Simultaneous diagonalization of self adjoint matrices If U and T are self adjoint linear operators on finite dimensional real I nner product space which commute with each other then there is an orthonormal basis for V consisting of vectors that are eigenvectors of both U and T.
Hi!
Please help me with this problem.
It is problem no 14 , chapter 6.4 in Linear Algebra Book by Friedberg, Insel and Spence.
Hint that is given is to use the an eigenspace of T is U invariant as well. But I don't know how to proceed further to get a basis consisting of eigenvectors of both U and T.
Does T and U have same set of eigenvectors? I am really not getting what is going on.
 A: If $U$ and $T$ are commuting linear operators, they will preserve each other's eigenspaces. The proof is short: If $Uv = \lambda v$, then $U(Tv) = T(Uv) = T( \lambda v) = \lambda(Tv)$. Since in your case $U$ and $T$ are self-adjoint, the operators will be diagonalisable, with real eigenvalues.
So, first take the eigenspace $V_\lambda$ of $U$ corresponding to the eigenvalue $\lambda$. Since $V_\lambda$ is stable under $T$, the restriction  $T|_{V_\lambda}$ is a self-adjoint operator on $V_\lambda$. Now, by the spectral theorem $V_\lambda$ has an orthonormal basis of eigenvectors $T|_{V_\lambda}$, say $(v_1, \ldots, v_r)$ corresponding to eigenvalues $\mu_1, \ldots, \mu_r$ respectively. Going back up to the whole space $V$, we have produced the first part of a simultaneous diagonalisation: $Uv_i = \lambda v_{i}$ and $T v_{i} = \mu_i v_i$ for each $i$.
Now, repeat this process for every different eigenspace of $U$, and collect together each group of vectors you find. Finally, use the fact that $U$ is self-adjoint (which we haven't really used yet) to conclude that this collection is mutually orthogonal, and spans $V$.
