Question

Let \(W\) be a subspace of \(\mathbb{R}^{5}\) defined by $$ \mathcal{W}=\left\\{\left(x_{1}, \ldots, x_{5}\right) \in \mathbb{R}^{5} \mid x_{1}=3 x_{2}+x_{3}, x_{2}=x_{5}, \text { and } x_{4}=2 x_{3}\right\\} $$ Find a basis for \(\mathcal{W}\).

Short answer

The basis for the subspace \(\mathcal{W}\) is \{(3,1,0,0,1), (1,0,1,2,0)\}.

Step-by-step solution

Identify the Variables

The given conditions in the problem indicate that the values of \(x_1\), \(x_2\), \(x_4\), and \(x_5\) are determined by the values of \(x_2\) and \(x_3\), which we will call free variables. So let us take \(x_2 = s\) and \(x_3 = t\).

Find the Equations for the Other Variables

We can use the given conditions to find the equations for \(x_1\), \(x_2\), \(x_4\), and \(x_5\) in terms of \(s\) and \(t\): \(x_1 = 3s + t\), \(x_2 = s\), \(x_4 = 2t\), \(x_5 = s\).

Form the Vectors

We can form vectors using these values: \((x_1, x_2, x_3, x_4, x_5) = \((3s + t, s, t, 2t, s)\). This can be written as two separate vectors: \(s(3,1,0,0,1) + t(1,0,1,2,0)\). Thus, the vectors (3,1,0,0,1) and (1,0,1,2,0) form a basis for the subspace \(\mathcal{W}\).

Verify the Basis

To verify that these vectors form a basis for \(\mathcal{W}\), first ensure that they are in \(\mathcal{W}\). They must satisfy the given conditions, which they do since they were created based on these conditions. Next, show that they are independent (none is not a multiple of another), and that they span \(\mathcal{W}\) (any vector in \(\mathcal{W}\) can be written as a linear combination of them). These vectors satisfy these properties.

Key concepts

Understanding Subspaces in Linear Algebra

When we refer to subspaces in linear algebra, we're talking about a specific division of vector spaces. A subspace is essentially a set of vectors that satisfy two main conditions: it must contain the zero vector, and it must be closed under addition and scalar multiplication. In simpler terms, if you take any vectors from this subspace and combine them or stretch/shrink them (through scalar multiplication), they will still belong to the same subspace.

Looking at the given exercise, \(W\) is defined as a subspace of \(\mathbb{R}^{5}\). It's described with certain linear equations that all the vectors in this subspace have to satisfy. To find a basis for \(W\), which is one of the core tasks in understanding subspaces, you examine combinations of vectors that are both independent (no vector in the set can be expressed as a linear combination of the others) and capable of spanning the entire subspace (you can reach any vector in the subspace using linear combinations from the basis).

In the provided solution, these principles were applied. By identifying the free variables and using the given conditions, the basis vectors were determined and verified. These basis vectors are crucial as they form the 'building blocks' for every other vector in the subspace.

The Role of Vector Spaces

Vector spaces are at the heart of linear algebra. They are collections of objects, called vectors, that can be scaled and added together following certain rules. To become a vector space, the set must satisfy a handful of properties, such as the existence of a zero vector (which acts as an identity element for addition), the commutative and associative properties for vector addition, and distributive properties for scalar multiplication, among others.

It's important to understand that subspaces are special examples of vector spaces. They inherit all the properties of vector spaces but are defined with additional conditions. In our exercise with \(\mathbb{R}^{5}\), which is a well-known vector space, the subspace \(W\) is a smaller set within \(\mathbb{R}^{5}\) that still satisfies all vector space properties but is confined to meeting the specific equations that describe it. By understanding the structure of vector spaces, students can better grasp how subspaces like \(W\) function within the broader context of linear algebra.

Linear Combinations and Their Significance

Linear combinations come into play when we deal with vector spaces and subspaces. A linear combination is formed when you add together several vectors, each multiplied by a scalar coefficient. It's a fundamental concept because it defines how vectors in a space can be combined to reach other vectors within that space.

The ability to express vectors as linear combinations of others is what defines the span of a set of vectors. In the context of our exercise, the goal was to express any vector in subspace \(W\) as a linear combination of the basis vectors. If you can do this, it means the basis vectors span \(W\). Moreover, verifying that the basis vectors are also linearly independent is crucial – otherwise, there would be redundancies, and the set wouldn't truly be a basis.

The skill of forming and understanding linear combinations is thus key to solving numerous problems in linear algebra, such as finding bases for subspaces, solving systems of linear equations, and performing more abstract operations within vector spaces.