Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 8B / a

Let $Mv=\lambda v$ with $v\in {\mathbb{C}}^n$ nonzero. For a real symmetric matrix, $M^{\ast }=M$, where the star denotes the conjugate transpose. Consequently $v^{\ast }Mv$ equals its complex conjugate and is real. The eigenvalue equation givessince $v^{\ast }v>0$. Allowing a complex eigenvector in this argument is essential: reality of an eigenvalue must be proved, rather than assumed in advance.

Write a complex eigenvector as $v=u+iw$, where $u,w\in {\mathbb{R}}^n$. Since $M$ and its eigenvalue $\lambda$ are real,At least one of $u,w$ is nonzero, and it is a real eigenvector for the same eigenvalue. More generally every complex eigenvector is a complex linear combination of real ones: its real and imaginary parts lie in the real eigenspace. Thus each eigenspace has a basis chosen over the reals. The conclusion does not assert that every arbitrary complex eigenvector is itself real.

Choose real eigenvectors $u,v$ with distinct eigenvalues $\lambda ,\mu$. Symmetry givesTherefore $(\lambda -\mu )u^{T}v=0$, and $\boxed{u\cdot v=0}$. For complex choices, replace transpose by conjugate transpose; the reality proved in part (i) gives the same conclusion for the Hermitian inner product. Thus different eigenspaces are orthogonal.