Problem 1 - 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. $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{ll} 4 & 2 \\ 5 & 1 \end{array}\right) ; \quad \mathbf{K}_{1}=\left(\begin{array}{r} 5 \\ -2 \end{array}\right), \quad \mathbf{K}_{2}=\left(\begin{array}{l} 2 \\ 5 \end{array}\right) \\ &\mathbf{K}_{3}=\left(\begin{array}{r} -2 \\ 5 \end{array}\right) \end{aligned} $$

Step-by-Step Solution

Answer

\(\mathbf{K}_3\) is an eigenvector with eigenvalue \(-1\).

Step 1: Understand Eigenvectors and Eigenvalues

An eigenvector of a matrix \(\mathbf{A}\) is a non-zero vector \(\mathbf{K} \) such that when \(\mathbf{A}\) multiplies \(\mathbf{K}\), the product is a scalar multiple of \(\mathbf{K}\), i.e., \(\mathbf{A} \cdot \mathbf{K} = \lambda \cdot \mathbf{K}\), where \(\lambda\) is the eigenvalue.

Step 2: Multiply Matrix A and Vector K1

Multiply \(\mathbf{A} = \begin{pmatrix} 4 & 2 \ 5 & 1 \end{pmatrix}\) with \(\mathbf{K}_1 = \begin{pmatrix} 5 \ -2 \end{pmatrix}\) to determine if \(\mathbf{K}_1\) is an eigenvector. Perform the multiplication:\[\mathbf{A} \cdot \mathbf{K}_1 = \begin{pmatrix} 4 & 2 \ 5 & 1 \end{pmatrix} \cdot \begin{pmatrix} 5 \ -2 \end{pmatrix} = \begin{pmatrix} 4 \times 5 + 2 \times (-2) \ 5 \times 5 + 1 \times (-2) \end{pmatrix} = \begin{pmatrix} 16 \ 23 \end{pmatrix}.\]

Step 3: Check if K1 is an Eigenvector

Compare the result \(\begin{pmatrix} 16 \ 23 \end{pmatrix}\) with \(\lambda \cdot \mathbf{K}_1\). If \(\mathbf{K}_1\) were an eigenvector, there would exist a \(\lambda\) such that \(\begin{pmatrix} 16 \ 23 \end{pmatrix} = \lambda \begin{pmatrix} 5 \ -2 \end{pmatrix}\). No common \(\lambda\) satisfies both components, so \(\mathbf{K}_1\) is not an eigenvector.

Step 4: Multiply Matrix A and Vector K2

Now, multiply \(\mathbf{A}\) with \(\mathbf{K}_2 = \begin{pmatrix} 2 \ 5 \end{pmatrix}\):\[\mathbf{A} \cdot \mathbf{K}_2 = \begin{pmatrix} 4 & 2 \ 5 & 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \ 5 \end{pmatrix} = \begin{pmatrix} 4 \times 2 + 2 \times 5 \ 5 \times 2 + 1 \times 5 \end{pmatrix} = \begin{pmatrix} 18 \ 15 \end{pmatrix}.\]

Step 5: Check if K2 is an Eigenvector

Check if \(\begin{pmatrix} 18 \ 15 \end{pmatrix}\) is a scalar multiple of \(\begin{pmatrix} 2 \ 5 \end{pmatrix}\):To find \(\lambda\), set \(\lambda \cdot 2 = 18\), so \(\lambda = 9\). Then check the second component: \(\lambda \cdot 5 = 45 eq 15\). Hence, \(\mathbf{K}_2\) is not an eigenvector.

