Problem 49 - Differential Equations and Linear Algebra

Question

Prove that $$ \operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}=\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\} $$ if and only if \(v_{3}\) can be written as a linear combination of \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\)

Step-by-Step Solution

Answer

To prove that \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\} = \operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) if and only if \(v_3\) can be written as a linear combination of \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), we showed the following: 1) If \(v_3 = a\mathbf{v}_1 + b\mathbf{v}_2\), we proved that any vector in the \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) can be written as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\), and thus the two spans are equal. 2) If the spans are equal, we showed that \(\mathbf{v}_3\) can be written as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\). Therefore, the biconditional statement is proven true.

Step 1: Define the given vectors and spans

First, let's define our vectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) and the two spans: \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) and \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\).

Step 2: Prove the first direction of the implication

Assume that \(v_3\) can be written as a linear combination of \(v_1\) and \(v_2\). This means there exist scalars \(a, b\) such that: \[ \mathbf{v}_3 = a\mathbf{v}_1 + b\mathbf{v}_2 \] Now to show that \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\} = \operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\), we need to prove that any vector in the left-hand side span can also be written as a linear combination of the vectors in the right-hand side span. Let \(\mathbf{w}\) be an arbitrary vector in the left-hand side span, so \(\mathbf{w} = p\mathbf{v}_1 + q\mathbf{v}_2 + r\mathbf{v}_3\), for some scalars \(p, q, r\). Substitute the expression for \(\mathbf{v}_3\) found earlier: \(\mathbf{w} = p\mathbf{v}_1 + q\mathbf{v}_2 + r(a\mathbf{v}_1 + b\mathbf{v}_2)\). By rearranging the terms, we get: \(\mathbf{w} = (p+ra)\mathbf{v}_1 + (q+rb)\mathbf{v}_2\), which shows that \(\mathbf{w}\) can be written as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\). Therefore, \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is a subset of \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\). However, it is clear that \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is always a subset of \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\), since the latter span includes all the vectors in the former span. Hence, the two spans are equal.

Step 3: Prove the second direction of the implication

Assume that \(\operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\} = \operatorname{span}\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\). As \(v_3\) is in the left-hand side span, it must also be in the right-hand side span because of the assumption. This means there exist scalars \(a, b\) such that: \[ \mathbf{v}_3 = a\mathbf{v}_1 + b\mathbf{v}_2 \] Hence, \(\mathbf{v}_3\) can be written as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\) under this assumption.

Step 4: Conclude the proof

We have proven both directions of the implication: if \(v_3\) can be written as a linear combination of \(v_1\) and \(v_2\), then the spans are equal, and if the spans are equal, then \(v_3\) can be written as a linear combination of \(v_1\) and \(v_2\). Thus, the biconditional statement is proven true.

Key Concepts

Vector Spaces

A vector space is a fundamental concept in linear algebra, forming the environment where vectors live and operate. Imagine vector spaces as a greenhouse for vectors, providing a set of rules they must obey.
These rules are formally known as axioms, and they ensure that operations such as vector addition or scalar multiplication make sense.

• Vectors can be added together, and the sum is still within the vector space. For instance, if \( \mathbf{u} \) and \( \mathbf{v} \) are vectors within a specific vector space, their sum \( \mathbf{u} + \mathbf{v} \) is in that space too.
• Vectors can be multiplied by scalars (numbers), which means if you take a vector and "stretch" or "shrink" it with a scalar, it remains in the same space.

Vector spaces can have various dimensions, with 3-D spaces being the most intuitive for us, like our physical universe with width, height, and depth. But vector spaces could extend beyond 3D to accommodate different applications.

Span

The "span" of a set of vectors describes all the possible vectors you can get through linear combinations of those vectors. Think of it as all the destinations you can reach if you have specific vectors as your stepping stones.
In simpler terms, given vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \), their span consists of all vectors of the form: \[ a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \ldots + a_n\mathbf{v}_n,\]where \( a_1, a_2, \ldots, a_n \) are scalars.

• If the span of your given vectors fills up the whole vector space, they can produce any vector in that space.
• If not, the span might only represent a subspace, a smaller part of the entire space.

It's important to know that the span gives insight into how powerful or flexible your set of vectors is in generating new directions or vectors.

Linear Dependence

Linear dependence is a key concept to determine whether a set of vectors can account for each other without introducing something new to the space. A set of vectors is said to be linearly dependent if at least one of the vectors is expressible as a linear combination of the others.
Say you have vectors \( \mathbf{v}_1, \mathbf{v}_2, \) and \( \mathbf{v}_3 \). If \( \mathbf{v}_3 \) can be written as \( a\mathbf{v}_1 + b\mathbf{v}_2 \) for some scalars \( a \) and \( b \), then these vectors are linearly dependent. In essence, \( \mathbf{v}_3 \) doesn't bring anything "new" to the table because it's something you can already reach using \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).

• Linear dependence implies redundancy within your set of vectors, meaning not all vectors are needed to form the span.
• If vectors are linearly independent, none of them can be written as a combination of others, hence, each vector contributes a unique direction or dimension.

Understanding linear dependence and independence helps realize the potential or limitations of your vector set in spanning new dimensions or directions.