are in ${ W }$), their sum is also 0:

${ 0 + 0 = 0 }$

Thus, we have shown that ${ (x_1 + x_2) + 2(y_1 + y_2) - (z_1 + z_2) = 0 }$, which means that the vector ${ extbf{u} + extbf{v} }$ satisfies the equation defining ${ W }$ and therefore belongs to ${ W }$. This rigorously proves that ${ W }$ is closed under addition.

Closure under Scalar Multiplication

The second condition for a subspace is closure under scalar multiplication. This means that if we take any vector in ${ W }$ and multiply it by a scalar (a real number), the resulting vector must also be in ${ W }$. Mathematically, if ${ extbf{u} }$ is in ${ W }$ and ${ c }$ is a scalar, then ${ c extbf{u} }$ must also be in ${ W }$. Let's demonstrate this for our subspace ${ W }$. Suppose we have a vector ${ extbf{u} = (x, y, z) }$ in ${ W }$, which means it satisfies the equation ${ x + 2y - z = 0 }$. Now, let ${ c }$ be any scalar (a real number). Let's multiply the vector ${ extbf{u} }$ by the scalar ${ c }$:

${ c extbf{u} = c(x, y, z) = (cx, cy, cz) }$

To check if ${ c extbf{u} }$ is in ${ W }$, we need to verify if its components satisfy the defining equation of ${ W }$. Let's substitute the components of ${ c extbf{u} }$ into the equation:

${ cx + 2(cy) - cz }$

We can factor out the scalar ${ c }$ from this expression:

${ c(x + 2y - z) }$

Since ${ extbf{u} = (x, y, z) }$ is in ${ W }$, we know that ${ x + 2y - z = 0 }$. Therefore, the expression simplifies to:

${ c(0) = 0 }$

Thus, we have shown that ${ cx + 2(cy) - cz = 0 }$, which means that the vector ${ c extbf{u} }$ satisfies the equation defining ${ W }$ and therefore belongs to ${ W }$. This rigorously proves that ${ W }$ is closed under scalar multiplication.

Conclusion: W as a Subspace

Having demonstrated that ${ W }$ is closed under both addition and scalar multiplication, we can definitively conclude that ${ W }$ is a subspace of ${ R^3 }$. This is a significant result, as it allows us to apply the powerful tools and theorems of linear algebra specifically to the set ${ W }$. Understanding ${ W }$ as a subspace opens doors to further analysis, such as determining its dimension and finding a basis, which we will explore in the next section.

Finding a Basis for W

Now that we have established ${ W }$ as a subspace of ${ R^3 }$, our next objective is to find a basis for ${ W }$. A basis is a set of linearly independent vectors that span the subspace. In simpler terms, it's a minimal set of vectors that can be combined (using linear combinations) to generate any vector in the subspace. The number of vectors in a basis is called the dimension of the subspace. Finding a basis is crucial because it provides a concise and efficient way to represent all vectors within the subspace.

Expressing z in terms of x and y

Recall that ${ W }$ is defined by the equation ${ x + 2y - z = 0 }$. We can rearrange this equation to express ${ z }$ in terms of ${ x }$ and ${ y }$:

${ z = x + 2y }$

This equation provides a key insight into the structure of vectors in ${ W }$. Any vector ${ (x, y, z) }$ in ${ W }$ can be written as ${ (x, y, x + 2y) }$. This representation allows us to express the vector as a linear combination of two other vectors.

Decomposing a Vector in W

Let's decompose a general vector ${ (x, y, x + 2y) }$ in ${ W }$ into a linear combination of two vectors. We can rewrite the vector as:

${ (x, y, x + 2y) = (x, 0, x) + (0, y, 2y) }$

Now, we can factor out the scalars ${ x }$ and ${ y }$ from each vector:

${ (x, 0, x) + (0, y, 2y) = x(1, 0, 1) + y(0, 1, 2) }$

This decomposition is crucial. It shows that any vector in ${ W }$ can be written as a linear combination of the vectors ${ (1, 0, 1) }$ and ${ (0, 1, 2) }$. This means that the set ${ \{(1, 0, 1), (0, 1, 2)\} }$ spans the subspace ${ W }$. In other words, these two vectors are sufficient to generate any vector in ${ W }$ through linear combinations.

Verifying Linear Independence

The next step is to verify that the vectors ${ (1, 0, 1) }$ and ${ (0, 1, 2) }$ are linearly independent. Linear independence means that neither vector can be written as a scalar multiple of the other. To formally check this, we set up the following equation:

${ c_1(1, 0, 1) + c_2(0, 1, 2) = (0, 0, 0) }$

where ${ c_1 }$ and ${ c_2 }$ are scalars. This equation represents a linear combination of the two vectors equaling the zero vector. If the only solution to this equation is ${ c_1 = 0 }$ and ${ c_2 = 0 }$, then the vectors are linearly independent. Let's expand the equation:

${ (c_1, 0, c_1) + (0, c_2, 2c_2) = (0, 0, 0) }$

${ (c_1, c_2, c_1 + 2c_2) = (0, 0, 0) }$

This gives us a system of linear equations:

${ c_1 = 0}$ ${ c_2 = 0}$ ${ c_1 + 2c_2 = 0}$

From the first two equations, we immediately see that ${ c_1 = 0 }$ and ${ c_2 = 0 }$. The third equation is also satisfied with these values. Therefore, the only solution is the trivial solution, which confirms that the vectors ${ (1, 0, 1) }$ and ${ (0, 1, 2) }$ are indeed linearly independent.

Determining the Basis

Since we have shown that the vectors ${ (1, 0, 1) }$ and ${ (0, 1, 2) }$ span ${ W }$ and are linearly independent, we can conclude that they form a basis for ${ W }$. A basis is a set of linearly independent vectors that span the entire vector space (or subspace). In this case, the set ${ \{(1, 0, 1), (0, 1, 2)\} }$ is a basis for the subspace ${ W }$. This means that any vector in ${ W }$ can be uniquely expressed as a linear combination of these two basis vectors. Furthermore, since the basis contains two vectors, the dimension of ${ W }$ is 2. This aligns with our initial understanding that ${ W }$ represents a plane in ${ R^3 }$, which is a two-dimensional subspace.

Conclusion

In this article, we have conducted a thorough exploration of the subspace ${ W }$ within ${ R^3 }$, defined by the equation ${ x + 2y - z = 0 }$. We rigorously demonstrated that ${ W }$ is closed under both vector addition and scalar multiplication, thereby confirming its status as a subspace of ${ R^3 }$. Furthermore, we embarked on a journey to discover a basis for ${ W }$, successfully identifying the vectors ${ (1, 0, 1) }$ and ${ (0, 1, 2) }$ as a basis. This means that any vector in ${ W }$ can be uniquely represented as a linear combination of these two vectors. The dimension of ${ W }$ is 2, which is consistent with its geometric interpretation as a plane in three-dimensional space. This comprehensive analysis provides a solid understanding of the structure and properties of ${ W }$, highlighting the importance of concepts like subspaces, closure, linear independence, and bases in the broader context of linear algebra. The techniques and principles explored here are applicable to a wide range of vector spaces and subspaces, making them essential tools for mathematicians, scientists, and engineers alike.