91Ó°ÊÓ

Eigenvalues are essential in determining whether a matrix is diagonalizable. An eigenvalue is a special scalar, denoted by \( \lambda \), associated with a square matrix. To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This process involves:

• Subtracting \( \lambda \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix, from the matrix \( \mathbf{A} \).
• Calculating the determinant of the resulting matrix.
• Solving the resulting polynomial equation for \( \lambda \).

In our example, solving the polynomial \( \lambda^2 + 3\lambda - 10 = 0 \) results in finding two eigenvalues: \( \lambda_1 = 2 \) and \( \lambda_2 = -5 \). These values are crucial for further calculations of eigenvectors and confirming diagonalizability.

Once we have the eigenvalues of a matrix \( \mathbf{A} \), the next step is to find the corresponding eigenvectors. An eigenvector is a non-zero vector \( \mathbf{v} \) that, when multiplied by the matrix \( \mathbf{A} \), results in a scalar multiple of itself, specifically the eigenvalue.
This can be expressed as the equation \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = \mathbf{0} \). To find the eigenvector, for each eigenvalue, we:

• Substitute the eigenvalue \( \lambda \) into the matrix equation \( (\mathbf{A} - \lambda \mathbf{I}) \).
• Set up the resulting homogeneous system of linear equations.
• Solve for the non-trivial solution \( \mathbf{v} \).

In our problem, for \( \lambda_1 = 2 \), the corresponding eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 13 \ 11 \end{pmatrix} \). For \( \lambda_2 = -5 \), the eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 13 \ 4 \end{pmatrix} \). These vectors are used to form the matrix \( \mathbf{P} \) in the diagonalization process.

The determinant is a value that can be calculated from a square matrix and is denoted as \( \det \mathbf{A} \). It provides important characteristics of the matrix, such as invertibility and the solutions to the characteristic equation. If the determinant of a matrix is zero, the matrix is not invertible.
In the process of finding eigenvalues, the determinant is used specifically in the characteristic equation, where \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \) must be solved. This step is crucial to ensuring the matrix is diagonalizable through its eigenvalues.
In our example, we calculated \[ (-9-\lambda)(6-\lambda) - (-2)(13) = \lambda^2 + 3\lambda - 10 \]. The determinant of this characteristic expression is equated to zero for solving \( \lambda \) values, further leading to the eigenvalues of the matrix.

The characteristic equation is a critical tool in linear algebra for uncovering the properties of a matrix. It is derived from the formula \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where you seek values of \( \lambda \) (eigenvalues) that satisfy the determinant equation.

• Create \( \mathbf{A} - \lambda \mathbf{I} \) by subtracting a multiple of the identity matrix from \( \mathbf{A} \).
• Compute the determinant of the resulting matrix.
• Set the determinant equal to zero and solve the polynomial for \( \lambda \).

For example, in our matrix \( \mathbf{A} = \begin{pmatrix} -9 & 13 \ -2 & 6 \end{pmatrix} \), subtracting \( \lambda \mathbf{I} \) forms the basis of our characteristic equation. Solving \( \lambda^2 + 3\lambda - 10 = 0 \) gives us the eigenvalues that are necessary to diagonalize the matrix, showing us the paths taken to find the diagonal matrix \( \mathbf{D} \).