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.

  • 2
    $\begingroup$ If $A=I$ then $\begin{bmatrix}1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\0\end{bmatrix}$ are both eigenvectors... $\endgroup$ Jun 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. $\endgroup$
    – KCd
    Jun 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. $\endgroup$
    – Peter
    Jun 13, 2021 at 10:04
  • $\begingroup$ $v$ and $2v$ are eigenvectors for $\lambda$ as soon as $v$ is. $\endgroup$
    – user239203
    Jun 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 $\endgroup$ Jun 13, 2021 at 10:07

1 Answer 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.

  • $\begingroup$ okay I understand thank you, sir $\endgroup$ Jun 13, 2021 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.