Problem 15 - Differential Equations and Linear Algebra

Question

Show that \(\mathbf{v}_{1}=(1,1), \mathbf{v}_{2}=(-1,2), \mathbf{v}_{3}=(1,4)\) span \(\mathbb{R}^{2} .\) Do \(\mathbf{v}_{1}, \mathbf{v}_{2}\) alone span \(\mathbb{R}^{2}\) also?

Step-by-Step Solution

Answer

In short, both the sets \(\{\mathbf{v}_{1}, \mathbf{v}_{2}\}\) and \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\}\) span \(\mathbb{R}^{2}\). Since the determinant of the matrix formed by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) is not zero, they are linearly independent and form a basis for \(\mathbb{R}^{2}\). Adding \(\mathbf{v}_{3}\) to the set does not change the fact that they span the entire space of \(\mathbb{R}^{2}\).

Step 1: Check if the given vectors are linearly independent

To check the linear independence of the vectors, we can use the determinant test. That is, we calculate the determinant of the matrix formed by these vectors. If the determinant is not zero, then the vectors are linearly independent. First, let's create a matrix A with these vectors as columns: \[A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 2 & 4 \end{bmatrix}\] Now, let's calculate the determinant of the square matrix formed by any two vectors. We can choose the first two columns since we are specifically interested in checking if just \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) can span \(\mathbb{R}^{2}\) also: \[\det\left(\begin{bmatrix} 1 & -1 \\ 1 & 2 \end{bmatrix}\right) = (2) - (-1) = 2 + 1 = 3\] The determinant is not zero, which means \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent.

Step 2: Determine if the given vectors span \(\mathbb{R}^{2}\)

Since \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent, they can form a basis for \(\mathbb{R}^{2}\). Therefore, they can span the space \(\mathbb{R}^{2}\). Now let's analyze if the all the three given vectors: \(\mathbf{v}_{1}, \mathbf{v}_{2}\), and \(\mathbf{v}_{3}\) span \(\mathbb{R}^{2}\). Since \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) already span \(\mathbb{R}^{2}\), adding more vectors to the set won't change the fact that they generate the entire space of \(\mathbb{R}^{2}\). Therefore, \(\mathbf{v}_{1}, \mathbf{v}_{2}\), and \(\mathbf{v}_{3}\) also span \(\mathbb{R}^{2}\). In conclusion, both the sets \(\{\mathbf{v}_{1}, \mathbf{v}_{2}\}\) and \(\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\}\) span \(\mathbb{R}^{2}\).

Key Concepts

Linear Independence

Understanding linear independence is crucial when studying linear algebra. A set of vectors is considered linearly independent if no vector in the set can be written as a linear combination of the others. In simpler terms, think of it like a recipe where each ingredient (vector) cannot be made from mixing the others – it's unique!

To check for linear independence, we can calculate the determinant of a square matrix formed by the vectors. If the determinant is non-zero, just like a unique ingredient, it tells us that each vector adds something new to the mix and none is redundant. This concept is essential because it helps us to understand whether our set of vectors can span a space without overlapping efforts.

Determinant

The determinant of a matrix is much like a secret decoder—it reveals hidden properties of the matrix. Particularly, when the matrix represents a transformation in space, the determinant can tell you if the transformation preserves the space (non-zero determinant) or squashes it down to a lower dimension (zero determinant).

Consider the matrix formed by the vectors \(\textbf{v}_1\) and \(\textbf{v}_2\), and its determinant being 3. This non-zero value confirms that our vectors are linearly independent and their transformation doesn’t crush the plane to a line or a point. Calculating the determinant is a straightforward and powerful way to test the muscle of our vectors in preserving space dimensions.

Basis for Vector Space

A basis for vector space is the skeleton that supports the entire structure of that space. Think of it as the minimal set of building blocks from which any vector in the space can be constructed. In our case, if a set of vectors can span a vector space, they're already giving us the vibes of a potential basis.

Once it’s clear that a set of vectors like \(\textbf{v}_1\) and \(\textbf{v}_2\) is linearly independent and spans \(\textbb{R}^2\), it’s like declaring, “Yes, we have found our sturdy foundation!” These vectors form the basis for the space, meaning with these vectors, we can reach any point in the space through linear combination, just like using LEGO blocks to build any structure you can imagine.