4
$\begingroup$

I'm trying to prove the following:

If $S\colon V\to V$ and $T\colon V\to V$ are unitary linear transformations on unitary space $V$ ($\dim V=n$, $n$ is finite), such that $ST=TS$, then they have a joint eigenvector basis (aka there is a basis of $V$ composed of eigenvectors of both $S$ and $T$ - not necessarily of the same eigenvalue per each).

Can anyone help me out? I've tried rephrasing the 'matrix equivalent' of the theorem, but I didn't get much further.

Thanks!

  • 0
    BTW there is a really beautiful abstract proof of a somewhat more general statement here: http://planetmath.org/encyclopedia/CommutingMatrices.html2011-08-13

2 Answers 2

4

Based on what you've already covered in class, there is an orthonormal basis with respect to which $S$ has matrix

$A= \begin{pmatrix} \lambda_1 I_{k_1} & 0 & \cdots & 0 \\ 0 & \lambda_2 I_{k_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_m I_{k_m} \end{pmatrix},$

where $k_i$ is the dimension of the eigenspace for the eigenvalue $\lambda_i$ of $S$. If $B$ is the matrix of $T$ with respect to this basis, then because $AB=BA$ you have

$B= \begin{pmatrix} B_{1} & 0 & \cdots & 0 \\ 0 & B_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & B_{m} \end{pmatrix},$

where $B_i$ is a $k_i$-by-$k_i$ matrix (e.g., see here). Since each $B_i$ is unitary, each can be unitarily diagonalized. Note that doing so leaves $A$ unchanged.

  • 0
    Thank you very much! I'll 'delve into this' tomorrow.2011-08-12
2

Jonas' answer was excellent and helped me a lot, but today I thought of a different direction and it'd be nice if you fellows could help me tell whether it works:

Let $V_1, V_2, ..., V_k$ be the eigenspaces pertaining to eigenvalues $\lambda_1,\lambda_2,...\lambda_k$ of S. Since S is T-invariant, we know that $T(V_i)\subseteq V_i$, which means the reduction of T to the eigenspace $V_i$, $T_i:V_i\to V_i$, is also a unitary transform. Subsequently $T_i$ has an orthonormal eigenvector basis in $V_i$. Because S is unitary, the bases found for the $T_i$s contain vectors orthonormal to each other, and so their union would be an orthonormal eigenvector basis of T, which, because S is unitary, would also be an eigenvector basis of S.

Is this proof valid-looking? Thanks!

  • 0
    @iroiroaru: This is the sort of proof I was trying to point you towards in my earlier hint, and is correct: succinctly, $T$ has an orthonormal basis of eigenvectors on each eigenspace of $S$. Put all these together, and you get an orthonormal basis which consists of vectors which are simultaneously eigenvectors for $T$ and for $S$.2011-08-13