Question
Obtain the eigenvalues and the orthonormal eigenvectors for the following real symmetric matrix:
Answer :
Word Count : 807
To find the eigenvalues and orthonormal eigenvectors of the given real symmetric matrix: \[ A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ \sin\theta & -\cos\theta & 0 \\ 0 & 0 & -1 \end{bmatrix}, \] we proceed step by step. --- ### Step 1: Find the eigenvalues The eigenvalues \(\lambda\) are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0, \] where \(I\) is the identity matrix. Substituting \(A\) and \(I\), we get: \[ A - \lambda I = \begin{bmatrix} \cos\theta - \lambda & \sin\theta & 0 \\ \sin\theta & -\cos\theta - \lambda & 0 \\ 0 & 0 & -1 - \lambda \end{bmatrix}. \] The determinant of this matrix is: \[ \det(A - \lambda I) = \begin{vmatrix} \cos\theta - \lambda & \sin\theta & 0 \\ \sin\theta & -\cos\theta - \lambda & 0 \\ 0 & 0 & -1 - \lambda \end{vmatrix}. \] Expanding the determinant along the third row, we get: \[ \det(A - \lambda I) = (-1 - \lambda) \cdot \begin{vmatrix} \cos\theta - \lambda & \sin\theta \\ \sin\theta & -\cos\theta - \lambda \end{vmatrix}. \] The \(2 \times 2\) determinant is: \[ \begin{vmatrix} \cos\theta - \lambda & \sin\theta \\ \sin\theta & -\cos\theta - \lambda \end{vmatrix} _____ _______ __________ _______ ______ _________ ________ ___.
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To find the eigenvalues and orthonormal eigenvectors of the given real symmetric matrix: \[ A = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ \sin\theta & -\cos\theta & 0 \\ 0 & 0 & -1 \end{bmatrix}, \] we proceed step by step. --- ### Step 1: Find the eigenvalues The eigenvalues \(\lambda\) are found by solving the characteristic equation: \[ \det(A - \lambda I) = 0, \] where \(I\) is the identity matrix. Substituting \(A\) and \(I\), we get: \[ A - \lambda I = \begin{bmatrix} \cos\theta - \lambda & \sin\theta & 0 \\ \sin\theta & -\cos\theta - \lambda & 0 \\ 0 & 0 & -1 - \lambda \end{bmatrix}. \] The determinant of this matrix is: \[ \det(A - \lambda I) = \begin{vmatrix} \cos\theta - \lambda & \sin\theta & 0 \\ \sin\theta & -\cos\theta - \lambda & 0 \\ 0 & 0 & -1 - \lambda \end{vmatrix}. \] Expanding the determinant along the third row, we get: \[ \det(A - \lambda I) = (-1 - \lambda) \cdot \begin{vmatrix} \cos\theta - \lambda & \sin\theta \\ \sin\theta & -\cos\theta - \lambda \end{vmatrix}. \] The \(2 \times 2\) determinant is: \[ \begin{vmatrix} \cos\theta - \lambda & \sin\theta \\ \sin\theta & -\cos\theta - \lambda \end{vmatrix} _____ _______ __________ _______ ______ _________ ________ ___.
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