91Ó°ÊÓ

In linear algebra, an eigenvalue is a crucial concept associated with a square matrix. To put it simply, if you have a matrix \( \mathbf{A} \), an eigenvalue \( \lambda \) is a scalar such that there exists a non-zero vector \( \mathbf{v} \) (called an eigenvector) satisfying the equation \( \mathbf{A}\mathbf{v} = \lambda\mathbf{v} \). This means multiplying \( \mathbf{A} \) and \( \mathbf{v} \) results in a vector that is a scaled version of \( \mathbf{v} \).
This scaling factor is what we call the eigenvalue. A matrix can have multiple eigenvalues. In the exercise, the given eigenvalues are \( \lambda_1 = -7 \) and \( \lambda_2 = 4 \). These distinct eigenvalues suggest that the matrix can be diagonalized. Distinct here means that the eigenvalues are different from one another, which is crucial because a matrix with distinct eigenvalues is always diagonalizable.

Once you understand eigenvalues, the next step is to comprehend eigenvectors. An eigenvector for a matrix \( \mathbf{A} \) corresponding to an eigenvalue \( \lambda \) is a non-zero vector \( \mathbf{v} \) that satisfies the relationship \( \mathbf{A}\mathbf{v} = \lambda\mathbf{v} \). In layman's terms, when \( \mathbf{A} \) acts on \( \mathbf{v} \), it doesn't alter the direction of \( \mathbf{v} \), only its magnitude, multiplying it by \( \lambda \).
Eigenvectors are important because they help in simplifying matrix operations, especially in diagonalization. For diagonalization, the matrix \( \mathbf{P} \) consists of these eigenvectors arranged as columns. With distinct eigenvalues, distinct eigenvectors are guaranteed, ensuring a proper and valid \( \mathbf{P} \). The matrix \( \mathbf{P} \) in our problem, \( \begin{bmatrix} 13 & 1 \ 2 & 1 \end{bmatrix} \), is constructed from these eigenvectors.

A crucial aspect of diagonalization is the invertibility of the matrix \( \mathbf{P} \). An invertible matrix, also known as a non-singular matrix, is one where you can calculate its inverse (\( \mathbf{P}^{-1} \)). This means if you multiply the matrix by its inverse, you get the identity matrix. A matrix is invertible if and only if its determinant is non-zero.
In the exercise, the given matrix \( \mathbf{P} = \begin{bmatrix} 13 & 1 \ 2 & 1 \end{bmatrix} \) must be invertible for the diagonalization process to work. We verify this by calculating the determinant of \( \mathbf{P} \), which is \( 13 \times 1 - 2 \times 1 = 11 \). Since the determinant is 11, which is not zero, \( \mathbf{P} \) is indeed invertible. This confirms that diagonalization is possible.

The determinant is a special number that can be calculated from a square matrix. It provides important information about the matrix, such as whether it's invertible, and plays a significant role in solving linear systems, among other applications.
To compute the determinant of a 2x2 matrix \( \mathbf{P} = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the formula is \( ad - bc \). In our exercise, for the matrix \( \mathbf{P} = \begin{bmatrix} 13 & 1 \ 2 & 1 \end{bmatrix} \), the determinant is calculated as \( 13 \times 1 - 1 \times 2 = 11 \).
Since the determinant is 11 (not zero), this indicates that \( \mathbf{P} \) is invertible, further supporting its role in diagonalization. Remember, a determinant can tell us a lot about properties of matrices, including solving for eigenvalues.