Eigenvalues of Symmetric Matrix are Orthogonal

Theorem
Let $K$ be a ring.

Let $A$ be a symmetric matrix over $K$.

Let $\lambda_1, \lambda_2$ be distinct eigenvalues of $A$.

Let $\mathbf v_1, \mathbf v_2$ be eigenvectors of $A$ corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$ respectively.

Let $\innerprod \cdot \cdot$ be the dot product on $K$.

Then $\mathbf v_1$ and $\mathbf v_2$ are orthogonal with respect to $\innerprod \cdot \cdot$.

Proof
We have:


 * $A \mathbf v_1 = \lambda_1 \mathbf v_1$

and:


 * $A \mathbf v_2 = \lambda_2 \mathbf v_2$

We also have:

and:

We therefore have:


 * $0 = \paren {\lambda_2 - \lambda_1} \innerprod {\mathbf v_1} {\mathbf v_2}$

Since $\lambda_1 \ne \lambda_2$, we have:


 * $\innerprod {\mathbf v_1} {\mathbf v_2} = 0$

hence $\mathbf v_1$ and $\mathbf v_2$ are orthogonal with respect to $\innerprod \cdot \cdot$.