Problem 69 - Calculus for Biology and Medicine

Question

Let $$ A=\left[\begin{array}{rr} -1 & 1 \\ 0 & 2 \end{array}\right] $$ (a) Show that $$ \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \quad \text { and } \quad \mathbf{u}_{2}=\left[\begin{array}{l} 1 \\ 3 \end{array}\right] $$ are eigenvectors of \(A\) and that \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are linearly independent. (b) Represent $$ \mathbf{x}=\left[\begin{array}{r} 1 \\ -3 \end{array}\right] $$ as a linear combination of \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\). (c) Use your results in (a) and (b) to compute \(A^{20} \mathbf{x}\).

Step-by-Step Solution

Answer

\(A^{20} \mathbf{x} = 2\mathbf{u}_1 - 2^{20} \mathbf{u}_2\).

Step 1: Verify Eigenvectors

To prove that \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are eigenvectors of \( A \), we must show that there exist scalars \( \lambda_1 \) and \( \lambda_2 \) such that \( A \mathbf{u}_1 = \lambda_1 \mathbf{u}_1 \) and \( A \mathbf{u}_2 = \lambda_2 \mathbf{u}_2 \).\\( A \mathbf{u}_1 = A \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} -1 \ 0 \end{pmatrix} = (-1) \begin{pmatrix} 1 \ 0 \end{pmatrix} = \mathbf{u}_1 \). Thus, \( \lambda_1 = -1 \).\\( A \mathbf{u}_2 = A \begin{pmatrix} 1 \ 3 \end{pmatrix} = \begin{pmatrix} 1 \ 6 \end{pmatrix} = 2 \begin{pmatrix} 1 \ 3 \end{pmatrix} = 2\mathbf{u}_2 \). Thus, \( \lambda_2 = 2 \).

Step 2: Check Linear Independence

To check if \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are linearly independent, check if the determinant of the matrix formed by these vectors is non-zero.\\[\det \begin{pmatrix} 1 & 1 \ 0 & 3 \end{pmatrix} = (1)(3) - (1)(0) = 3 eq 0.\]Since the determinant is non-zero, \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are linearly independent.

Step 3: Express \( \mathbf{x} \) as Combination of Eigenvectors

To express \( \mathbf{x} = \begin{pmatrix} 1 \ -3 \end{pmatrix} \) as a linear combination of \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), solve \( c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 = \mathbf{x} \), where \( c_1 \) and \( c_2 \) are scalars.\\[\begin{pmatrix} 1 \times c_1 + 1 \times c_2 \ 0 \times c_1 + 3 \times c_2 \end{pmatrix} = \begin{pmatrix} 1 \ -3 \end{pmatrix}\]This gives the equations:\1. \( c_1 + c_2 = 1 \)\2. \( 3c_2 = -3 \)\Solving, \( c_2 = -1 \) and substituting into equation 1, \( c_1 = 2 \). Thus, \( \mathbf{x} = 2\mathbf{u}_1 - \mathbf{u}_2 \).

Step 4: Compute \( A^{20} \mathbf{x} \) Using Eigenvectors

From above, \( \mathbf{x} = 2\mathbf{u}_1 - \mathbf{u}_2 \). Use the eigenvalue property to compute: \( A^{20} \mathbf{x} = A^{20} (2\mathbf{u}_1 - \mathbf{u}_2) = 2 A^{20} \mathbf{u}_1 - A^{20} \mathbf{u}_2 \).\Since \( A \mathbf{u}_1 = -1 \mathbf{u}_1 \) and \( A \mathbf{u}_2 = 2 \mathbf{u}_2 \), raise these to the 20th power:\\[A^{20} \mathbf{u}_1 = (-1)^{20} \mathbf{u}_1 = \mathbf{u}_1, \quad A^{20} \mathbf{u}_2 = 2^{20} \mathbf{u}_2.\]Thus: \[A^{20} \mathbf{x} = 2 \cdot \mathbf{u}_1 - 2^{20} \mathbf{u}_2.\]

Key Concepts

Linear Independence

Understanding linear independence is crucial in linear algebra. Two vectors are linearly independent if no scalar multiple of one vector can result in the other. In simpler terms, neither vector can be expressed as a "stretch" or "shrink" of the other.

To verify linear independence between two vectors, we often use the concept of the determinant. In the given exercise, you have vectors \(\mathbf{u}_1 = \begin{pmatrix} 1 \ 0 \end{pmatrix} \) and \( \mathbf{u}_2 = \begin{pmatrix} 1 \ 3 \end{pmatrix} \).

These can be vertically stacked to form a matrix:

• Calculate the determinant of the formed matrix.
• If the determinant is non-zero, the vectors are linearly independent.

For the matrix \( \begin{pmatrix} 1 & 1 \ 0 & 3 \end{pmatrix} \), the determinant is calculated as \( (1)(3) - (1)(0) = 3 \), which is clearly non-zero. Therefore, \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) are confirmed to be linearly independent vectors.

Linear Combination

A linear combination involves expressing one vector as the weighted sum of others. In the context of eigenvectors, you are often required to represent a given vector as a combination of particular eigenvectors.

The challenge in the exercise involves expressing \( \mathbf{x} = \begin{pmatrix} 1 \ -3 \end{pmatrix} \) as a linear combination of the eigenvectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \). This means finding scalars \( c_1 \) and \( c_2 \) such that:

• \( c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 = \mathbf{x} \)
• Match the components resulting in a system of equations.

The equations from the exercise are:
1. \( c_1 + c_2 = 1 \)
2. \( 3c_2 = -3 \)
By solving these, we find \( c_2 = -1 \) and substitute back to get \( c_1 = 2 \). Thus, \( \mathbf{x} \) is expressed as a linear combination: \( \mathbf{x} = 2\mathbf{u}_1 - \mathbf{u}_2 \).

This form demonstrates the flexibility and importance of linear combinations in mathematical expressions and calculations.

Matrix Determinant

The matrix determinant is a special number that can provide insights into various properties of the matrix, such as invertibility and linear dependence.

When you are given a 2x2 matrix like \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is found using the formula:

• \( \,ad - bc\)

In our exercise example, this method was used to determine the linear independence of the eigenvectors by setting up the matrix \( \begin{pmatrix} 1 & 1 \ 0 & 3 \end{pmatrix} \).

• If the determinant is zero, the matrix is singular, implying linear dependence.
• A non-zero determinant indicates the matrix is invertible, confirming linear independence.

The determinant offers a powerful way to discern the characteristics and behavior of matrices and their associated vectors in a straightforward manner.