Problem 33 - Differential Equations and Linear Algebra

Question

Let \(\mathbf{v}_{1}=(1,1)\) and \(\mathbf{v}_{2}=(-1,1)\) (a) Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) spans \(\mathbb{R}^{2}\). (b) Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is linearly independent. (c) Conclude from (a) \(o r\) (b) that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for \(\mathbb{R}^{2}\). What theorem in this section allows you to draw this conclusion from either (a) or (b), without proving both?

Step-by-Step Solution

Answer

In summary, we have shown that the given set of vectors \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) both spans \(\mathbb{R}^{2}\) and is linearly independent. By the Basis Theorem, this implies that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for \(\mathbb{R}^{2}\).

Step 1: Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) spans \(\mathbb{R}^{2}\)

In order to show that these vectors span \(\mathbb{R}^{2}\), we need to show that for any arbitrary vector \(\mathbf{w} = (x,y) \in \mathbb{R}^{2}\), there exist scalars \(c_1\) and \(c_2\) such that \(\mathbf{w} = c_1\mathbf{v}_{1} + c_2\mathbf{v}_{2}\). Let's start by setting up the equation: \[\mathbf{w} = c_1(1,1) + c_2(-1,1)\] Now we need to find \(c_1\) and \(c_2\) in terms of \(x\) and \(y\) to show they exist. We can write the above equation as a system of two linear equations: \[ \begin{cases} c_1 - c_2 = x \\ c_1 + c_2 = y \end{cases} \] We can easily solve for \(c_1\) and \(c_2\) by adding the two equations: \[2c_1 = x + y\] So, \(c_1 = \dfrac{x+y}{2}\). Subtracting the second equation from the first gives: \[2c_2 = y - x\] So, \(c_2 = \dfrac{y-x}{2}\). Since we have found scalars \(c_1\) and \(c_2\) in terms of \(x\) and \(y\), we can conclude that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) spans \(\mathbb{R}^{2}\).

Step 2: Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is linearly independent

To show that these vectors are linearly independent, we need to prove that the only solution to the equation \(c_1\mathbf{v}_{1} + c_2\mathbf{v}_{2} = \mathbf{0}\) is \(c_1 = 0\) and \(c_2 = 0\). The equation can be written as a system of two linear equations: \[ \begin{cases} c_1 - c_2 = 0 \\ c_1 + c_2 = 0 \end{cases} \] Subtracting the first equation from the second one gives: \[2c_2 = 0\] So, \(c_2 = 0\). Now substituting \(c_2=0\) into the first equation, we have: \[c_1 - 0 = 0\] So, \(c_1 = 0\). Since the only solution to this system is \(c_1 = 0\) and \(c_2 = 0\), we can conclude that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is linearly independent.

Step 3: Conclude that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for \(\mathbb{R}^{2}\)

We have shown that the given vectors span \(\mathbb{R}^{2}\) and are linearly independent. A set of vectors that spans a vector space and is linearly independent is a basis for that vector space. Therefore, we can conclude that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for \(\mathbb{R}^{2}\). The theorem that allows us to draw this conclusion from either (a) or (b) is the "Basis Theorem". The Basis Theorem states that in a finite-dimensional vector space, any linearly independent set of vectors can be extended to a basis for that vector space, and any spanning set of vectors can be reduced to a basis for that vector space. In this specific case, we have shown that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a spanning set of vectors and is linearly independent, so by the Basis Theorem, they form a basis for \(\mathbb{R}^{2}\).

Key Concepts

Spanning Vectors

Spanning vectors are a key concept in linear algebra, as they help us understand how a set of vectors can relate to a vector space. For a set of vectors to span a vector space, it means that any vector in that space can be expressed as a linear combination of the set. In the case of the vectors \(\mathbf{v}_1 = (1,1)\) and \(\mathbf{v}_2 = (-1,1)\), we can say they span \(\mathbb{R}^{2}\) if any vector \(\mathbf{w} = (x, y)\) in \(\mathbb{R}^{2}\) can be formed using \(\mathbf{v}_1\) and \(\mathbf{v}_2\).

To demonstrate this, consider the equation \(\mathbf{w} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2\). By setting this up as a system of equations, we get:

• \(c_1 - c_2 = x\)
• \(c_1 + c_2 = y\)

By solving this system, we can find values for \(c_1\) and \(c_2\), meaning that any \(x\) and \(y\) can be achieved. Thus, the set \(\{\mathbf{v}_1, \mathbf{v}_2\}\) spans \(\mathbb{R}^{2}\).

This property is crucial for applications such as geometric transformations, where it's important that a set of vectors can describe every direction within a space.

Linear Independence

Linear independence is about determining whether a set of vectors is related in a way that allows none of them to be written as a linear combination of the others. This concept is essential because it ensures that each vector in the set contributes something unique to the space. If a set of vectors is linearly independent, the only solution to the equation \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = \mathbf{0}\) is for all the coefficients \(c_1\) and \(c_2\) to be zero.

In our exercise, \(\mathbf{v}_1 = (1,1)\) and \(\mathbf{v}_2 = (-1,1)\), we check their independence by setting up the equations:

• \(c_1 - c_2 = 0\)
• \(c_1 + c_2 = 0\)

Solving these equations, we find that both \(c_1 = 0\) and \(c_2 = 0\). Therefore, they are linearly independent.

This property shows that the vectors do not lie on the same line, providing them the capacity to form the foundation of a vector space, like building blocks, without any redundancy.

Basis Theorem

The Basis Theorem is a vital part of linear algebra that bridges the concepts of spanning and linear independence. This theorem states that a set of vectors that both spans a vector space and is linearly independent is a basis for that space. A basis provides a coordinate system for the vector space, allowing any vector to be uniquely represented as a combination of the basis vectors.

Using our vectors \(\mathbf{v}_1 = (1,1)\) and \(\mathbf{v}_2 = (-1,1)\), we demonstrated they span \(\mathbb{R}^{2}\) and are linearly independent. Applying the Basis Theorem, we conclude they are a basis for \(\mathbb{R}^{2}\).

Understanding this concept allows us to reconfigure and manipulate spaces in dimensional analyses and transforms, much like creating a reliable blueprint for constructing geometric shapes or solving systems of equations. The comprehensive role of the Basis Theorem highlights how interconnected these linear algebra concepts are, emphasizing their underlying importance in both theoretical and applied contexts.