91Ó°ÊÓ

First, we will find the eigenvalues of matrix A. In order to do this, we need to solve the characteristic equation, which is given by: \[|A - \lambda I| = 0\] For the given matrix A: \begin{equation} \begin{vmatrix} 0 - \lambda & 1 \\ 1 & 0 - \lambda \end{vmatrix} = (\lambda^2 -1) = 0 \end{equation} Solving the equation, we get two eigenvalues: \[\lambda_1 = 1\] \[\lambda_2 = -1\]

Now we find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = 1\): \((A - \lambda_1 I)u_1 = 0\) \[\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}u_{11}\\u_{12}\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\] From this equation, we can see that \(u_{11} = u_{12}\), so we'll choose a simple solution: \(u_1 = \begin{pmatrix}1\\1\end{pmatrix}\) For \(\lambda_2 = -1\): \((A - \lambda_2 I)u_2 = 0\) \[\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix}u_{21}\\u_{22}\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\] From this equation, we can see that \(u_{21} = -u_{22}\), so we'll choose a simple solution: \(u_2 = \begin{pmatrix}1\\-1\end{pmatrix}\)

Now, we need to normalize both eigenvectors to obtain orthonormal eigenvectors. For \(u_1\): \[\|u_1\| = \sqrt{1^2+1^2} = \sqrt{2}\] \[\hat{u}_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}\] For \(u_2\): \[\|u_2\| = \sqrt{1^2+(-1)^2} = \sqrt{2}\] \[\hat{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\]

Finally, we can express A as a sum of scaled outer products of orthonormal eigenvectors: \[ A = \lambda_1 \hat{u}_1 \hat{u}_1^T + \lambda_2 \hat{u}_2 \hat{u}_2^T = \] \[ 1 \cdot \left(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}\right)\left(\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\end{pmatrix}\right) - 1 \cdot \left(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\right)\left(\frac{1}{\sqrt{2}}\begin{pmatrix}1&-1\end{pmatrix}\right) = \] \[ \frac{1}{2}\begin{pmatrix}1\\1\end{pmatrix}\begin{pmatrix}1&1\end{pmatrix} - \frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}\begin{pmatrix}1&-1\end{pmatrix} = \] \[ \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} -\frac{1}{2} \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} =\] \[ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \] So, the result is: \[A = \lambda_{1} \mathbf{u}_{1} \mathbf{u}_{1}^{T}+\lambda_{2} \mathbf{u}_{2} \mathbf{u}_{2}^{T} = \frac{1}{2}\begin{pmatrix}1\\1\end{pmatrix}\begin{pmatrix}1&1\end{pmatrix} - \frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}\begin{pmatrix}1&-1\end{pmatrix}\]