As we know by the famous theorem "eigenvectors corresponding to distinct eigenvalues are orthogonal for a real symmetric matrix" can this result be also true for the same eigenvalues My intuition says yes. i.e. for a real symmetric matrix eigenvectors are orthogonal whether the eigenvalue is distinct or the same. Note we have one counter-example for if the matrix is not real Symmetrix then eigenvectors are not orthogonal if the eigenvalues are the same. \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix} but in real case I'm confused, so solution is appearsiable.
$\begingroup$
$\endgroup$
5
-
2$\begingroup$ If $A=I$ then $\begin{bmatrix}1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\0\end{bmatrix}$ are both eigenvectors... $\endgroup$– David C. UllrichJun 13, 2021 at 10:03
-
3$\begingroup$ Do not confuse if eigenvectors must be orthogonal and if they can be chosen to be orthogonal. All nonzero vectors in $\mathbf R^2$ are eigenvectors of the identity matrix but they are not all orthogonal. $\endgroup$– KCdJun 13, 2021 at 10:03
-
$\begingroup$ If an eigenvalue of a real symmetric matrix has multiplicity $k$ , then we can find $k$ pairwise orthogonal eigenvectors corresponding with this eigenvalue. $\endgroup$– PeterJun 13, 2021 at 10:04
-
$\begingroup$ $v$ and $2v$ are eigenvectors for $\lambda$ as soon as $v$ is. $\endgroup$– user239203Jun 13, 2021 at 10:05
-
$\begingroup$ @KCd means all eigenvectors space may not be orthogonal for the real symmetric matrix but we can choose some orthogonal eigenvectors space. right sir $\endgroup$– Mathematics learnerJun 13, 2021 at 10:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
If an eigenspace has dimension $>1$, then any basis of that subspace consists of eigenvectors. Clearly, these can be picked to be orthogonal, but they need not be orthogonal.
answered Jun 13, 2021 at 10:03
-
$\begingroup$ okay I understand thank you, sir $\endgroup$ Jun 13, 2021 at 10:08