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Eigenvalues are fundamental in understanding how a matrix behaves under linear transformations. They are scalar values that indicate how vectors are stretched or compressed when acted upon by a matrix. To find the eigenvalues of a matrix, we use the characteristic polynomial, which is derived from the matrix in the form of \( \det(A - \lambda I) = 0 \). Here, \( A \) is your matrix, \( I \) is the identity matrix of the same size, and \( \lambda \) represents the eigenvalues we seek. For the matrix \( A = \begin{bmatrix} 1 & \sqrt{2} \ \sqrt{2} & 0 \end{bmatrix} \), the characteristic polynomial is formed as follows: \( \det \begin{bmatrix} 1-\lambda & \sqrt{2} \ \sqrt{2} & -\lambda \end{bmatrix} = -\lambda^2 - \lambda - 2 = 0 \). When solved, this quadratic equation yields the eigenvalues \( \lambda_1 = -2 \) and \( \lambda_2 = 1 \).

• The roots of the characteristic polynomial are the eigenvalues.
• Eigenvalues tell us how much vectors are scaled during the transformation.

Once you have eigenvalues, the next step is to find the corresponding eigenvectors. These vectors are non-zero vectors that change only by a scalar factor when the matrix is applied. To find them, we solve the equation \( (A - \lambda I)\mathbf{v} = 0 \) for each eigenvalue.

Consider \( \lambda_1 = -2 \): substitute back into \( (A-\lambda I)\mathbf{v} = 0 \) which gives us the equation \[ \begin{bmatrix} 3 & \sqrt{2} \ \sqrt{2} & 2 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]. Solving this, we get one possible eigenvector \( \mathbf{v}_1 = \begin{bmatrix} -\sqrt{2} \ 1 \end{bmatrix} \).

Repeat for \( \lambda_2 = 1 \) to find \( \mathbf{v}_2 = \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix} \).

• Each eigenvalue corresponds to an eigenvector.
• Eigenvectors, along with their eigenvalues, define the direction and magnitude of the transformation effect.

Orthonormalization is a process necessary when forming an orthogonal matrix from eigenvectors. This step ensures that the eigenvectors are both orthogonal (perpendicular) and normalized (unit length).

For the eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} -\sqrt{2} \ 1 \end{bmatrix} \) and \( \mathbf{v}_2 = \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix} \), normalize them by dividing by their magnitudes.

The magnitude \( \| \mathbf{v}_1 \| = \sqrt{(-\sqrt{2})^2 + 1^2} = \sqrt{3} \) leads to the normalized vector: \[ \frac{1}{\sqrt{3}} \begin{bmatrix} -\sqrt{2} \ 1 \end{bmatrix} \].

Similarly, \( \| \mathbf{v}_2 \| = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3} \) results in the normalized vector: \[ \frac{1}{\sqrt{3}} \begin{bmatrix} \sqrt{2} \ 1 \end{bmatrix} \].

• Orthogonal matrices are composed of orthonormal vectors.
• These vectors make up the columns of the orthogonal matrix \( Q \).

The characteristic polynomial is a crucial tool for finding the eigenvalues of a matrix. It essentially provides a polynomial equation whose roots are the eigenvalues of the matrix.

To derive this polynomial for a given matrix \( A \), we use \( \det(A - \lambda I) = 0 \). For our example, matrix \( A = \begin{bmatrix} 1 & \sqrt{2} \ \sqrt{2} & 0 \end{bmatrix} \), we first form \( A - \lambda I = \begin{bmatrix} 1-\lambda & \sqrt{2} \ \sqrt{2} & -\lambda \end{bmatrix} \). Then, calculate the determinant: \[ (1-\lambda)(-\lambda) - (\sqrt{2})(\sqrt{2}) = -\lambda^2 - \lambda - 2 \].

This equation, \(-\lambda^2 - \lambda - 2 = 0 \), is solved to find the eigenvalues.

• The characteristic polynomial simplifies finding eigenvalues.
• It transforms an algebra problem into a calculus problem by using determinants.