Problem 55 - Calculus for Biology and Medicine

Question

In Problems , find the eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) and corresponding eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) for each matrix A. Determine the equations of the lines through the origin in the direction of the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\), and graph the lines together with the eigenvectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) and the vectors \(A \mathbf{v}_{1}\) and \(A \mathbf{v}_{2}\). $$ A=\left|\begin{array}{ll} 2 & 1 \\ 2 & 3 \end{array}\right| $$

Step-by-Step Solution

Answer

Eigenvalues: \(\lambda_1 = 4, \lambda_2 = 1\). Eigenvectors: \(\mathbf{v}_1 = \begin{pmatrix} 0.5 \\ 1 \end{pmatrix}, \mathbf{v}_2 = \begin{pmatrix} -1 \\ 1 \end{pmatrix}\). Lines are \( y=2x \) and \( y=-x \).

Step 1: Calculate the Characteristic Polynomial

To find the eigenvalues, we start by calculating the characteristic polynomial of the matrix \( A \). The characteristic polynomial is given by \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. For the matrix \( A = \begin{pmatrix} 2 & 1 \ 2 & 3 \end{pmatrix} \), we have: \[ \det \begin{pmatrix} 2 - \lambda & 1 \ 2 & 3 - \lambda \end{pmatrix} = (2-\lambda)(3-\lambda) - (2)(1) = \lambda^2 - 5\lambda + 4 \]

Step 2: Find the Eigenvalues

Solve the characteristic polynomial \( \lambda^2 - 5\lambda + 4 = 0 \) for \( \lambda \). \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} = \frac{5 \pm 3}{2} \] The solutions are \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \).

Step 3: Find Eigenvectors for \( \lambda_1 \)

Solve \((A - \lambda_1I)\mathbf{v} = 0\) for \(\mathbf{v}\) when \(\lambda_1 = 4\):\[ \begin{pmatrix} 2-4 & 1 \ 2 & 3-4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -2 & 1 \ 2 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] Solving these equations, we find any vector \(\mathbf{v}_1 = \begin{pmatrix} \frac{1}{2}y \ y \end{pmatrix}\) for \(y eq 0\). Choose \(y = 1\), then \(\mathbf{v}_1 = \begin{pmatrix} 0.5 \ 1 \end{pmatrix} \).

Step 4: Find Eigenvectors for \( \lambda_2 \)

Solve \((A - \lambda_2I)\mathbf{v} = 0\) for \(\mathbf{v}\) when \(\lambda_2 = 1\): \[ \begin{pmatrix} 1 & 1 \ 2 & 2 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \] The solution to these is any vector \( \mathbf{v}_2 = \begin{pmatrix} -y \ y \end{pmatrix} \), so choosing \(y = 1\), we have \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \).

Step 5: Determine Equations of Lines

The lines through the origin in the direction of the eigenvectors are given by parametric equations. For \( \mathbf{v}_1 = \begin{pmatrix} 0.5 \ 1 \end{pmatrix} \), the equation is: \[ y = 2x \] For \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \), the equation is: \[ y = -x \]

Step 6: Plot Eigenvectors and Axes

Plot vectors \(\mathbf{v}_1\), \(\mathbf{v}_2\), and the transformed vectors \(A\mathbf{v}_1\) and \(A\mathbf{v}_2\). To find \(A\mathbf{v}_1 = \begin{pmatrix} 2.5 \ 3.5 \end{pmatrix}\) and \(A\mathbf{v}_2 = \begin{pmatrix} 1 \ 1 \end{pmatrix}\). Graph these vectors along with the lines \( y = 2x \) and \( y = -x \) on a Cartesian plane with corresponding axes.

Key Concepts

Eigenvalues

In linear algebra, eigenvalues are crucial in understanding the transformations described by a matrix. They give insight into the behavior and characteristics of the matrix in question, particularly when it acts on vectors. Supposing you have a square matrix \( A \), an eigenvalue \( \lambda \) is a scalar that satisfies the equation \( A \mathbf{v} = \lambda \mathbf{v} \) for some non-zero vector \( \mathbf{v} \). To find the eigenvalues, we need to solve the characteristic equation, which is derived by setting the determinant of \( A - \lambda I \) to zero, where \( I \) is the identity matrix.

For example, for our matrix \( A = \begin{pmatrix} 2 & 1 \ 2 & 3 \end{pmatrix} \), the characteristic polynomial is \( \det(A - \lambda I) = \lambda^2 - 5\lambda + 4 \). Solving this polynomial equation yields the eigenvalues \( \lambda_1 = 4 \) and \( \lambda_2 = 1 \).

Eigensolutions can tell us a lot about stability, vibrations, and even image data processing. Identifying eigenvalues is often the first step in performing eigen decomposition, which can simplify many types of matrix computations.

Eigenvectors

Eigenvectors are the vectors that correspond to eigenvalues. If \( \lambda \) is an eigenvalue of matrix \( A \), then the eigenvector \( \mathbf{v} \) associated with \( \lambda \) satisfies the equation \( (A - \lambda I)\mathbf{v} = \mathbf{0} \), where \( \mathbf{0} \) is the zero vector.

Once you've determined an eigenvalue, you can insert it back into this equation to solve for its corresponding eigenvector. For \( \lambda_1 = 4 \), the equation \( (A - 4I) \mathbf{v} = \mathbf{0} \) yields the solution \( \mathbf{v}_1 = \begin{pmatrix} 0.5 \ 1 \end{pmatrix} \), assuming we conveniently choose \( y = 1 \). Similarly, for \( \lambda_2 = 1 \), solving \( (A - 1I) \mathbf{v} = \mathbf{0} \) results in \( \mathbf{v}_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix} \).

Eigenvectors are significant because they maintain their direction under the transformation described by \( A \), only being stretched or shrunk by a factor of the respective eigenvalue. They form a basis for understanding how the matrix operates on vector spaces and are used in numerous applications including stability analysis in engineering and principal component analysis in statistics.

Characteristic Polynomial

The characteristic polynomial is a critical component in finding eigenvalues. It is derived from the determinant of the matrix \( A - \lambda I \), where \( \lambda \) is a scalar and \( I \) is the identity matrix. The solutions to the resulting characteristic polynomial give the eigenvalues of the matrix.

For our matrix \( A = \begin{pmatrix} 2 & 1 \ 2 & 3 \end{pmatrix} \), the characteristic polynomial comes about from calculating \( \det \begin{pmatrix} 2-\lambda & 1 \ 2 & 3-\lambda \end{pmatrix} = (2-\lambda)(3-\lambda) - 2 \). This results in \( \lambda^2 - 5\lambda + 4 = 0 \).

The characteristic polynomial not only helps determine eigenvalues but also reflects key properties of the matrix, such as its invertibility and its determinant. It’s a polynomial that mirrors the connections between linear transformations and roots, and the real or complex nature of its roots indicates whether the resulting eigenvalues will be real or complex.