Problem 34 - Differential Equations and Linear Algebra

Question

Use the result of Problem 32 to determine the sum and the product of the eigenvalues of the given matrix \(A\). $$A=\left[\begin{array}{rrr} 2 & 0 & 5 \\ 0 & -1 & 1 \\ 3 & -4 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The eigenvalues of the matrix A are \(\lambda_1 = 1\), \(\lambda_2 = -1\), and \(\lambda_3 = 3\). The sum of the eigenvalues is \(3\), and the product of the eigenvalues is \(-3\).

Step 1: Write down the matrix A minus lambda times the identity matrix

Original matrix A is: $$A=\left[\begin{array}{rrr} 2 & 0 & 5 \\\ 0 & -1 & 1 \\\ 3 & -4 & 2 \end{array}\right]$$ The matrix \((A-\lambda I)\) can be written as: $$A-\lambda I =\left[\begin{array}{ccc} 2-\lambda & 0 & 5\\ 0 & -1-\lambda & 1\\ 3 & -4 & 2-\lambda \end{array}\right]$$

Step 2: Calculate the determinant of (A - lambda*I)

To find the eigenvalues, we need to determine the determinant of \((A-\lambda I)\) and set it equal to zero. $$\det(A-\lambda I) = \det\left(\left[\begin{array}{ccc} 2-\lambda & 0 & 5\\ 0 & -1-\lambda & 1\\ 3 & -4 & 2-\lambda \end{array}\right]\right)$$ Calculate the determinant: \(\det(A-\lambda I)= ((2-\lambda)[(-1-\lambda)(2-\lambda) - (-4)(1)]) - (0) + (5)[0 - (3)(-4)]\)

Step 3: Compute the characteristic equation

Now we simplify the determinant expression to obtain the characteristic equation: $$ ((2-\lambda)[(-1-\lambda)(2-\lambda) - (-4)(1)]) - (0) + (5)[0 - (3)(-4)] = (\lambda^3 - 3\lambda^2-8) = (\lambda-1)(\lambda + 1)(\lambda - 3)$$

Step 4: Solve for eigenvalues (lambda)

Set the equation to zero and find the roots (eigenvalues): $$\lambda^3 - 2\lambda^2 - \lambda - 6 = (\lambda-1)(\lambda + 1)(\lambda - 3) = 0$$ Solving for \(\lambda\), we get 3 eigenvalues: \(\lambda_1 = 1\) \(\lambda_2 = -1\) \(\lambda_3 = 3\)

Step 5: Calculate the sum and product of eigenvalues

Now that we have found the eigenvalues, we can compute the sum and the product. The sum of the eigenvalues can be calculated as the sum of the diagonal of matrix A (trace of matrix A): \(\lambda_1 + \lambda_2 + \lambda_3 = 2 + (-1) + 2 = 3\) The product of the eigenvalues can be obtained as the determinant of matrix A: \(\lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 1 \cdot (-1) \cdot 3 = -3\) So, the sum and product of eigenvalues are: Sum: 3 Product: -3

Key Concepts

Matrix Determinant

The matrix determinant is a crucial concept in linear algebra. It is a scalar value that can be calculated from a square matrix. Determinants are used to solve systems of linear equations, understand matrix properties, and, notably, uncover eigenvalues. In the context of eigenvalues, the determinant of a matrix subtracting a scalar (eigenvalue) times the identity matrix must equal zero. This is what provides the eigenvalues.To find the determinant of a 3x3 matrix like our example, we can use the cofactor expansion method. For the matrix \(A - \lambda I\), the determinant is expressed as:

• \(\det(A-\lambda I) = (a(ei-fh) - b(di-fg) + c(dh-eg))\)

This formula will give us a polynomial when set to zero. Solving this polynomial will reveal the eigenvalues.

Characteristic Equation

A characteristic equation is obtained from the determinant of \(A - \lambda I\). It is a polynomial equation derived by setting the determinant equal to zero. The roots of this polynomial are the eigenvalues of the matrix.The characteristic equation allows us to find eigenvalues without directly solving the matrix equations. In our example, the characteristic equation is:

• \(\lambda^3 - 3\lambda^2 - 8 = 0\)

By solving this polynomial equation, we find the eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = -1\), and \(\lambda_3 = 3\). These eigenvalues have significant implications in transformations and the stability of systems.

Trace of a Matrix

The trace of a matrix is the sum of its diagonal elements. It offers a simple way to find properties of the matrix, including the sum of its eigenvalues. For a matrix \(A\), the trace is expressed as:

• \( \text{Trace}(A) = a_{11} + a_{22} + \, ... \, + a_{nn} \)

The relationship between the trace and eigenvalues states that the trace equals the sum of the eigenvalues.For our matrix \(A\):

• \(\text{Trace}(A) = 2 + (-1) + 2 = 3\)

Therefore, the sum of the eigenvalues \(\lambda_1 + \lambda_2 + \lambda_3 = 3\), which confirms our calculation.

Determinant Calculation

Calculating the determinant involves breaking down the matrix into smaller components until we reach a manageable size. For our purpose, we used the cofactor method previously mentioned. The calculation is detailed and must follow step-by-step progressions to avoid errors.For our matrix \(A - \lambda I\), by expanding along the first row:

• \(\det(A - \lambda I) = ((2-\lambda)((-1-\lambda)(2-\lambda) - (-4)(1))) + 0 + (5)(0 - 3 \cdot -4)\)

Solving this gives the determinant polynomial which simplifies down to \(\lambda^3 - 3\lambda^2 - 8\). This determinant set to zero leads to finding the roots, known as eigenvalues. Determinants thereby reveal key information about the matrix's eigenproperties.