# Symmetric Matrix , Eigenvectors are not orthogonal to the same eigenvalue.

I know that the symmetric matrix has orthogonal eigenvectors corresponding eigenvalues. Also, the eigenvectors from the same eigenvalue are linearly independent.

I need an example of symmetric matrix such that the eigenvectors from the same eigenvalue are not orthogonal.

Also, I know that the eigenvalues of the symmetric matrix are real. But , this statement true iff the entries are real numbers.Right? Since the matrix 2x2 $$A=\begin{bmatrix} 1 &i \\ i &1 \end{bmatrix}$$

has non real eigenvalues

• If $x_1$ and $x_2$ are eigenvectors with the same eigenvalue $\lambda$, the vector $x_1+x_2$ will also be an eigenvector with the same eigenvalue $\lambda$, and it won't be orthogonal to either $x_1$ or $x_2$.
– Paul
Apr 19, 2017 at 19:44

More generally, any combination of two eigenvectors with the same eigenvalue $\lambda$ is itself an eigenvector (with eigenvalue $\lambda$); even if your two original eigenvectors are orthogonal, a linear combinations thereof will not be orthogonal to either one.
For the second question, a complex-valued matrix has real eigenvalues iff the matrix is Hermitian, which is to say that it is equal to the conjugate of its transpose: $A^\dagger = (A^T)^* = A$. So while your $A$ is not Hermitian, the matrix $$B = \begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix}$$ is, and has two real eigenvalues (0 & 2).