How can I prove that two eigenvectors are orthogonal?

Question

I obtained 6 eigenpairs of a matrix using eigs of Matlab. How can I demonstrate that these eigenvectors are orthogonal to each other? I am almost sure that I normalized in the right way modulus and phase but they do not seem to be orthogonal. The matrix should be normal. The matrix comes from the discretization of the Euler-Bernoulli beam problem for a beam of length 1 with hinged free boundary conditions: $$ \frac{\partial u }{\partial t} + \gamma \frac{\partial^4 y}{\partial x^4} = 0,\\ \frac{\partial y}{\partial t} - u = 0.\\$$ Then, the eigenproblem can be written as:

$$ \lambda \left[ \begin{matrix} I & 0 \\ 0 & I \end{matrix} \right] \left\{ \begin{matrix} y \\ u \end{matrix} \right\} = \left[ \begin{matrix} 0 & I \\ -\gamma B & 0 \end{matrix} \right] \left\{ \begin{matrix} y \\ u \end{matrix} \right\},$$ where $I$ is the identity matrix and $B$ is the bilaplacian operator discretized using finite difference.

Answer 1

If the matrix is normal (i.e., $A^HA=AA^H$), you should indeed get orthonormal eigenvectors both theoretically or numerically. You can check this by numerically by taking the matrix V built from columns of eigenvectors obtained from [V,D] = eigs(A) and computing V'*V, which should give you (very close to) the identity matrix.

The fact that you are not observing orthogonality most likely is due to the matrix not being normal (which you can also check numerically, e.g., by norm(A'*A-A*A','fro')). This is not unsurprising: Although your differential operator (in particular, the bilaplacian) is self-adjoint, this need not be the case for its discretization. In particular, most ways of modifying the stiffness matrix (in your case, $B$) to incorporate Dirichlet boundary conditions destroys the symmetry.

Answer 2

When you are dealing with complex valued vectors, the inner product is probably defined as $(u,v)=u_1^*v_1+...+u_n^*v_n$, where * indicates the complex conjugate. For example, the vector $u=(1,i)$ is not orthogonal to $v=(-i,1)$, because $(u,v)=1(-i)+(i)^*(1)=-2i$. On the other hand, $u$ is orthogonal to $w=(i,1)$. You cannot just use the ordinary "dot product" to show complex vectors are orthogonal.

Consider the test matrix $\left(\begin{matrix}1& -i \\ i& 1\end{matrix}\right)$. This matrix is Hermitian and it has distinct eigenvalues 2 and 0 corresponding to the eigenvectors $u$ and $w$ respectively. You can use this test matrix and its eigenvectors in Matlab to verify you calculating an inner product rather than what I call the ordinary "dot product."

Also note, the inner product is defined as above in physics. That's why I said "probably defined as". Mathematicians are more likely to define the inner product on complex vector spaces as $(u,v)=u_1v_1^*+...+u_nv_n^*$, which is just the complex conjugate of the one I defined above.