Given the matrix \(A\): \[ A = \begin{pmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix} \]
We want to find the eigenvalues (\(\lambda\)), by solving the characteristic polynomial \(\det(A - \lambda I) = 0\).
The characteristic polynomial of \(A\) is obtained by: \[ \det \begin{pmatrix} 1-\lambda & -1 & 1 \\ -1 & 1-\lambda & -1 \\ 1 & -1 & 1-\lambda \end{pmatrix} = (1-\lambda)^3 - 3(1-\lambda) = 0 \]
This simplifies to: \[ (1-\lambda)^3 - 3(1-\lambda) = (1-\lambda)(\lambda^2 - 2\lambda - 2) = 0 \]
This equation gives us the eigenvalues directly: \[ \lambda_1 = 1, \quad \lambda_2 = 1 + \sqrt{3}, \quad \lambda_3 = 1 - \sqrt{3} \]
To find the eigenvectors for each \(\lambda_i\), we solve \((A - \lambda_i I)x = 0\). For \(\lambda_1 = 1\): \[ (A - I)x = \begin{pmatrix} 0 & -1 & 1 \\ -1 & 0 & -1 \\ 1 & -1 & 0 \end{pmatrix}x = 0 \]
The solutions to this system are the eigenvectors, and the eigenspace for each \(\lambda_i\) is the span of these eigenvectors. For \(\lambda_1 = 1\), the eigenvectors correspond to any scalar multiples of: \[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \]
To find the eigenvectors corresponding to \(\lambda_2\), we solve the equation \((A - \lambda_2 I)x = 0\). This simplifies to solving the system:
\[ \begin{pmatrix} -\sqrt{3} & -1 & 1 \\ -1 & -\sqrt{3} & -1 \\ 1 & -1 & -\sqrt{3} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
Choosing \(v_2 = 1\) and \(v_3 = \sqrt{3}\), we find \(v_1 = 1\). Thus, an eigenvector corresponding to \(\lambda_2\) is:
\[ \mathbf{v}_{\lambda_2} = \begin{pmatrix} 1 \\ 1 \\ \sqrt{3} \end{pmatrix} \]
Similarly, for \(\lambda_3\), solving \((A - \lambda_3 I)x = 0\) leads to:
\[ \begin{pmatrix} \sqrt{3} & -1 & 1 \\ -1 & \sqrt{3} & -1 \\ 1 & -1 & \sqrt{3} \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
Choosing \(v_2 = 1\) and \(v_3 = -\sqrt{3}\) gives \(v_1 = 1\). Therefore, an eigenvector for \(\lambda_3\) is:
\[ \mathbf{v}_{\lambda_3} = \begin{pmatrix} 1 \\ 1 \\ -\sqrt{3} \end{pmatrix} \]
These vectors provide the directions of the eigenspaces corresponding to their respective eigenvalues.
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