Problem 26 - Differential Equations and Linear Algebra

Question

(a) Let \(\mathbf{v}_{1}=(1,1)\) and \(\mathbf{v}_{2}=(1,-1) .\) Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for \(\mathbb{R}^{2}.\) (b) Let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation satisfying $$ T\left(\mathbf{v}_{1}\right)=(2,3), \quad T\left(\mathbf{v}_{2}\right)=(-1,1) $$ where \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are the basis vectors given in (a). Find \(T\left(x_{1}, x_{2}\right)\) for an arbitrary vector \(\left(x_{1}, x_{2}\right)\) in \(\mathbb{R}^{2} .\) What is \(T(4,-2) ?\)

Step-by-Step Solution

Answer

The basis {\(\mathbf{v}_{1}, \mathbf{v}_{2}\)} for \(\mathbb{R}^{2}\) is linearly independent and spans \(\mathbb{R}^2\). The transformation \(T(x_1, x_2) = (x_1 + x_2, x_1 + 2x_2)\), and \(T(4, -2) = (2, 0)\).

Step 1: Check for linear independence

To check if the vectors \(\mathbf{v}_{1}=(1,1)\) and \(\mathbf{v}_{2}=(1,-1)\) are linearly independent, we'll set up the following linear combination: $$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{0}.$$ That is: $$c_1 (1, 1) + c_2 (1, -1) = (0, 0).$$ Solving for \(c_1\) and \(c_2\), we have: \begin{align*} c_1 + c_2 &= 0, \\ c_1 - c_2 &= 0. \end{align*} It is clear that \(c_1 = c_2 = 0\). Since the only solution is the trivial solution, the vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are linearly independent.

Step 2: Show that the vectors span \(\mathbb{R}^2\)

To show that vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) span \(\mathbb{R}^2\), we need to show that any vector in \(\mathbb{R}^2\) can be written as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\). Let a vector \(\mathbf{x}=(x_1, x_2) \in \mathbb{R}^2\). We want to find scalars \(c_1\) and \(c_2\) such that: $$c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{x}.$$ That is: $$c_1 (1, 1) + c_2 (1, -1) = (x_1, x_2).$$ Solving for \(c_1\) and \(c_2\), we have: \begin{align*} c_1 + c_2 &= x_1, \\ c_1 - c_2 &= x_2. \end{align*} From the above system of equations, it follows that: \begin{align*} c_1 &= \frac{x_1 + x_2}{2}, \\ c_2 &= \frac{x_1 - x_2}{2}. \end{align*} Because we can find a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\) for any vector in \(\mathbb{R}^2\), the vectors span \(\mathbb{R}^2\). Since the vectors are linearly independent and span \(\mathbb{R}^2\), {\(\mathbf{v}_{1}, \mathbf{v}_{2}\)} is a basis for \(\mathbb{R}^{2}\). For part (b):

Step 3: Find the transformation \(T(x_1, x_2)\)

Using the basis vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\), we can write any vector in \(\mathbb{R}^2\) as: $$\mathbf{x} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 .$$ Then, applying the transformation \(T\), we get: $$T(\mathbf{x}) = c_1 T(\mathbf{v}_1) + c_2 T(\mathbf{v}_2).$$ Substituting the given transformations and values of \(c_1\) and \(c_2\), we have: \begin{align*} T(\mathbf{x}) &= \frac{x_1 + x_2}{2} (2, 3) + \frac{x_1 - x_2}{2} (-1, 1) \\ &= \left(\frac{4x_1 + 4x_2 - 2x_1 + 2x_2}{4}, \frac{6x_1 - 2x_1 + 6x_2 + 2x_2}{4}\right) \\ &= (x_1 + x_2, x_1 + 2x_2). \end{align*} So, the transformation \(T(x_1, x_2) = (x_1 + x_2, x_1 + 2x_2)\).

Step 4: Find \(T(4, -2)\)

Using the transformation we found in Step 3, we can find \(T(4, -2)\) as follows: $$T(4, -2) = (4 + (-2), 4 + 2(-2)) = (2, 0).$$ So, \(T(4, -2) = (2, 0)\).

Key Concepts

Linear Independence

Linear independence is a crucial concept in linear algebra. It refers to a set of vectors that do not linearly depend on each other. Essentially, no vector in the set can be expressed as a combination of the other vectors. If we take the vectors \( \mathbf{v}_{1} = (1, 1) \) and \( \mathbf{v}_{2} = (1, -1) \), they are linearly independent because the only solutions to the equation

• \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{0} \)
• are \( c_1 = 0 \) and \( c_2 = 0 \).

We check this by setting up and solving the system of equations:

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

The trivial solution proves linear independence, meaning these two vectors offer unique directions in the 2D plane.

Linear Transformation

A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In this context, we have a linear transformation \( T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \) defined by its action on the basis vectors:

• \( T(\mathbf{v}_1) = (2, 3) \)
• \( T(\mathbf{v}_2) = (-1, 1) \)

This transformation can be applied to any vector \( (x_1, x_2) \) in \(\mathbb{R}^2\) by expressing it as a combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \):

• \( T(x_1, x_2) = \left( \frac{x_1 + x_2}{2} (2, 3) + \frac{x_1 - x_2}{2} (-1, 1)\right) \)
• which simplifies to \( (x_1 + x_2, x_1 + 2x_2) \).

This approach highlights how transformations affect vectors, mapping them into a new space while retaining their linear structure.

Basis for Vector Spaces

A basis for a vector space is a set of vectors that both span the space and are linearly independent. For \( \mathbb{R}^2 \), the vectors \( \mathbf{v}_{1} = (1, 1) \) and \( \mathbf{v}_{2} = (1, -1) \) form a basis. This means any vector in \( \mathbb{R}^2 \) can be represented as a linear combination of these vectors. A basis is essential because it provides a concise way of describing the entire vector space. Knowing the basis allows you to:

• Reconstruct any vector using a combination of the basis vectors.
• Ensure uniqueness in representation due to the linear independence of the basis vectors.

This unique combination of spanning and independence defines the usability and flexibility of the vector space's basis.

Span of Vectors

The span of a set of vectors is the collection of all possible linear combinations of those vectors. To demonstrate that \(\mathbf{v}_{1} = (1, 1)\) and \(\mathbf{v}_{2} = (1, -1)\) span \(\mathbb{R}^{2}\), we show they can form any vector \(\mathbf{x} = (x_1, x_2)\) in that space. Specifically, you can find coefficients \( c_1 \) and \( c_2 \) such that:

• \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{x} \)
• which leads to the system: \( c_1 + c_2 = x_1 \) and \( c_1 - c_2 = x_2 \).

Solving this system, you find:

• \( c_1 = \frac{x_1 + x_2}{2} \)
• \( c_2 = \frac{x_1 - x_2}{2} \).

Since any vector \((x_1, x_2)\) can be represented this way, the span of these vectors is indeed the entire 2D plane \(\mathbb{R}^2\). Understanding the span is fundamental in analyzing how vectors can cover and define a space.