Let \(A \in \mathbb{R}^{n \times n}\) be a symmetric matrix and \(\lambda_1 \neq \lambda_2 \in \mathbb{R}\) be two distinct eigenvalues of \(A\) with corresponding eigenvectors \(v_1, v_2\).
Then \(v_1\) and \(v_2\) are orthogonal. Proof Included
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Let \(A \in \mathbb{R}^{n \times n}\) be a symmetric matrix and \(\lambda_1 \neq \lambda_2 \in \mathbb{R}\) be two distinct eigenvalues of \(A\) with corresponding eigenvectors \(v_1, v_2\).
Then \(v_1\) and \(v_2\) are orthogonal. Proof Included
\(\lambda_1 v_1 ^\top v_2 = (Av_1)^\top v_2\) \( = v_1^\top A ^\top v_2 = \) \(v_1^\top (Av_2)\) \( = \lambda_2 v_1^\top v_2\)
Field-by-field Comparison
| Field | Before | After |
|---|---|---|
| Text | <div>Let \(A \in \mathbb{R}^{n \times n}\) be a symmetric matrix and \(\lambda_1 {{c2::\neq}} \lambda_2 \in \mathbb{R}\) be two {{c2::distinct}} eigenvalues of \(A\) with corresponding eigenvectors \(v_1, v_2\).</div><div>Then \(v_1\) and \(v_2\) {{c1::are orthogonal}}. <i>Proof Included</i></div> | |
| Extra | \(\lambda_1 v_1 ^\top v_2 = (Av_1)^\top v_2\) \( = v_1^\top A ^\top v_2 = \) \(v_1^\top (Av_2)\) \( = \lambda_2 v_1^\top v_2\) |