Problem 2 - Differential Equations and Linear Algebra

Question

Let \(S=\left\\{\mathbf{x} \in \mathbb{R}^{2}: \mathbf{x}=(2 k,-3 k), k \in \mathbb{R}\right\\}\) (a) Show that \(S\) is a subspace of \(\mathbb{R}^{2}\). (b) Make a sketch depicting the subspace \(S\) in the Cartesian plane.

Step-by-Step Solution

Answer

a) S is a subspace of ℝ² since it satisfies the following properties: 1. The zero vector (0,0) is in S when k = 0. 2. S is closed under vector addition: (2k₁, -3k₁) + (2k₂, -3k₂) = (2(k₁ + k₂), -3(k₁ + k₂)). 3. S is closed under scalar multiplication: c(2k, -3k) = (2ck, -3ck). b) The subspace S can be depicted in the Cartesian plane as a straight line passing through the origin and the point (2, -3). Create a sketch of a Cartesian plane, plot the point (2, -3), and draw a line through the origin and this point.

Step 1: a) Checking Subspace Properties

To show that S is a subspace of ℝ², we need to verify that it satisfies the following properties: 1. The zero vector is in S. 2. If two vectors u and v are in S, their sum (u + v) is also in S. 3. If a vector u is in S and c is a scalar, their product (cu) is also in S.

Step 2: 1. Checking the Zero Vector

We are given that S consists of all vectors (2k, -3k) with k in ℝ. We want to find a value of k that results in the zero vector (0,0): (2k, -3k) = (0, 0) To find such a value for k, set each component equal to 0: 2k = 0 => k = 0 -3k = 0 => k = 0 Since k = 0 gives us the zero vector (0,0), the first property is satisfied.

Step 3: 2. Checking Closure under Vector Addition

Let u = (2k₁, -3k₁) and v = (2k₂, -3k₂) be two arbitrary vectors in S. We want to show that their sum, u + v, is also in S: u + v = (2k₁ + 2k₂, -3k₁ - 3k₂) = (2(k₁ + k₂), -3(k₁ + k₂)) Since (k₁ + k₂) is a real number (as the sum of two real numbers), the resulting vector (2(k₁ + k₂), -3(k₁ + k₂)) is also in S. Therefore, the second property is satisfied.

Step 4: 3. Checking Closure under Scalar Multiplication

Let u = (2k, -3k) be an arbitrary vector in S and let c be an arbitrary scalar. We want to show that the product cu is also in S: cu = c(2k, -3k) = (2ck, -3ck) Since ck is a real number (as the product of two real numbers), the resulting vector (2ck, -3ck) is also in S. Therefore, the third property is satisfied. Since S satisfies all three properties, it is a subspace of ℝ².

Step 5: b) Sketching the Subspace in the Cartesian Plane

To sketch the subspace in the Cartesian plane, notice that the vectors of S consist of elements (2k, -3k). Since we're in ℝ², we can rewrite the elements of S as the scalar multiple of a single vector: (2k, -3k) = k(2, -3) This means the subspace S consists of scalar multiples of the vector (2, -3). The subspace will be a straight line passing through the origin and the point (2, -3). Draw a Cartesian plane with the x-axis and y-axis, and plot the point (2, -3). Then draw a line through this point and the origin to represent the subspace S.

Key Concepts

Vector Addition

In the context of vector spaces, vector addition is a fundamental operation. It's similar to adding numbers, but instead, you're combining vectors. When you add two vectors, you add their corresponding components. For instance, if you have two vectors in \( \mathbb{R}^2 \) represented as \( u = (u_1, u_2) \) and \( v = (v_1, v_2) \), the sum \( u + v \) is \( (u_1 + v_1, u_2 + v_2) \). This operation is crucial in verifying subspace criteria as it tells us whether the vector space is closed under addition.
In the case of the subspace \( S \) defined in the exercise, we verified that the sum of any two vectors \( (2k_1, -3k_1) \) and \( (2k_2, -3k_2) \) also results in a vector of the form \( (2k, -3k) \). Hence, the subspace \( S \) retains this property of closure, which is necessary for \( S \) to be a valid subspace of \( \mathbb{R}^2 \).

Vector Scalar Multiplication

Scalar multiplication involves taking a vector and multiplying it by a scalar, which is a single number. For a vector \( (x, y) \), multiplying by a scalar \( c \) results in a new vector \( (cx, cy) \). This property is essential for vector spaces because it must hold true for them to be valid, which means there should be closure under scalar multiplication.
In the exercise, we checked if multiplying a vector from \( S \) represented as \( (2k, -3k) \) by any scalar \( c \) results in another vector of the same form. The multiplication gives us \( (2ck, -3ck) \), showing that it creates another vector in \( S \). This ensures that the subspace \( S \) endorses scalar multiplication, making it a valid subspace of \( \mathbb{R}^2 \).

Zero Vector

The zero vector is a special vector where all components are zero. In two-dimensional space, this is simply \( (0,0) \). The presence of a zero vector is crucial in any vector space because it serves as an additive identity.
To confirm if a set is a subspace, it must contain the zero vector. This means you should be able to find a \( k \) such that \( (2k, -3k) = (0, 0) \). As shown in the solution steps, for \( k = 0 \), we indeed get the vector \( (0,0) \).
This inclusion of the zero vector ensures that the space is not just closed under addition and scalar multiplication, but also includes this identity element, solidifying it as a subspace.

Cartesian Plane

The Cartesian plane is a two-dimensional surface defined by an x-axis and a y-axis. It helps visualize relationships between variables, especially geometric ones. In the context of our exercise, understanding how the subspace looks involves seeing it as a subset of the Cartesian plane.
Our exercise involves a set \( S \) defined in terms of vectors \( (2k, -3k) \), which can be rewritten as scalar multiples of a base vector \( (2, -3) \). This implies that the subspace looks like a line in the Cartesian plane.
To sketch \( S \), you'd plot the line that extends through the origin \( (0,0) \) and passes through the point \( (2, -3) \). By drawing this line, you're essentially representing the infinite collection of vectors that make up the subspace, each represented as a scalar multiple of \( (2, -3) \). As it slices through the origin, it aligns with the definition of a subspace, which is a key realization about how subspaces appear in a geometric context.