Does every symmetric real matrix have an eigenbasis of real vectors? Let $A$ be an $n\times n$ matrix that is real and symmetric: $A^T=A$.
We know that any such matrix is unitarily diagonalisable and has real eigenvalues.
Is it always possible to find an eigenbasis for $A$ made of only real vectors?
If $A$ has unit rank, this is always the case: $A=\lambda \,vv^T, \lambda\in\mathbb R,$ implies $v=e^{i\phi}v'$ with $\phi\in\mathbb R$ and $v'\in\mathbb R^n$. This because $(vv^T)_{ij}=v_i \bar v_j$ while $(vv^T)_{ji}=(vv^T)_{ij}^*$, thus if $v_i=|v_i|e^{i\phi_i}$, then $v_i \bar v_j=|v_i v_j| e^{i(\phi_i-\phi_j)}$, and if $A$ is real then we must have $e^{i(\phi_i-\phi_j)}\in\mathbb R$, and therefore $v_j =|v_j|e^{i\pi n_{ij}}e^{i\phi_i}$ for all $i,j$, for some $n_{ij}\in\mathbb Z$. We conclude that $v$ is real, up to a global phase.
In higher dimensions, we can have real symmetric matrices with complex eigenvectors. A trivial case being the $2\times 2$ identity, which can be written as $I=P_{+i}+P_{-i}$ with $P_{\pm i}\equiv v_{\pm i}v_{\pm i}^\dagger$ and $\sqrt2 \, v_{\pm i}\equiv (1,i)^T$. Still, the identity obviously always also admits an eigenbasis of real vectors.
What about the general case? Is there an example of a real symmetric matrix for which there is no eigenbasis of real matrices?
 A: If $A$ is real and $\lambda$ is a real eigenvalue with geometric multiplicity $k$ over $\mathbb C$, then it also has geometric multiplicity $k$ over $\mathbb R$. This is because the geometric multiplicity is simply $n$ minus the rank of $A-\lambda I$ -- and the rank can be computed by for example Gaussian elimination, which goes exactly the same no matter what we take the ambient field to be.
Therefore we can find a real basis for the eigenspace of $\lambda$, comprising $k$ vectors. Concatenating all of those bases gives a real eigenbasis for $A$.
A: The orthogonal projection $P_{\lambda}$ onto the eigenspace with eigenvalue $\lambda$ is the matrix limit
$$
         P_{\lambda}=\lim_{\mu\rightarrow\lambda\\ \mu\in\mathbb{R}}(\mu-\lambda)(\mu I-A)^{-1}
$$
This is a real matrix because the matrices on the right are real. And
$$
                 AP_{\lambda}=\lambda P_{\lambda}.
$$
Every column of $P_{\lambda}$ is either $0$ or is a non-zero vector with real values that is an eigenvector with eigenvalue $\lambda$ of $A$.
A: If it’s diagonalizable, then the geometric multiplicity of every eigenvalue equals the algebraic multiplicity, so there’s always an eigenvector to any eigenvalue. If I have a complex eigenvector for a real eigenvalue, then the real part and imaginary part of that eigenvector will be purely real eigenvectors for that eigenvalue.
But also, the fact that symmetric matrices are orthogonally diagonalizable (by a real orthogonal matrix) tells you that in fact, you can pick an orthonormal basis of real eigenvectors for your eigenvalues, and the eigenvectors will just be the columns of the orthogonal matrix you use to diagonalize.
