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Eigenvectors of a symmetric matrix with different eigenvalues are orthogonal.

Created over 8 years ago, updated 10 days ago

Theorem. Eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are orthogonal.

Let $A$ be a real symmetric matrix. If $Av_1 = \lambda_1 v_1$ and $Av_2 = \lambda_2 v_2$ with $\lambda_1 \neq \lambda_2$, then $v_1 \cdot v_2 = 0$.

Proof. We have:
$$\lambda_1 (v_1 \cdot v_2) = (Av_1)^T v_2 = v_1^T A^T v_2 = v_1^T A v_2 = v_1^T (\lambda_2 v_2) = \lambda_2 (v_1 \cdot v_2)$$
Since $\lambda_1 \neq \lambda_2$, we must have $v_1 \cdot v_2 = 0$.

Example. For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 1$ with eigenvectors $v_1 = (1, 1)^T$ and $v_2 = (1, -1)^T$, which are orthogonal.