# eigenvectors of a real symmetric matrix are always orthogonal

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.

• If $A=I$ then $\begin{bmatrix}1\\1\end{bmatrix}$ and $\begin{bmatrix}1\\0\end{bmatrix}$ are both eigenvectors... Jun 13, 2021 at 10:03
• 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.
– KCd
Jun 13, 2021 at 10:03
• If an eigenvalue of a real symmetric matrix has multiplicity $k$ , then we can find $k$ pairwise orthogonal eigenvectors corresponding with this eigenvalue. Jun 13, 2021 at 10:04
• $v$ and $2v$ are eigenvectors for $\lambda$ as soon as $v$ is.
– user239203
Jun 13, 2021 at 10:05
• @KCd means all eigenvectors space may not be orthogonal for the real symmetric matrix but we can choose some orthogonal eigenvectors space. right sir Jun 13, 2021 at 10:07

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.