What does it mean to say that the eigenvectors of a symmetric matrix "can be chosen to be" orthogonal? Symmetric matrices ($ A = A^T$) are always diagonalizable, which means that their eigenvectors are orthogonal. The book I am reading is careful to say:

...its eigenvectors can always be chosen to be orthogonal (emphasis mine).

I must be missing something because I don't see why one cannot simply say that the eigenvectors corresponding to the different eigenvalues are orthogonal to each other.
 A: The idea that "eigenvectors for distinct eigenvalues are orthogonal" is approximately ok, but potentially fatally flawed, since, as noted by others, eigenspaces can have dimensions greater than $1$...
A way to dodge this is by "eigenspaces for distinct eigenvalues are orthogonal".
That way one avoids inadvertently claiming irrelevantly silly things about eigenvectors in a greater-than-one-dimensional eigenspace.
Although, yes, having eigenspaces of dimension $>1$ is "non-generic", it is easily possible in non-pathological examples.
A: "Diagonalizable" does not mean "diagonalizable with orthogonal eigenvectors". Consider the linear transformation with eigenvalues $1$ and $2$ corresponding to eigenvectors $(1,0)$ and $(1,1)$.
Of course the corresponding matrix won't be symmetric.
When the matrix is symmetric, then eigenvectors corresponding to different eigenvalues are orthogonal. But eigenvalues may be multiple. The identity matrix is symmetrical. Every basis is a basis of eigenvectors. You can, however, choose s basis that is orthogonal.
