Problem 2 - 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 \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{rr} -4 & -5 \\ 8 & 10 \end{array}\right) $$

Step-by-Step Solution

Answer

Yes, \(\mathbf{A}\) is diagonalizable with \(\mathbf{P} = \begin{bmatrix}-5 & 1 \\ 4 & -2\end{bmatrix}\) and \(\mathbf{D} = \begin{bmatrix}0 & 0 \\ 0 & 6\end{bmatrix}\).

Step 1: Find Eigenvalues

To determine if the matrix \(\mathbf{A}\) is diagonalizable, we first need to find its eigenvalues. The eigenvalues \(\lambda\) are found by solving the characteristic equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). For the given matrix, this translates to the following determinant equation:\[\det\left(\begin{bmatrix}-4 - \lambda & -5 \8 & 10 - \lambda\end{bmatrix}\right) = 0\]Simplifying the determinant, we get:\[(-4 - \lambda)(10 - \lambda) - (-5)(8) = 0\]\[\lambda^2 - 6\lambda = 0\]\[\lambda(\lambda - 6) = 0\]Thus, the eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = 6\).

Step 2: Check Diagonalizability

A matrix is diagonalizable if the algebraic multiplicity of each eigenvalue equals its geometric multiplicity. We have two distinct eigenvalues, which implies that the matrix is diagonalizable because distinct eigenvalues imply linearly independent eigenvectors corresponding to each eigenvalue.

Step 3: Find Eigenvectors

Next, we need to find the eigenvectors corresponding to each eigenvalue. This requires solving the equation \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0}\) for each eigenvalue:For \(\lambda_1 = 0\):\[\begin{bmatrix}-4 & -5 \8 & 10\end{bmatrix} \begin{bmatrix}x \y\end{bmatrix} = \begin{bmatrix}0 \0\end{bmatrix}\]Simplifying, we have:\[-4x - 5y = 0 \8x + 10y = 0\]From the first equation, solving for \(x\) in terms of \(y\), we find \(x = \frac{-5}{4}y\). Thus, an eigenvector for \(\lambda_1 = 0\) is \(\begin{bmatrix}-5 \4\end{bmatrix}\).For \(\lambda_2 = 6\):\[\begin{bmatrix}-10 & -5 \8 & 4\end{bmatrix} \begin{bmatrix}x \y\end{bmatrix} = \begin{bmatrix}0 \0\end{bmatrix}\]Simplifying, we have:\[-10x - 5y = 0 \8x + 4y = 0\]From the first equation, solving for \(x\) in terms of \(y\), we find \(x = -\frac{1}{2}y\). Thus, an eigenvector for \(\lambda_2 = 6\) is \(\begin{bmatrix}1 \-2\end{bmatrix}\).

Step 4: Form the Matrix \(\mathbf{P}\)

Matrix \(\mathbf{P}\) is formed by placing the eigenvectors as columns. Therefore, we have:\[\mathbf{P} = \begin{bmatrix}-5 & 1 \4 & -2\end{bmatrix}\]

Step 5: Find the Diagonal Matrix \(\mathbf{D}\)

The diagonal matrix \(\mathbf{D}\) is formed using the eigenvalues on the diagonal, in the same order as their corresponding eigenvectors in \(\mathbf{P}\):\[\mathbf{D} = \begin{bmatrix}0 & 0 \0 & 6\end{bmatrix}\]

Step 6: Verify \(\mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P}\)

To verify the diagonalization, we can calculate \(\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). First, find \(\mathbf{P}^{-1}\) by calculating the inverse of \(\mathbf{P}\):\(\mathbf{P}^{-1} = \frac{1}{(-5)(-2) - (1)(4)} \begin{bmatrix}-2 & -1 \-4 & -5\end{bmatrix} = \frac{1}{6} \begin{bmatrix}-2 & -1 \-4 & -5\end{bmatrix}\).Now, compute \(\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\) and verify it equals \(\mathbf{D}\). This step involves matrix multiplication and should yield \(\begin{bmatrix}0 & 0 \0 & 6\end{bmatrix}\).

Key Concepts

Eigenvalues

In linear algebra, an eigenvalue is a special scalar associated with a linear transformation represented by a square matrix. Eigenvalues are crucial for understanding the properties of a matrix, especially in the context of diagonalization. To find the eigenvalues of a matrix, we use the characteristic equation. This is formed by taking the determinant of the matrix minus an identity matrix scaled by \(\lambda\), where \(\lambda\) is the eigenvalue.

For example, if we have a matrix \(\mathbf{A}\), the characteristic equation is \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). Solving this equation for \(\lambda\) gives us the eigenvalues. In our example, the matrix \(\mathbf{A}\) had eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 6\).

Eigenvalues can tell us a lot about a system, such as stability and energy levels in physical systems, which are only a few of their applications.

Eigenvectors

An eigenvector is a non-zero vector that changes only by a scalar factor when that linear transformation is applied to it. Once we have the eigenvalues, we find the corresponding eigenvectors to fully understand the transformation. This involves solving the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) for each eigenvalue, where \(\mathbf{v}\) is the eigenvector.

The process involves plugging in each eigenvalue into the equation \( (\mathbf{A} - \lambda \mathbf{I}) \) and finding the null space, or the set of vectors that result in the zero vector when multiplied with this matrix.

• For \(\lambda_1 = 0\), the eigenvector was found to be \[\begin{bmatrix} -5 \ 4 \end{bmatrix} \].
• For \(\lambda_2 = 6\), it was \[\begin{bmatrix} 1 \ -2 \end{bmatrix} \].

Each eigenvalue usually leads to a unique eigenvector, which is important for understanding how the matrix can be represented in a simpler form.

Characteristic Equation

The characteristic equation of a matrix is central to the process of finding eigenvalues. It is derived by setting the determinant of the matrix minus \(\lambda\mathbf{I}\) to zero, where \(\mathbf{I}\) is the identity matrix and \(\lambda\) are the eigenvalues.

Visually, for a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the characteristic equation is \(\det\begin{bmatrix} a-\lambda & b \ c & d-\lambda \end{bmatrix} = 0\). Solving this determinant involves simple algebraic manipulation and is typically a polynomial in \(\lambda\), where the roots of this polynomial are the eigenvalues.

For our specific example, we arrived at the equation \(\lambda^2 - 6\lambda = 0\), leading to the eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 6\). This equation forms the foundation for determining whether a matrix can be simplified through diagonalization.

Diagonal Matrix

A diagonal matrix is a matrix that has non-zero elements only on its main diagonal, with all other elements being zero. In the context of matrix diagonalization, we aim to transform a given matrix into a diagonal matrix using its eigenvalues and corresponding eigenvectors.

The primary goal is to represent a matrix \(\mathbf{A}\) as \(\mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P}\), where \(\mathbf{D}\) is the diagonal matrix, and \(\mathbf{P}\) is the matrix formed by placing the eigenvectors as columns.

A diagonal matrix is straightforward to work with because matrix operations become simple. For instance, exponentiation and finding the determinant can be done directly using the diagonal elements.

• In our solution example, the matrix \(\mathbf{D}\) had the eigenvalues \(0\) and \(6\) placed on its main diagonal, making it \[\begin{bmatrix} 0 & 0 \ 0 & 6 \end{bmatrix} \].

Diagonal matrices are powerful in simplifying complex matrix calculations.