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 2×2 $$A=\begin{bmatrix}
1 &i \\
i &1
\end{bmatrix}$$
has non real eigenvalues
Best Answer
For your first question, the identity matrix does the trick: any two vectors, orthogonal or not, are eigenvectors with eigenvalue 1.
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).