If a real symmetric matrix has repeated eigenvalues, why does it still have n linearly independent eigenvectors? I know that if $A=A^T$ is real, then the eigenvalues are real, and that eigenvectors corresponding to different eigenvalues are orthogonal. Is there an easy way to see what happens in the case of a repeated eigenvalue, or does this require some sophisticated arguments (that I guess will come later in my book)?
 A: The best argument I know of is actually completely agnostic about whether the eigenvalues are repeated. It depends on the following simple calculation:
Suppose $x$ is an eigenvector of $A$, $A$ is symmetric, and $y$ is orthogonal to $x$. Then
$$(x,Ay)=(Ax,y)=(\lambda x,y)=\lambda (x,y)=0.$$
Thus $Ay$ is also orthogonal to $x$.
Now we know that if $x$ is an eigenvector of $A$, then the orthogonal complement of the span of $x$ is an invariant subspace under $A$. In other words, the restriction of $A$ to the orthogonal complement of the span of $x$ is an endomorphism (a linear map from a vector space to itself). Since everything is finite dimensional, this map must have an eigenvalue. You can also show that this restriction is symmetric again. But now we can do the procedure again, reducing the dimension by $1$ at each step, until we get to dimension $1$ where the result is trivial.
An interesting feature of this argument is that we do not extract the entire eigenspace of the eigenvalue $\lambda$ at once. This means we do not have to directly prove that the algebraic multiplicity of $\lambda$ and the geometric multiplicity of $\lambda$ coincide. Instead, maybe we get that eigenvalue again during the construction, maybe we don't. The procedure doesn't care either way.
Incidentally, in the case of a repeated eigenvalue, we can still choose an orthogonal eigenbasis: to do that, for each eigenvalue, choose an orthogonal basis for the corresponding eigenspace. (This procedure does that automatically.) The only funny thing about this case is that there exist non-orthogonal eigenbases.
