Given a basis of complex eigenvectors for, say, a $2 \times 2$ symmetric matrix $A$ (which hence has real eigenvalues).
Can one generate a basis of real(-valued) eigenvectors from the real and imaginary parts of the given basis of complex eigenvectors?
Say $v = ((a_1 + b_1i),(a_2 + b_2i))$ is a complex eigenvector with real eigenvalue $\lambda_1$ and $w = ((c_1 + d_1i), (c_2 + d_2i))$ is a complex eigenvector with real eigenvalue $\lambda_2$
and these two complex eigenvectors $v$ and $w$ are orthogonal.
Then $(a1,a2)$ and $(b1,b2)$ are real eigenvectors with real eigenvalue $\lambda_1$ (for matrix $A$) And $(c1,c2)$ and $(d1,d2)$ are real eigenvectors with real eigenvalue $\lambda_2$ (for matrix $A$)
Can these real-valued eigenvectors be used to construct an orthogonal basis of real-valued eigenvectors for A? How does the construction proceed?
If $x$ is a complex eigenvector of a real matrix $A$ for a real eigenvector $\lambda$, i.e. $A x = \lambda x$, then taking real and imaginary parts in this equation we find that $\text{Re}(x)$ and $\text{Im}(x)$, if nonzero, are also eigenvectors of $A$ for eigenvalue $\lambda$. Since $x$ is nonzero, at least one of $\text{Re}(x)$ and $\text{Im}(x)$ is nonzero, and of course $x$ is in the linear span of $\text{Re}(x)$ and $\text{Im}(x)$.
If you have a basis consisting of complex eigenvectors, take their real and imaginary parts and you still have a set that spans the whole space. Take a maximal linearly independent subset and you have a basis. It might not be orthogonal: eigenvectors for different eigenvalues are automatically orthogonal, but eigenvectors for the same eigenvalue might not be. So you might use the Gram-Schmidt process to find an orthonormal basis.