Problem 53 - 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 \(A \mathbf{v}_{2}\). $$ A=\left[\begin{array}{ll} -4 & 2 \\ -3 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

Eigenvalues are \(\lambda_1 = -1\), \(\lambda_2 = -2\). Eigenvectors are \(\mathbf{v}_1 = \begin{pmatrix}2\\3\end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix}1\\1.5\end{pmatrix}\). Lines are \(y = \frac{3}{2}x\) and \(y = 1.5x\).

Step 1: Setting Up the Characteristic Equation

To find the eigenvalues, solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(A\) is the given matrix and \(I\) is the identity matrix of the same order. For the matrix \(A=\begin{pmatrix}-4 & 2\-3 & 1\end{pmatrix}\), the characteristic equation is \(\det\left(\begin{pmatrix}-4 & 2\-3 & 1\end{pmatrix} - \lambda \begin{pmatrix}1 & 0\0 & 1\end{pmatrix}\right) = 0\).

Step 2: Solve Determinant for Eigenvalues

Compute the determinant: \(\det\left(\begin{pmatrix}-4-\lambda & 2 \ -3 & 1-\lambda\end{pmatrix}\right) = (-4-\lambda)(1-\lambda) - (2)(-3) = \lambda^2 + 3\lambda + 2\). Set this equal to zero and solve the quadratic equation \(\lambda^2 + 3\lambda + 2 = 0\) to find \(\lambda_1\) and \(\lambda_2\).

Step 3: Solving the Quadratic Equation

Factor the quadratic equation: \((\lambda + 1)(\lambda + 2) = 0\). This gives the roots: \(\lambda_1 = -1\) and \(\lambda_2 = -2\), which are the eigenvalues of the matrix \(A\).

Step 4: Finding the Eigenvectors

For each eigenvalue \(\lambda\), find the corresponding eigenvector \(\mathbf{v}\) by solving \((A - \lambda I)\mathbf{v} = 0\). For \(\lambda_1 = -1\), solve \(\begin{pmatrix}-3 & 2 \ -3 & 2\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}\). A particular solution is \(\mathbf{v}_1 = \begin{pmatrix}2\3\end{pmatrix}\).

Step 5: Finding the Second Eigenvector

Repeat the process for \(\lambda_2 = -2\): solve \(\begin{pmatrix}-2 & 2 \ -3 & 3\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}0\0\end{pmatrix}\). A particular solution is \(\mathbf{v}_2 = \begin{pmatrix}1\1.5\end{pmatrix}\).

Step 6: Equations of Lines Through Origin

The line through the origin in the direction of \(\mathbf{v}_1 = \begin{pmatrix}2\3\end{pmatrix}\) is \(y = \frac{3}{2}x\). For \(\mathbf{v}_2 = \begin{pmatrix}1\1.5\end{pmatrix}\), the line is \(y = 1.5x\).

Step 7: Graphing the Lines and Vectors

Graph the lines \(y = \frac{3}{2}x\) and \(y = 1.5x\) along with their respective vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\). Also graph \(A\mathbf{v}_1\) and \(A\mathbf{v}_2\) which are obtained by applying the matrix \(A\) to each eigenvector.

Key Concepts

Eigenvectors Unveiled

In the realm of linear algebra, eigenvectors play a pivotal role. An eigenvector of a matrix is a special vector whose direction remains unchanged when a linear transformation is applied, via multiplication by the matrix. This unique property makes eigenvectors incredibly useful in diverse fields like data science, physics, and engineering.

• To find an eigenvector, one needs to solve the equation \( (A - \lambda I) \mathbf{v} = 0\), where \( A\) is a given matrix, \( \lambda\) represents an eigenvalue, and \( I\) is the identity matrix of the same dimension as \( A\).
• The vector \( \mathbf{v}\) is our eigenvector for the eigenvalue \( \lambda\).

Consider our matrix \( A\) from the exercise, and suppose we aim to find its eigenvectors. By first finding the eigenvalues, as I'll describe in the following sections, you can substitute these values back into the equation above to find that the directions of the vectors \( \begin{pmatrix} 2 \ 3 \end{pmatrix}\) and \( \begin{pmatrix} 1 \ 1.5 \end{pmatrix}\) remain unchanged under transformation by \( A\).
These vectors correspond to each eigenvalue and lie on lines through the origin.

Unlocking the Characteristic Equation

The characteristic equation is fundamental when determining the eigenvalues of a matrix. This equation arises from determining the conditions under which a matrix minus a scalar multiple of the identity matrix becomes non-invertible, i.e., has a determinant of zero.

• Start with the matrix \( A\) and subtract \( \lambda I\), which gives \( A - \lambda I\).
• Calculate the determinant of this result, setting it equal to zero: \( \det(A - \lambda I) = 0\).

The equation you derive is what we call the characteristic equation. For our specific exercise, this becomes \( \lambda^2 + 3\lambda + 2 = 0\). Solving this equation yields the eigenvalues, showing crucial steps in graphing and understanding the behavior of matrices.

Solving the Quadratic Equation

Quadratic equations frequently appear when computing eigenvalues, particularly for 2x2 matrices. A quadratic equation is a polynomial equation of degree two and typically looks like \( ax^2 + bx + c = 0\). To find the solutions or "roots," you can use various methods such as factoring, completing the square, or applying the quadratic formula.

• For the factoring method, observe the equation: If it can be written as \( (x+p)(x+q) = 0\), then roots are \( x = -p\) and \( x = -q\).
• In the step-by-step solution, the equation \( \lambda^2 + 3\lambda + 2 = 0\) factors to \( (\lambda + 1)(\lambda + 2) = 0\), revealing roots \( \lambda_1 = -1\) and \( \lambda_2 = -2\).

Each root here is a specific eigenvalue critical for further calculations.

Engaging in Matrix Graphing

Matrix graphing introduces a visual approach to comprehending eigenvectors and their corresponding lines through the origin. Through this method, we can visually verify and emphasize the direction of these vectors provided by eigenvalues.

• To graph, start by plotting the eigenvectors. Using our solutions, plot \( \mathbf{v}_1 = \begin{pmatrix} 2 \ 3 \end{pmatrix}\) and \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 1.5 \end{pmatrix}\).
• Draw straight lines along these vectors, using the slope of each vector. This step helps make the lines described in the original problem: \( y = \frac{3}{2}x\) for \( \mathbf{v}_1\) and \( y = 1.5x\) for \( \mathbf{v}_2\).
• Graph the lines together with vectors \( A\mathbf{v}_1\) and \( A\mathbf{v}_2\) to observe how matrix \( A\) uniquely affects them.

This visual representation not only solidifies understanding but also demonstrates the unchanging direction of eigenvectors after transformation, a cornerstone in linear algebra.