Detail in the proof that eigenvectors are orthogonal

Question

Show that, for a symmetric matrix $\textbf{A}$, if $\textbf{x}_1$ and $\textbf{x}_2$ are two eigenvectors of $\textbf{A}$ such that their respective eigenvalues are distinct and non-zero then $\textbf{x}_1^\top\textbf{x}_2 = 0$

Solution

because $\textbf{A}$ is symmetric we know that $$ (\textbf{x}_2^\top\textbf{A}\textbf{x}_1)^\top = \textbf{x}_1^\top\textbf{A}\textbf{x}_2 $$ $$ \textbf{x}_1^\top\textbf{A}\textbf{x}_2 = \textbf{x}_2^\top\textbf{A}\textbf{x}_1\tag{1} $$ from the eigenvalue/eigenvector definition we know that: \begin{align*} \textbf{x}_1^\top\textbf{A}\textbf{x}_2 &= \textbf{x}_1^\top\lambda_2\textbf{x}_2\\ \textbf{x}_2^\top\textbf{A}\textbf{x}_1 &= \textbf{x}_2^\top\lambda_1\textbf{x}_1 \end{align*} Hence \begin{align*} \textbf{x}_1^\top\lambda_2\textbf{x}_2&=\textbf{x}_2^\top\lambda_1\textbf{x}_1\\ \implies \lambda_2\textbf{x}_1^\top\textbf{x}_2&=\lambda_1\textbf{x}_2^\top\textbf{x}_1\\ \implies 0&= \lambda_2\textbf{x}_1^\top\textbf{x}_2-\lambda_1\textbf{x}_2^\top\textbf{x}_1\\ &= \lambda_2\textbf{x}_1^\top\textbf{x}_2-\lambda_1\textbf{x}_1^\top\textbf{x}_2\\ &= \underbrace{(\lambda_2-\lambda_1)}_{\ne 0 \text{ by definition}}\textbf{x}_1^\top\textbf{x}_2\\ \implies 0&= \textbf{x}_1^\top\textbf{x}_2 \quad \square \end{align*}

My question

• which property of linear algebra do we have to invoke to justify (1)?

Answer 1

The transpose of a scalar is the same scalar, so we can drop the $\top$ in the left hand side. Then swap LHS and RHS.

Answer 2

Whenever $AB$ is defined we have $(AB)^{T}=B^{T}A^{T}$. Apply this twice to derive 1). [Also $(A^{T})^{T}=A$].