(real symmetric matrices) does orthogonality of eigenvectors (distinct eigenvalues) depend on choice of basis?

For some reason I cannot wrap my head around this one. My schoolbook begins the chapter on eigendecomposition of real symmetric matrices by stating that eigenvectors from distinct eigenspaces are orthogonal. The ensuing proof made use of transposes, row vectors and matrix multiplication. That's all fine but in the end it seemed to suggest we're dealing in cartesian coordinates (where dotting two vectors implies summing the product of their corresponding components). Is there a fault in my reasoning or does eigendecomposition of real symmetric matrices depend on our initial basis?

Picture: example

In the above example, if the u's behave like i and j, then the alpha's are certainly orthogonal with respect to the dot product. But say we didn't know this. As in most exercices in the book, I'm thrown squares with numbers in each entry and the only indication is to 'show the spectral decomposition'. Must I make the assumption that I'm dealing with cartesian coordinates?

• The transpose/adjoint operator (needed to check that the matrix of eigenvectors is orthogonal and that the original matrix is symmetric) depends on some choice of basis. It also defines an inner product. – reuns May 14 at 5:18