91影视

Given matrix A is \[A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{bmatrix}\] To find the eigenvalues, we need to solve the characteristic equation, `det(A - 位I) = 0`, where I is the identity matrix and 位 are the eigenvalues we're solving for. Let's calculate the determinant: \(|A-\lambda I|=\begin{vmatrix} 1-\lambda & 2 & 1 \\ 2 & 4-\lambda & 2 \\ 1 & 2 & 1-\lambda \end{vmatrix}\) Expanding the determinant, we get the characteristic polynomial: \((1-\lambda)[(4-\lambda)(1-\lambda) - 4] - 2[2(1-\lambda) - 2] + [2 - 4(1-\lambda)] = -\lambda^{3} + 6\lambda^{2} - 6\lambda\) We can factor out a \(-\lambda\) from the polynomial: \(-\lambda(\lambda^{2} - 6\lambda + 6) = 0\) The eigenvalues are: 位鈧� = 0, 位鈧� = 3, and 位鈧� = 3 Now, let's find the eigenvectors corresponding to each eigenvalue. For 位鈧� = 0: Solve (A - 位鈧両)v = 0 for v: \(\begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) We get the eigenvector: v鈧� = [1, -0.5, 0] For 位鈧� = 3 and 位鈧� = 3 (here we have a repeated eigenvalue, so we only need to find one linearly independent eigenvector): Solve (A - 位鈧侷)v = 0 for v: \(\begin{bmatrix} -2 & 2 & 1 \\ 2 & 1 & 2 \\ 1 & 2 & -2 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) We get two linearly independent eigenvectors: v鈧� = [1, 1, 0] and v鈧� = [-1, 0, 1]

We need to normalize the eigenvectors we found in step 1: Normalized v鈧� = \(\frac{1}{\sqrt{1^2 + (-0.5)^2 + 0^2}}\)[1, -0.5, 0] = [2/3, -1/3, 0] Normalized v鈧� = \(\frac{1}{\sqrt{1^2 + 1^2 + 0^2}}\)[1, 1, 0] = [1/\(\sqrt{2}\), 1/\(\sqrt{2}\), 0] Normalized v鈧� = \(\frac{1}{\sqrt{(-1)^2 + 0^2 + 1^2}}\)[-1, 0, 1] = [-1/\(\sqrt{2}\), 0, 1/\(\sqrt{2}\)]

Now we'll form the orthogonal matrix S by using the normalized eigenvectors as its columns: \[S = \begin{bmatrix} 2/3 & 1/\sqrt{2} & -1/\sqrt{2} \\ -1/3 & 1/\sqrt{2} & 0 \\ 0 & 0 & 1/\sqrt{2} \end{bmatrix}\]

To verify that the resulting matrix S is indeed orthogonal, we need to show that S岬�S = I: \[S岬�S = \begin{bmatrix} 2/3 & -1/3 & 0 \\ 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 0 & 1/\sqrt{2} \end{bmatrix} \begin{bmatrix} 2/3 & 1/\sqrt{2} & -1/\sqrt{2} \\ -1/3 & 1/\sqrt{2} & 0 \\ 0 & 0 & 1/\sqrt{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\] S岬�S = I, so the matrix S is orthogonal. Therefore, we have found the orthogonal matrix S that satisfies the requirement of the exercise: \[S = \begin{bmatrix} 2/3 & 1/\sqrt{2} & -1/\sqrt{2} \\ -1/3 & 1/\sqrt{2} & 0 \\ 0 & 0 & 1/\sqrt{2} \end{bmatrix}\]