$AA^T=A^TA$ and othogonallity of eigenvectors In his Unit 3 Exam Review, Strang gives the theorem that a matrix $A$ has orthogonal eigenvectors if and only if $AA^T=A^TA$, and then lists symmetric, antisymmetric and orthogonal matrices as meeting this condition.
However, I know that in the symmetric case it's the eigenspaces that are guaranteed to be orthogonal, not the eigenvectors.
Is it correct to assume that he meant to say eigenspaces and not eigenvectors?
EDIT
In his lecture on Symmetric Matrices and Positive Definateness Strang says:

So -- I have a -- you should say "why?" and I'll at least answer why
  for case one, maybe case two, the checking the Eigen -- that the
  eigenvectors are perpendicular, I'll leave to, the -- to the book. But
  let's just realize what -- well, first I have to say, it -- it could
  happen, like for the identity matrix -- there's a symmetric matrix.
  Its eigenvalues are certainly all real, they're all one for the
  identity matrix. What about the eigenvectors? Well, for the identity,
  every vector is an eigenvector. So how can I say they're
  perpendicular? What I really mean is the -- they -- this word are
  should really be written can be chosen perpendicular. That is, if we
  have -- it's the usual case. If the eigenvalues are all different,
  then each eigenvalue has one line of eigenvectors and those lines are
  perpendicular here. But if an eigenvalue's repeated, then there's a
  whole plane of eigenvectors and all I'm saying is that in that plain,
  we can choose perpendicular ones. So that's why it's a can be chosen
  part, is -- this is in the case of a repeated eigenvalue where there's
  some real, substantial freedom. But the typical case is different
  eigenvalues, all real, one dimensional eigenvector space, Eigen
  spaces, and all perpendicular.

 A: I'm assuming Strang is allowing complex eigenvectors. The statement "$A$ has orthogonal eigenvectors" is a bit imprecise; a better statement would be "You can choose a basis of pairwise orthogonal eigenvectors". If you take his statement literally then it is correct but trivial, since a large enough normal matrix ($AA^*=A^* A$) will have some pair of orthogonal eigenvectors.
The eigenspaces for distinct eigenvalues are orthogonal, but that's different from saying there exists an orthonormal basis consisting of eigenvectors. The eigenspaces might not sum to the whole space.
A: For normal matrices, the eigenvectors corresponding to the same eigenvalue are linear independent, not necessary orthogonal. However, since any linear combination of two eigenvectors with the same eigenvalue is still an eigenvector associated with that same eigenvalue (i.e. the eigenvectors are not unique), we can always construct orthogonal eigenvectors through Gram-Schmidt procedure. In other words, if a normal matrix has eigenvalues with multiplicity, there are infinite sets of eigenvectors, some of which are orthogonal.
A: Whenever you have an eigenspace of dimension greater than $1$, you can of course find non-orthogonal bases of that space if you want to; there is nothing a matrix could undertake to prevent that (except not having such eigenspaces). However you can choose an orthonormal basis in the eigenspace. If you do that for every eigenspace and those eigenspaces are orthogonal, then you will have an orthonormal basis for their direct sum, which in this case is the whole space.
A: I assume all the matrices below are real. For a symmetric matrix, the eigenvectors are orthogonal. The proof is relatively straight forward. Let $A$ be the symmetric matrix. Consider two distinct eigenvalues say $\lambda$ and $\mu$. Let $x$ and $y$ be the associated eigenvectors.
Hence, we have that $$Ax = \lambda x; \,\,\,\, Ay = \mu y$$
We hence have that
$$y^TAx = \lambda y^Tx; \,\,\,\, x^TAy = \mu x^T y$$
But note that $y^TAx = y^T(Ax) = (Ax)^T y= x^T A^T y = x^T Ay$. Hence, we get that
$$\lambda y^Tx = \mu x^Ty$$ Again $y^Tx = x^Ty$ and since we chose $\lambda \neq \mu$. We have that $x^Ty = 0$. Hence, the eigenvectors corresponding to distinct eigenvalues are orthogonal.
One of the key properties of a symmetric matrix is that it is diagonalizable. Hence, if we have an eigenvalue with multiplicity $2$, there are two vectors $x$ and $y$ such that $Ax = \lambda x$ and $A y = \lambda y$, where $x \neq y$. But now any linear combination of $x$ and $y$ is also an eigenvector with the same eigenvalue. Hence, any vector in the space spanned by $x$ and $y$ is an eigenvector. Now choose any two orthogonal vectors in this space to be your eigenvectors.
