Problem 35 - Differential Equations and Linear Algebra

Question

Let \(A=\left[\begin{array}{rr}1 & 2 \\ 2 & -2\end{array}\right].\) (a) Determine all eigenvalues of \(A\) (b) Reduce \(A\) to row-echelon form, and determine the eigenvalues of the resulting matrix. Are these the same as the eigenvalues of \(A ?\)

Step-by-Step Solution

Answer

The eigenvalues of the original matrix \(A\) are \(\lambda_1 = 3\) and \(\lambda_2 = -2\). However, the row-echelon form matrix \(B\) does not have any real eigenvalues. So, the eigenvalues of \(A\) and its row-echelon form matrix \(B\) are not the same.

Step 1: Find the eigenvalues of the given matrix A

To find the eigenvalues of matrix \(A\), we need to solve the characteristic polynomial equation, given by \(\det(A - \lambda I)=0\), where \(\lambda\) is the eigenvalue and \(I\) is the identity matrix. The given matrix \(A\) is: \[A = \left[\begin{array}{rr}1 & 2 \\\ 2 & -2\end{array}\right]\] Now, let's find \((A - \lambda I)\): \[A - \lambda I=\left[\begin{array}{rr}(1-\lambda) & 2 \\\ 2 & (-2-\lambda)\end{array}\right]\] Now, we will find the determinant of \((A - \lambda I)\), denoted by \(\det(A - \lambda I)\) : \[\det(A - \lambda I)=(1-\lambda)(-2-\lambda)-2\cdot2\] Expanding and simplifying the determinant, \[\det(A - \lambda I) =-2(1-\lambda)-\lambda(1-\lambda) - 4\] \[\det(A - \lambda I) = -2+\lambda -2\lambda +\lambda^2 - 4\] \[\det(A - \lambda I) = \lambda^2-\lambda-6\] Now, let's solve the equation \(\det(A - \lambda I) = 0\) for \(\lambda\): \[\lambda^2-\lambda-6=0\] Factoring the quadratic equation, we have: \[(\lambda - 3)(\lambda + 2) = 0\] So, the eigenvalues of \(A\) are \(\lambda_1 = 3\) and \(\lambda_2 = -2\).

Step 2: Reduce the given matrix to row-echelon form

To reduce the given matrix \(A\) into row-echelon form, we will use Gaussian elimination method. Start with the original matrix \(A\): \[A = \left[\begin{array}{rr}1 & 2 \\\ 2 & -2\end{array}\right]\] First, we'll subtract 2 times first row from second row: \\ \( Row_2 \leftarrow Row_2 - 2\cdot Row_1 \) The resulting matrix is: \[\left[\begin{array}{cc}1 & 2 \\\ 0 & -6\end{array}\right]\] The given matrix is now in its row-echelon form.

Step 3: Find the eigenvalues of the row-echelon form matrix

Now we will determine the eigenvalues of the row-echelon form matrix to compare with those of the original matrix \(A\). Let the row-echelon form matrix be \(B\): \[B= \left[\begin{array}{rr}1 & 2 \\\ 0 & -6\end{array}\right]\] Calculating the characteristic polynomial as we did in step 1, and solving for \(\lambda\), we have: \[\det(B - \lambda I)=\lambda^2+6\lambda\] This characteristic polynomial has no real roots (since the discriminant is negative), hence, the row-echelon form matrix \(B\) does not have any real eigenvalues.

Step 4: Compare the eigenvalues

Now let's compare the eigenvalues of \(A\) and \(B\). The eigenvalues of original matrix \(A\) are \(\lambda_1 = 3\) and \(\lambda_2 = -2\). The row-echelon form matrix \(B\) does not have any real eigenvalues. Thus, the eigenvalues of the original matrix \(A\) and its row-echelon form matrix \(B\) are not the same.

Key Concepts

Characteristic Polynomial

Understanding the characteristic polynomial is crucial when dealing with eigenvalues in linear algebra. This polynomial is obtained by computing the determinant of the matrix subtracted by the product of the eigenvalue and the identity matrix, denoted as \(A - \lambda I\). In simpler terms, the characteristic polynomial yields a scalar equation whose roots are the eigenvalues of matrix \(A\).

When solving the characteristic polynomial, one seeks values of \(\lambda\) that satisfy the equation \(\det(A - \lambda I) = 0\). Through this process, characteristic values or eigenvalues are obtained which reveal significant properties about the transformations that the matrix can represent. These values can tell us if the matrix can be diagonalized, indicate the matrix's invertibility, and also offer insights into geometric transformations associated with the matrix.

Determinants

The determinant of a matrix is a special number that can provide a lot of information about the matrix itself. For a square matrix, the determinant helps us understand whether the matrix is invertible or singular. If the determinant of a matrix is zero, then the matrix does not have an inverse. Furthermore, the determinant is used when computing characteristic polynomials, as seen in the previous section.

To compute the determinant of a 2x2 matrix like \(A\), one can use the formula \(ad - bc\), where \(a\), \(b\), \(c\), and \(d\) are the elements of the matrix \(\left[\begin{array}{rr}a & b \ c & d\end{array}\right]\). This straightforward method directly applies to finding the characteristic polynomial. It's important to note that the determinant concept scales to larger matrices, though the computation becomes more complex.

Row-echelon Form

Row-echelon form is a version of a matrix that simplifies solving systems of linear equations. This form has two main features: all nonzero rows are above any rows of all zeroes, and the leading coefficient (also known as the pivot) of a nonzero row is always to the right of the leading coefficient of the row above it. One achieves row-echelon form through a process called Gaussian elimination, which involves adding or subtracting multiples of some rows from others to create a staircase pattern of zeros below the pivots.

By transforming the original matrix into its row-echelon form, you can easily see the solutions to the corresponding system of equations or realize that no solution exists. However, as demonstrated in the exercise, transforming a matrix to row-echelon form can change its eigenvalues — a critical insight for students to note when analyzing the effects of such operations on different matrix properties.