Problem 18 - Advanced Engineering Mathematics

Question

In Problems, 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rrr} 0 & 0 & 1 \\ 1 & 0 & -3 \\ 0 & 1 & 3 \end{array}\right) $$

Step-by-Step Solution

Answer

\(\mathbf{A}\) is not diagonalizable over the real numbers due to complex eigenvalues.

Step 1: Calculate eigenvalues of \(\mathbf{A}\)

To check if \(\mathbf{A}\) is diagonalizable, we start by finding its eigenvalues. We need to solve the characteristic equation \(|\mathbf{A} - \lambda \mathbf{I}| = 0\), where \(\lambda\) is an eigenvalue and \(\mathbf{I}\) is the identity matrix of the same size as \(\mathbf{A}\). This leads to the equation:\[\left|\begin{array}{ccc}-\lambda & 0 & 1 \1 & -\lambda & -3 \0 & 1 & 3-\lambda\end{array}\right| = 0\]

Step 2: Solve the characteristic polynomial

Expanding the determinant, we have:\[-\lambda \left((-\lambda)(3-\lambda) - (-3)(1)\right) = 0\]Simplifying, we get:\[-\lambda(\lambda^2 - 3\lambda + 3) = 0\]Thus, the characteristic polynomial is \(-\lambda(\lambda^2 - 3\lambda + 3) = 0\), giving eigenvalues \(\lambda = 0\) and solving \(\lambda^2 - 3\lambda + 3 = 0\).

Step 3: Solve \(\lambda^2 - 3\lambda + 3 = 0\) for remaining eigenvalues

The quadratic \(\lambda^2 - 3\lambda + 3 = 0\) can be solved using the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1, b = -3, c = 3\):\[\lambda = \frac{3 \pm \sqrt{9 - 12}}{2} = \frac{3 \pm i\sqrt{3}}{2}\]These roots are complex, \(\lambda = \frac{3}{2} + \frac{i\sqrt{3}}{2}\) and \(\lambda = \frac{3}{2} - \frac{i\sqrt{3}}{2}\), indicating that \(\mathbf{A}\) is not diagonalizable over the real numbers.

Key Concepts

Eigenvalues

Eigenvalues are fundamental to understanding how matrices behave. They help identify a matrix's properties and are essential in determining whether a matrix can be diagonalized. In the context of matrices, eigenvalues are special scalars that make the equation \((\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0}\) true, where \(\mathbf{v}\) is a non-zero vector, \(\lambda\) is the eigenvalue, and \(\mathbf{I}\) is the identity matrix of the same size as \(\mathbf{A}\).

The process to find eigenvalues involves:

• Setting up the equation \(|\mathbf{A} - \lambda \mathbf{I}| = 0\).
• The term \(|\cdot|\) denotes the determinant of the matrix, which must equal zero.
• This leads to finding values of \(\lambda\) that satisfy this equality.

The solution of this equation provides the matrix's eigenvalues. These eigenvalues indicate the matrix's invariant directions, meaning they stretch or contract by the same factor \(\lambda\) when transformed by \(\mathbf{A}\). You need to calculate these to assess if diagonalization is feasible.

In our exercise, one eigenvalue is 0, while the others are complex numbers, which affect the ability to diagonalize in real number spaces.

Characteristic Polynomial

The characteristic polynomial plays a crucial role in finding eigenvalues. By definition, it is a polynomial derived from the determinant of the matrix \(\mathbf{A} - \lambda \mathbf{I}\). When we equate this polynomial to zero, we obtain the characteristic equation that the eigenvalues solve.

To compute the characteristic polynomial:

• Subtract \(\lambda\) (a scalar) from the diagonal elements of \(\mathbf{A}\).
• Calculate the determinant of this new matrix.
• Write the expression as a polynomial in terms of \(\lambda\).

In this process, the determinant's value is expanded into a polynomial equation. Solving this equation results in the eigenvalues, each representing a solution to the matrix identity problem.

For our matrix, the characteristic polynomial was \(-\lambda(\lambda^2 - 3\lambda + 3)\). The roots of this polynomial yield one real eigenvalue at \(\lambda = 0\), and two complex roots, indicating specific transformation characteristics of \(\mathbf{A}\).

Complex Numbers

Complex numbers might seem daunting, but they are indispensable in linear algebra, especially when dealing with matrices not diagonalizable over the reals. A complex number consists of a real and an imaginary part, usually expressed as \(a + bi\), where \(i\) is the imaginary unit \(\sqrt{-1}\).

The unique properties of complex numbers include:

• Allowing for solutions to polynomial equations that might not have real roots.
• Providing a broader platform for operations in mathematics and physics.
• Facilitating transformations represented by non-real eigenvalues.

In our scenario, the matrix had complex eigenvalues \(\lambda = \frac{3}{2} \pm \frac{i\sqrt{3}}{2}\). These values indicate that the transformations of \(\mathbf{A}\) involve rotations and scaling, rather than straightforward stretching along real coordinate axes.

Recognizing and interpreting complex eigenvalues help us understand non-diagonalizability over the real numbers. This concept is pivotal in fields like quantum mechanics and other areas that rely on complex transformations.