This concept is one of the first introduced in linear algebra university course. However, as it happens with most knowledge you were not ready to understand, just few facts left in my head.

The reason why I had to return to it was use of eigenvalues and eigenvectors in graph clustering. Their existence was guaranteed by well-known theorem, that eigenvectors of symmetric matrices are orthonormal basis of vector space.

But what is orthonormal? We need scalar product to introduce such concept. In turn, we need basis to define scalar product. It’s not a problem if we stay at matrix level of understanding vector spaces, cause when we have a matrix it means, that we’ve already chosen some basis. But what if we try to make one step up to abstract linear spaces and linear operators defined on them?

Symmetric matrix corresponds to self-adjoint operator $A$:
$(x, Ay) = (Ax, y)$

As we see, we need scalar product (and thus a basis) to introduce self-adjointness. So my question is: is self-adjointness property of linear operator or linear operator with respect to some basis of vector space?

The answer is easy. Let’s fix basis $E$. Operator is self-adjoint in basis $E$ if and only if it is distension in a set of directions, that can be obtained from $E$ with isometry (orthogonal transformations). Thus, any operator with system of eigenvectors $F$ unequivalent to $E$ (there’s no isometry to map $E$ to $F$) is not self-adjoint with respect to $E$ but is self-adjoint with respect to $F$!

So self-adjointness is property of operator with respect to basis.

Well, having done with one question we meet another one: what about orthogonal operators? Is this property basis independent or not?

The answer is negative again. Unfortunately, I don’t know beautiful explanation. But I’ll show a counterexample. If discussed property doesn’t depend on the basis, all the matrices of form $S^{-1}AS$ should be orthogonal is $A$ is orthogonal. But it doesn’t hold for:

$A = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

$S = \left( \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right)$

$S^{-1}AS = \left( \begin{array}{cc} -1 & 0 \\ 1 & 1 \end{array} \right)$