Problem 12 - Differential Equations and Linear Algebra

Question

Show that \(\mathbf{v}_{1}=(1,-5), \mathbf{v}_{2}=(6,3)\) span \(\mathbb{R}^{2},\) and express the vector \(\mathbf{v}=(x, y)\) as a linear combination of \(\mathbf{v}_{1}, \mathbf{v}_{2}\)

Step-by-Step Solution

Answer

Vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) span \(\mathbb{R}^2\) because their determinant is non-zero: \(\det(A) = 33\). An arbitrary vector \(\mathbf{v} = (x, y)\) can be expressed as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\) with scalars \(a = \frac{3x - 6y}{33}\) and \(b = \frac{5x + y}{33}\): \[\mathbf{v} = \left(\frac{3x - 6y}{33}\right) \mathbf{v}_1 + \left(\frac{5x + y}{33}\right) \mathbf{v}_2\]

Step 1: Check if \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent

Calculate the determinant of the matrix formed by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) as columns: \[A = \begin{bmatrix} 1 & 6 \\ -5 & 3 \end{bmatrix}\] \[\det(A) = (1)(3) - (6)(-5) = 3 + 30 = 33\] Since the determinant is non-zero, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are linearly independent, which implies that they span \(\mathbb{R}^2\).

Step 2: Express \(\mathbf{v} = (x, y)\) as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\)

To express \(\mathbf{v}\) as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\), we are looking for scalars \(a\) and \(b\) such that: \[\mathbf{v} = a\mathbf{v}_1 + b\mathbf{v}_2\] This can be written as a system of linear equations: \[\begin{cases} x = a \cdot 1 + b \cdot 6 \\ y = a \cdot (-5) + b \cdot 3 \end{cases}\] We already have the matrix \(A\) from Step 1, so we can solve for the unknowns \(a\) and \(b\) using the inverse of \(A\): \[A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 3 & -6 \\ 5 & 1 \end{bmatrix} = \frac{1}{33} \begin{bmatrix} 3 & -6 \\ 5 & 1 \end{bmatrix}\] Multiply \(A^{-1}\) by the vector \(\mathbf{v}\): \[\begin{bmatrix} a \\ b \end{bmatrix} = A^{-1} \mathbf{v} = \frac{1}{33} \begin{bmatrix} 3 & -6 \\ 5 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\] \[\begin{bmatrix} a \\ b \end{bmatrix} = \frac{1}{33} \begin{bmatrix} 3x - 6y \\ 5x + y \end{bmatrix} = \begin{bmatrix} \cfrac{3x - 6y}{33} \\ \cfrac{5x + y}{33} \end{bmatrix}\] Therefore, \(\mathbf{v} = (x, y)\) can be expressed as a linear combination of \(\mathbf{v}_1\) and \(\mathbf{v}_2\) with the scalars \(a = \frac{3x - 6y}{33}\) and \(b = \frac{5x + y}{33}\): \[\mathbf{v} = \left(\frac{3x - 6y}{33}\right) \mathbf{v}_1 + \left(\frac{5x + y}{33}\right) \mathbf{v}_2\]

Key Concepts

Linear Independence

Linear independence is an essential concept in linear algebra. It helps us understand if a set of vectors is sufficient to explore an entire space. Vectors are deemed linearly independent if no vector in the set can be written as a combination of others. It's like having distinct directions that don't overlap.
In our problem, we have vectors \( \mathbf{v}_{1} = (1, -5) \) and \( \mathbf{v}_{2} = (6, 3) \). To test if they are linearly independent, we compute the determinant of the matrix formed by these vectors:

• Write \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) as columns in a matrix: \( A = \begin{bmatrix} 1 & 6 \ -5 & 3 \end{bmatrix} \).
• Calculate the determinant: \( \det(A) = (1)(3) - (6)(-5) = 33 \).

A non-zero determinant (here \( 33 \)) confirms that the vectors are linearly independent. This means they cover the entire 2D space of \( \mathbb{R}^{2} \), offering a foundation to explore the whole plane.

Spanning Sets

A set of vectors spans a space if you can construct any vector within that space using a linear combination of the set. Essentially, spanning allows you to reach every point within a specific dimension using a set of vectors formulated from it.
The vectors \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) span \( \mathbb{R}^{2} \) because they are linearly independent. Thus, you can express any vector \( (x, y) \) in terms of these vectors. They're like tools or building blocks for creating any vector in the plane. As they span the whole space, they ensure thorough coverage of all possible combinations.

Linear Combinations

Understanding linear combinations involves combining different vectors through scalar multiplication and then adding them together. It's like mixing colors to get different shades.
To express the vector \( \mathbf{v} = (x, y) \) as a linear combination of \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \), we seek scalars \( a \) and \( b \):

• Set up an equation: \( \mathbf{v} = a \mathbf{v}_{1} + b \mathbf{v}_{2} \).
• Write it as two equations: \( x = a \cdot 1 + b \cdot 6 \) and \( y = a \cdot (-5) + b \cdot 3 \).

By substituting the values for \( a \) and \( b \) determined in the solution, any vector \( \mathbf{v} \) in \( \mathbb{R}^{2} \) becomes a combination of \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \) using specific coefficients based on its position in the space.

Inverse of a Matrix

The inverse of a matrix is like a mathematical undo button. When you multiply a matrix by its inverse, you return to the identity matrix, essentially "removing" the matrix's effect on another vector or matrix.
In this context, the inverse helps solve linear equations more easily. To express \( \mathbf{v} \) as described above, we determine the inverse of matrix \( A \), constructed from vectors \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \):

• Calculate using: \( A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 3 & -6 \ 5 & 1 \end{bmatrix} \).
• This gives: \( A^{-1} = \frac{1}{33} \begin{bmatrix} 3 & -6 \ 5 & 1 \end{bmatrix} \).

Multiplying \( A^{-1} \) with \( \mathbf{v} \) gets the scalars \( a \) and \( b \). This illustrates how vector \( \mathbf{v} \) can be broken down into components of the vectors \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \), offering a clearer perspective and solution.