
Orthonormal Basis
Let $\displaystyle B$ be a $\displaystyle nxn$ matrix and $\displaystyle V_{\lambda} =\{v \in \mathbb{C}^n : Bv=\lambda v\}$. Let $\displaystyle \{e_1^\lambda ,e_2^\lambda, ... ,e_{k(\lambda)}^\lambda\}$ be an orthonormal basis for $\displaystyle V_\lambda$ consisting of eigenvectors for $\displaystyle C_\lambda$ , where$\displaystyle C_\lambda$ is a selfadjoint linear transformation on $\displaystyle V_\lambda$.
Prove that $\displaystyle \{e_i^\lambda : \lambda$ is an eigenvalue for $\displaystyle B$ and $\displaystyle 1\le i\le k(\lambda)\}$ is an orthonormal basis for $\displaystyle \mathbb{C}^n$.

Since we are given that these vectors are an orthonormal basis for $\displaystyle V_\lambda$ and we want to prove it is an orthonormal basis for $\displaystyle C^n$, it would appear that we just want to prove that $\displaystyle V_\lambda= C^n$. Self adjoint transformations have several nice properities that you would want to prove first, if you haven't already:
All eigenvalues are real.
If u and v are eigenvectors corresponding to distinct eigenvalues, then u and v are orthogonal.
If $\displaystyle \{v_1,v_2, ..., v_i\}$ are eigenvectors and we restrict the transformation to the orthogonal complement of the span of $\displaystyle \{v_1,v_2, ..., v_i\}$ the restriction is still self adjoint.
That way, we can start with one eigenvector, the restict ourselves to the orthogonal complement of its span, find a second eigenvector, etc. That way you should be able to show that there are, in fact, n independent eigenvectors and so $\displaystyle V_\lambda= C^n$