Step 6: Multiply Matrix A and Vector K3

Finally, multiply \(\mathbf{A}\) with \(\mathbf{K}_3 = \begin{pmatrix} -2 \ 5 \end{pmatrix}\):\[\mathbf{A} \cdot \mathbf{K}_3 = \begin{pmatrix} 4 & 2 \ 5 & 1 \end{pmatrix} \cdot \begin{pmatrix} -2 \ 5 \end{pmatrix} = \begin{pmatrix} 4 \times (-2) + 2 \times 5 \ 5 \times (-2) + 1 \times 5 \end{pmatrix} = \begin{pmatrix} 2 \ -5 \end{pmatrix}.\]

Step 7: Check if K3 is an Eigenvector

Check if \(\begin{pmatrix} 2 \ -5 \end{pmatrix}\) is a scalar multiple of \(\begin{pmatrix} -2 \ 5 \end{pmatrix}\):Set \(\lambda \cdot (-2) = 2\), giving \(\lambda = -1\). Confirm with the second component: \(-\lambda \cdot 5 = -5\), which is true. So \(\mathbf{K}_3\) is an eigenvector with eigenvalue \(\lambda = -1\).

Key Concepts

Matrix multiplication

Matrix multiplication is a fundamental operation in linear algebra that is key to understanding eigenvectors and eigenvalues. To multiply a matrix by a vector, you take each row of the matrix and perform a dot product with the column vector. This results in a new vector.
For example, let's consider a matrix \( \mathbf{A} \) given by:

• \( \begin{pmatrix} 4 & 2 \ 5 & 1 \end{pmatrix} \)

If you multiply this by a vector \( \mathbf{K} \) as in \( \mathbf{K}_1 = \begin{pmatrix} 5 \ -2 \end{pmatrix} \), you carry out the multiplication for each row. Here’s how:

• First row: \((4 \times 5) + (2 \times -2) = 16\)
• Second row: \((5 \times 5) + (1 \times -2) = 23\)

So the resulting vector from the matrix multiplication is \( \begin{pmatrix} 16 \ 23 \end{pmatrix} \).
This operation helps to check if the vector is an eigenvector by comparing it to a scalar multiple of the original vector.

Eigenvalues

Eigenvalues are scalars that give vital information about a matrix, especially when investigating the eigenvectors of the matrix. When a matrix transforms an eigenvector, it does so by simply stretching or compressing it by a factor, which is the eigenvalue.
In the context of the exercise, an eigenvalue \( \lambda \) is the scalar for which there exists a nonzero vector \( \mathbf{K} \) satisfying:

• \( \mathbf{A} \cdot \mathbf{K} = \lambda \cdot \mathbf{K} \)

To find \( \lambda \), you compare the resulting vector from the matrix multiplication with the original vector scaled by \( \lambda \).
If we take \( \mathbf{K}_3 \), for example, after multiplication with \( \mathbf{A} \):

• The result is \( \begin{pmatrix} 2 \ -5 \end{pmatrix} \)

Find \( \lambda \) such that \( \lambda \times (-2) = 2 \) and \( \lambda \times 5 = -5 \). Here, \( \lambda = -1 \).
So, \( -1 \) is the eigenvalue corresponding to \( \mathbf{K}_3 \). This makes the eigenvector and eigenvalue relationship clear: eigenvectors are scaled versions of themselves without direction alteration concerning their matrix transformation.

Linear transformations

Linear transformations are functions that map vectors to vectors in a linear fashion, keeping the operations of vector addition and scalar multiplication intact. Matrices represent these transformations, and they are particularly handy because they simplify complex computations into simple matrix multiplications.
In terms of eigenvectors and eigenvalues, linear transformations using matrices reveal powerful characteristics. When we say a vector \( \mathbf{K} \) is an eigenvector of a matrix \( \mathbf{A} \), it's because \( \mathbf{A} \) transforms \( \mathbf{K} \) into a new vector that is a linear stretch of \( \mathbf{K} \). This is a special linear transformation where the direction of the vector uniquely stays the same, only the magnitude changes by a scalar factor, the eigenvalue.
For example, if \( \mathbf{A} \) transforms \( \mathbf{K}_3 \) such that:

• The resulting vector is \( \begin{pmatrix} 2 \ -5 \end{pmatrix} \)

The matrix is creating a linear transformation that scales \( \mathbf{K}_3 \) by \(-1\) times itself, indicating that the direction is effectively inverted, showcasing a brief understanding of how linear transformations exploit matrices to scale and rotate vectors in their space.