Problem 10 - Advanced Engineering Mathematics

Question

In Problems 1-20, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A P}\). $$ \left(\begin{array}{rr} 1 & 2 \\ -\frac{1}{2} & 1 \end{array}\right) $$

Step-by-Step Solution

Answer

Matrix \( \mathbf{A} \) is not diagonalizable over the real numbers.

Step 1: Find Eigenvalues

First, we need to find the eigenvalues of \( \mathbf{A} \). The characteristic equation is given by \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For \( \mathbf{A} = \begin{pmatrix} 1 & 2 \ -\frac{1}{2} & 1 \end{pmatrix} \), the characteristic equation is \[ \left| \begin{array}{cc} 1-\lambda & 2 \ -\frac{1}{2} & 1-\lambda \end{array} \right| = 0 \]. Solving the determinant, we get \((1-\lambda)^2 + 1 = 0\). This simplifies to \((1-\lambda)^2 = -1\), leading to complex eigenvalues, \( \lambda = 1 \pm i \). Since the eigenvalues are complex, the matrix \( \mathbf{A} \) is not diagonalizable over the real numbers.

Step 2: Conclusion

With complex eigenvalues \( \lambda = 1 \pm i \), the matrix \( \mathbf{A} \) is not diagonalizable over the real numbers. Diagonalizability requires real eigenvalues and enough eigenvectors to form a basis.

Key Concepts

Eigenvalues

When dealing with matrices, eigenvalues are an important concept. They are special scalars associated with a square matrix. To find the eigenvalues of a matrix like \[ \mathbf{A} = \begin{pmatrix} 1 & 2 \ -\frac{1}{2} & 1 \end{pmatrix}, \]you need to solve the characteristic equation. This involves calculating the determinant of \( \mathbf{A} - \lambda \mathbf{I} \), where \( \lambda \) represents the eigenvalues, and \( \mathbf{I} \) is the identity matrix of the same size as \( \mathbf{A} \). The result is a polynomial equation in terms of \( \lambda \), and the eigenvalues are the solutions to this equation.

To summarize:

• Identify the matrix \( \mathbf{A} \).
• Set up \( \mathbf{A} - \lambda \mathbf{I} \).
• Find the determinant.
• Solve for \( \lambda \).

In our example, solving for \( \lambda \) gives complex eigenvalues \( 1 \pm i \). Understanding eigenvalues is critical when exploring matrix diagonalization.

Eigenvectors

Eigenvectors work hand-in-hand with eigenvalues. They are vectors that, when the matrix \( \mathbf{A} \) is multiplied by, only scale by the associated eigenvalue. This essentially means \[ \mathbf{A} \mathbf{v} = \lambda \mathbf{v}, \]where \( \mathbf{v} \) is the eigenvector and \( \lambda \) is the eigenvalue.
Once you determine the eigenvalues, you can find corresponding eigenvectors by substituting each eigenvalue back into the equation \[(\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0}.\]
Here’s the step-by-step:

• Substitute an eigenvalue into \( \mathbf{A} - \lambda \mathbf{I} \).
• Solve the resulting homogeneous equation to find \( \mathbf{v} \).

Eigenvectors are essential for tasks like matrix diagonalization, as they form the columns of the matrix \( \mathbf{P} \), which diagonalizes \( \mathbf{A} \) under the right conditions.

Characteristic Equation

The characteristic equation is crucial when analyzing matrices to find eigenvalues. It arises from setting the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) equal to zero.In our example, the determinant \[\left|\begin{array}{cc} 1-\lambda & 2 \ -\frac{1}{2} & 1-\lambda \end{array}\right|=0\]gives us a polynomial equation in \( \lambda \).

This equation is\[(1-\lambda)^2 + 1 = 0.\]
The solutions to this polynomial, which are the roots of the equation, are the eigenvalues. Solving this gives the eigenvalues \( \lambda = 1 \pm i \). Recognizing the characteristic equation allows us to see how the behavior of the matrix \( \mathbf{A} \) changes with scaling factors.

Complex Numbers

Complex numbers come into play, especially when dealing with non-real solutions in mathematics. They have a real part and an imaginary part denoted typically as \( a + bi \), where \( i \) is the square root of \( -1 \). In our matrix example:

• The matrix has complex eigenvalues \( 1 \pm i \).
• These complex numbers point to the fact that the matrix \( \mathbf{A} \) cannot be diagonalized using real numbers.

Complex eigenvalues complicate diagonalization, as matrices need real eigenvalues and corresponding real eigenvectors to fully diagonalize. Openness to dealing with complex numbers broadens the scope of matrix operations and their applications in areas like quantum mechanics and signal processing.