$Show that eigenvectors of a symmetric matrix are orthogonal$

Question

$Show that eigenvectors of a symmetric matrix are orthogonal$

Let $A$ be a symmetric matrix and $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $A$, and $\xi_1$ and $\xi_2$ be the corresponding eigenvectors. Show that
\[\xi_1 \cdot \xi_2=0.\]

Linear Algebra Matrices Eigenvalues & Eigenvectors

Sophiam

26

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

Since $\lambda_1 \neq \lambda_2$, we have
\[(1) A \xi_1 =\lambda_1 \xi_1 \Rightarrow \xi_2^{T} A \xi_1=\lambda_1 \xi_2^{T} \cdot \xi_1\]
and
\[(2) A \xi_2 =\lambda_2 \xi_2 \Rightarrow \xi_1^{T} A \xi_2=\lambda_2 \xi_1^{T} \cdot \xi_2.\]
Since $A$ is symmetric, it follows from (1) and (2) that
\[\lambda_1 \xi_1^{T} \cdot \xi_2=(\lambda_1 \xi_2^{T} \cdot \xi_1 )^T= (\xi_2^{T} A \xi_1)^T= \xi_1^{T} A^T \xi_2 =\xi_1^{T} A \xi_2=\lambda_2 \xi_1^{T} \cdot \xi_2\]
\[\Rightarrow \lambda_1 \xi_1^{T} \cdot \xi_2 =\lambda_2 \xi_1^{T} \cdot \xi_2 \Rightarrow (\lambda_1-\lambda_2) \xi_1^{T} \cdot \xi_2 =0 \]
\[\Rightarrow \xi_1^{T} \cdot \xi_2 =0,\]
and thus $\xi_1$ and $\xi_2$ are orthogonal. Here we have used the properties of the matrix transpose (See: https://en.wikipedia.org/wiki/Transpose ).

Malindad

53

$Show that eigenvectors of a symmetric matrix are orthogonal$

Answer

Related Questions

Search