Problem 3 - Advanced Engineering Mathematics

Question

In Problems 1-6, determine which of the indicated column vectors are eigenvectors of the given matrix \(\mathbf{A}\). Give the corresponding eigenvalue. $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ll} 0 & 5 \\ 2 & 1 \end{array}\right) ; \quad \mathbf{K}_{1}=\left(\begin{array}{r} 5 \\ -2 \end{array}\right) \\ &\mathbf{K}_{2}=\left(\begin{array}{l} 1 \\ 0 \end{array}\right), \quad \mathbf{K}_{3}=\left(\begin{array}{c} -5 \\ 10 \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

\( \mathbf{K}_3 \) is an eigenvector with eigenvalue \( \lambda = -10 \).

Step 1: Understand Eigenvector Property

An eigenvector of a matrix \( \mathbf{A} \) is a non-zero vector \( \mathbf{v} \) such that \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \) for some scalar \( \lambda \), which is the corresponding eigenvalue.

Step 2: Verify \( \mathbf{K}_1 \) as Eigenvector of \( \mathbf{A} \)

Calculate the product of \( \mathbf{A} \) and \( \mathbf{K}_1 \): \[\mathbf{A} \mathbf{K}_1 = \begin{pmatrix}0 & 5 \ 2 & 1\end{pmatrix} \begin{pmatrix}5 \ -2\end{pmatrix} = \begin{pmatrix}0 \cdot 5 + 5 \cdot (-2) \ 2 \cdot 5 + 1 \cdot (-2)\end{pmatrix} = \begin{pmatrix}-10 \ 8\end{pmatrix}\]Now, verify if this is a scalar multiple of \( \mathbf{K}_1 \) by checking if there exists a \( \lambda \) such that:\( \lambda \begin{pmatrix}5 \ -2\end{pmatrix} = \begin{pmatrix}-10 \ 8\end{pmatrix} \).This equation does not hold for any scalar \( \lambda \), thus \( \mathbf{K}_1 \) is not an eigenvector.

Step 3: Verify \( \mathbf{K}_2 \) as Eigenvector of \( \mathbf{A} \)

Calculate the product of \( \mathbf{A} \) and \( \mathbf{K}_2 \): \[\mathbf{A} \mathbf{K}_2 = \begin{pmatrix}0 & 5 \ 2 & 1\end{pmatrix} \begin{pmatrix}1 \ 0\end{pmatrix} = \begin{pmatrix}0 \cdot 1 + 5 \cdot 0 \ 2 \cdot 1 + 1 \cdot 0\end{pmatrix} = \begin{pmatrix}0 \ 2\end{pmatrix}\]Check if this is a scalar multiple of \( \mathbf{K}_2 \) by determining \( \lambda \):\( \lambda \begin{pmatrix} 1 \ 0\end{pmatrix} = \begin{pmatrix}0 \ 2\end{pmatrix} \).This equation does not hold, hence \( \mathbf{K}_2 \) is not an eigenvector.

Step 4: Verify \( \mathbf{K}_3 \) as Eigenvector of \( \mathbf{A} \)

Calculate the product of \( \mathbf{A} \) and \( \mathbf{K}_3 \): \[\mathbf{A} \mathbf{K}_3 = \begin{pmatrix}0 & 5 \ 2 & 1\end{pmatrix} \begin{pmatrix}-5 \ 10\end{pmatrix} = \begin{pmatrix}0 \cdot (-5) + 5 \cdot 10 \ 2 \cdot (-5) + 1 \cdot 10\end{pmatrix} = \begin{pmatrix}50 \ 0\end{pmatrix}\]Determine if there is a scalar \( \lambda \) such that:\( \lambda \begin{pmatrix} -5 \ 10\end{pmatrix} = \begin{pmatrix}50 \ 0\end{pmatrix} \).Choosing \( \lambda = -10 \) satisfies both components: \(-10 \times -5 = 50\) and \(-10 \times 10 = 0\).Thus, \( \mathbf{K}_3 \) is an eigenvector with eigenvalue \( \lambda = -10 \).

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental concept in linear algebra, essential in dealing with vectors and matrices. To multiply a matrix \( \mathbf{A} \) with a vector \( \mathbf{v} \), you follow a methodical process:

• Consider each row of the matrix.
• Multiply the corresponding elements of the row with the elements of the vector.
• Add these products together for each row to get the resulting vector.

Here's an example with matrix \( \mathbf{A} \) and vector \( \mathbf{K}_1 \):\[\mathbf{A} \mathbf{K}_1 = \begin{pmatrix}0 & 5 \ 2 & 1\end{pmatrix} \begin{pmatrix}5 \ -2\end{pmatrix}\]Calculate for the first row: \((0 \cdot 5 + 5 \cdot (-2)) = -10\).
Calculate for the second row: \((2 \cdot 5 + 1 \cdot (-2)) = 8\).
This results in vector \( \begin{pmatrix}-10 \ 8\end{pmatrix} \).

Eigenvalue Calculation

Calculating eigenvalues involves finding scalars \( \lambda \) such that the matrix operation \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \) holds true. The scalar \( \lambda \) is the eigenvalue if it satisfies

• It converts \( \mathbf{A} \mathbf{v} \) into a scalar multiplication of \( \mathbf{v} \)
• Both sides of \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \) yield the same result.

For instance, with vector \( \mathbf{K}_3 = \begin{pmatrix} -5 \ 10\end{pmatrix} \) and matrix \( \mathbf{A} \):\[\mathbf{A} \mathbf{K}_3 = \begin{pmatrix}0 & 5 \ 2 & 1\end{pmatrix} \begin{pmatrix}-5 \ 10\end{pmatrix} = \begin{pmatrix}50 \ 0\end{pmatrix}\]Finding \( \lambda\begin{pmatrix} -5 \ 10\end{pmatrix} = \begin{pmatrix}50 \ 0\end{pmatrix}\), select \( \lambda = -10 \) making both components valid: \(-10 \cdot (-5) = 50\) and \(-10 \cdot 10 = 0\).
Thus, the eigenvalue \( \lambda = -10 \) is correct.

Linear Algebra Concepts

Linear algebra is an essential branch of mathematics that studies vectors, matrices, and linear transformations. Under this umbrella, eigenvectors and eigenvalues are key concepts.
An eigenvector is a non-zero vector whose direction remains unchanged after a linear transformation represented by a matrix is applied. Meanwhile, an eigenvalue is a scalar that represents how much the eigenvector is stretched or compressed.

• Eigenvectors provide fundamental insights into matrix behavior.
• They help simplify complex matrix operations.
• Applications range from solving differential equations to transforming data in machine learning.

In practical terms, knowing if a vector \( \mathbf{v} \) is an eigenvector of \( \mathbf{A} \) requires checking if the product \( \mathbf{A} \mathbf{v} \) is a scalar version of \( \mathbf{v} \). This understanding is pivotal in efficiently analyzing systems that can be represented in a matrix form.