Question

Let \(U\) and \(W\) be subspaces of \(V\) with bases \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) and \(\left\\{\mathbf{w}_{1}, \mathbf{w}_{2}\right\\}\) respectively. If \(U\) and \(W\) have only the zero vector in common, show that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}, \mathbf{w}_{1}, \mathbf{w}_{2}\right\\}\) is independent.

Short answer

The set \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \mathbf{w}_1, \mathbf{w}_2\}\) is linearly independent.

Step-by-step solution

Understand the Problem

We are given two subspaces, \(U\) and \(W\), with bases \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) and \(\{\mathbf{w}_1, \mathbf{w}_2\}\) respectively. The task is to show that the vectors in the set \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \mathbf{w}_1, \mathbf{w}_2\}\) are linearly independent given that \(U\) and \(W\) have only the zero vector in common.

Set Up the Independence Equation

To demonstrate independence, consider a linear combination of the vectors set to zero: \[c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + c_3\mathbf{u}_3 + d_1\mathbf{w}_1 + d_2\mathbf{w}_2 = \mathbf{0}.\] Our goal is to show that all coefficients \(c_1, c_2, c_3, d_1,\) and \(d_2\) must be zero.

Analyze Relation Among Vectors in U

Since \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) is a basis of \(U\), the vectors are linearly independent. Therefore, any linear combination of these vectors equal to zero implies \(c_1 = c_2 = c_3 = 0\). However, include \(d_1\mathbf{w}_1 + d_2\mathbf{w}_2\) and consider zero summation.

Analyze Relation Among Vectors in W

Similarly, since \(\{\mathbf{w}_1, \mathbf{w}_2\}\) is a basis of \(W\), these vectors are also independent. Any linear combination leading to zero means \(d_1 = d_2 = 0\) when substituted from subspace \(U\)'s basis contribution.

Consider Intersection Condition

The key condition given is that the intersection of \(U\) and \(W\) is only the zero vector. So, if the linear combination from Step 2 is zero, the only solution is \(c_1 = c_2 = c_3 = d_1 = d_2 = 0\).

Conclusion

Given the above analysis, the entire set of vectors \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \mathbf{w}_1, \mathbf{w}_2\}\) must be linearly independent because their coefficients are forced to be zero due to the basis and intersection conditions.

Key concepts

Vector Spaces

A vector space is a mathematical structure formed by a collection of vectors. These vectors combine according to specific rules of addition and scalar multiplication. Think of a vector space like a coordinate grid where each point on the grid represents a different vector. This grid expands infinitely in each direction based on its defined dimension.

In a vector space, certain criteria must hold:

• Vectors can be added together, and the sum is also a vector in the same space.
• Vectors can be multiplied by scalars (numbers), and the result remains within the space.
• The space contains a zero vector, acting as an additive identity.
• Addition of vectors is commutative and associative.
• Scalar multiplication is distributive over vector addition and scalar sums.

The dimension of a vector space, such as the one discussed with subspaces \(U\) and \(W\), indicates the number of vectors in its basis. These vectors span the space, meaning any vector in the space can be expressed as a combination of these basis vectors.

Basis of a Subspace

Understanding a basis is crucial to explicating vector spaces and subspaces. A basis of a subspace is a set of vectors that are both linearly independent and span the entire subspace. If you consider subspace \(U\), it means the vectors \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\) form a basis for \(U\). This indicates the following:

• The vectors are linearly independent, meaning no vector in the set can be written as a combination of others.
• They span \(U\), suggesting that any vector in \(U\) can be expressed as a linear combination of these basis vectors.

Having a basis is significant because it provides a 'coordinate system' for the subspace, allowing every vector in \(U\) to be uniquely represented via coefficients in a linear combination. In our exercise, it's pivotal because showing linear independence of the extended set \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3, \mathbf{w}_1, \mathbf{w}_2\}\) hinges on the basis properties of both \(U\) and \(W\).

Intersection of Subspaces

The intersection of two subspaces \(U\) and \(W\) entails finding all vectors that satisfy being a part of both subspaces simultaneously. In geometrical terms, think of intersections as the overlapping region of two linear spans. In our exercise, it is given that the intersection of \(U\) and \(W\) is solely the zero vector.

This scenario implies that there are no non-zero vectors common to both subspaces, which plays a fundamental role in demonstrating the linear independence of the combined set of vectors from \(U\) and \(W\). Why does this matter? Because it ensures that any linear combination equal to zero involving vectors from \(U\) and vectors from \(W\) separately yields trivial solutions—that is, solutions where all scalar coefficients are zero, thereby confirming linear independence of the union of vectors from \(U\) and \(W\).

This principle is vital in vector algebra, illustrating how constraints on intersections can inform us about the structure and independence of combined vector sets.