Given that $A = A^\top$, we know that matrix $A$ is symmetric. A crucial property of symmetric matrices is that their eigenvectors corresponding to distinct eigenvalues are orthogonal.

Steps to Show Orthogonality

Let $\lambda_1$ and $\lambda_2$ be eigenvalues of the matrix $A$, with corresponding eigenvectors $v_1$ and $v_2$, such that: $A v_1 = \lambda_1 v_1$ $A v_2 = \lambda_2 v_2$ We aim to show that $v_1$ and $v_2$ are orthogonal, i.e., $v_1^\top v_2 = 0$.

Proof:

1. Start with the eigenvalue equations: $A v_1 = \lambda_1 v_1 \quad \text{and} \quad A v_2 = \lambda_2 v_2$ Multiply the first equation on the left by $v_2^\top$ and the second equation on the left by $v_1^\top$: $v_2^\top A v_1 = \lambda_1 v_2^\top v_1 \quad (1)$ $v_1^\top A v_2 = \lambda_2 v_1^\top v_2 \quad (2)$

2. Using symmetry of $A$: Since $A = A^\top$, we know that: $v_2^\top A v_1 = v_1^\top A v_2$ Therefore, from equations $(1)$ and $(2)$, we have: $\lambda_1 v_2^\top v_1 = \lambda_2 v_1^\top v_2$ This simplifies to: $\lambda_1 v_2^\top v_1 = \lambda_2 v_2^\top v_1$

3. Factor out $v_2^\top v_1$: $(\lambda_1 - \lambda_2) v_2^\top v_1 = 0$

4. Conclusion: Since $\lambda_1 \neq \lambda_2$ (by assumption), it must be that: $v_2^\top v_1 = 0$ Hence, $v_1$ and $v_2$ are orthogonal.

Final Conclusion:

The eigenvectors $v_1$ and $v_2$, corresponding to the distinct eigenvalues $\lambda_1$ and $\lambda_2$ of the symmetric matrix $A$, are orthogonal.

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How do you compute eigenvalues and eigenvectors for a given matrix?
2. What are some real-world applications of symmetric matrices?
3. Why do symmetric matrices have real eigenvalues?
4. What happens if $A$ has repeated eigenvalues in terms of eigenvector orthogonality?
5. How does this property extend to complex-valued matrices?

Tip: In linear algebra, symmetric matrices are especially important because their eigenvectors form an orthonormal basis, simplifying many computations.