91Ó°ÊÓ

To find the eigenvalues of a matrix, we start by setting up the characteristic equation. This equation is derived from the matrix minus the eigenvalue times the identity matrix. In mathematical terms, this is represented as \(A - \lambda I\), where \A\ is our given matrix, \lambda\ is the eigenvalue, and \I\ is the identity matrix.

The characteristic equation is found by computing the determinant of \(A - \lambda I\) and setting it equal to zero. This yields an equation in \lambda\, which we can solve to find the eigenvalues. For example, if we have a 3x3 matrix \(\left(\begin{array}{rrr}2 & 0 & 2 \ 0 & 2 & 0 \ 2 & 0 & -1\end{array}\right)\), we set up the equation:

\[\text{det}(A - \lambda I) = 0\]

and proceed with finding the determinant.

The determinant is a scalar value that can be computed from a square matrix and provides important information about the matrix, including solutions to linear systems, and eigenvalues. To find the determinant of a matrix \(A - \lambda I\), for our example matrix, we get:

\A - \lambda I = \left(\begin{array}{rrr}2-\lambda & 0 & 2 \ 0 & 2-\lambda & 0 \ 2 & 0 & -1-\lambda\end{array}\right)\

The determinant is calculated by expanding this matrix. For our matrix, it simplifies to:

\[(2-\lambda)\left[ (2-\lambda)(-1-\lambda) \right] - 2\left[ 0 - 2(2-\lambda) \right]\]

which upon further simplification gives:

\[(2-\lambda)^2(\lambda + 3) = 0\]

• Each term in the equation is solved to find the eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\), and \(\lambda_3 = -3\).

After finding the eigenvalues, the next step is to find the corresponding eigenvectors. For each eigenvalue, we solve the equation \(A - \lambda I\)\textbf{v} = 0\) to find the vector \textbf{v}\ that satisfies this.

For our matrix and eigenvalue \(\lambda_1 = 2\), we have:
\A - 2I = \left(\begin{array}{rrr}0 & 0 & 2 \ 0 & 0 & 0 \ 2 & 0 & -3\end{array}\right)\

Solving the equation \(\left(\begin{array}{rrr}0 & 0 & 2 \ 0 & 0 & 0 \ 2 & 0 & -3\end{array}\right)\mathbf{v} = 0\) gives us the eigenvector \(\mathbf{v_1} = \left(\begin{array}{r}1 \ 0 \ 0\end{array}\right)\).

We perform similar steps for \(\lambda_2 = 2\) and get the eigenvector \(\mathbf{v_2} = \left(\begin{array}{r}0 \ 1 \ 0\end{array}\right)\).

For \(\lambda_3 = -3\), the matrix changes to:
\(A + 3I = \left(\begin{array}{rrr}5 & 0 & 2 \ 0 & 5 & 0 \ 2 & 0 & 2\end{array}\right)\

Solving \(\left(\begin{array}{rrr}5 & 0 & 2 \ 0 & 5 & 0 \ 2 & 0 & 2\end{array}\right)\mathbf{v} = 0\) results in the eigenvector \(\mathbf{v_3} = \left(\begin{array}{r}-2 \ 0 \ 5\end{array}\right)\). These eigenvectors help describe the directions in which the linear transformation acts by only scaling.