# Orthonormal Basis

• Jan 25th 2011, 01:03 AM
problem
Orthonormal Basis
Let $B$ be a $nxn$ matrix and $V_{\lambda} =\{v \in \mathbb{C}^n : Bv=\lambda v\}$. Let $\{e_1^\lambda ,e_2^\lambda, ... ,e_{k(\lambda)}^\lambda\}$ be an orthonormal basis for $V_\lambda$ consisting of eigenvectors for $C_\lambda$ , where $C_\lambda$ is a self-adjoint linear transformation on $V_\lambda$.
Prove that $\{e_i^\lambda : \lambda$ is an eigenvalue for $B$ and $1\le i\le k(\lambda)\}$ is an orthonormal basis for $\mathbb{C}^n$.
• Jan 25th 2011, 06:00 AM
HallsofIvy
Since we are given that these vectors are an orthonormal basis for $V_\lambda$ and we want to prove it is an orthonormal basis for $C^n$, it would appear that we just want to prove that $V_\lambda= C^n$. 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 $\{v_1,v_2, ..., v_i\}$ are eigenvectors and we restrict the transformation to the orthogonal complement of the span of $\{v_1,v_2, ..., v_i\}$ 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 $V_\lambda= C^n$