## Monday, 31 March 2008

### Schur's Lemma (2)

Here's the follow up post to Schur's Lemma, in case you were anxiously awaiting it.

The corollary says that if α=β, f has to be a multiple of the identity function (or unit matrix).

$\forall R \in G: f \circ \alpha(R) = \alpha(R) \circ f\Rightarrow \exists c \in \mathbb{C}: f = c \times id_V$

For the proof one looks at an eigenvalue λi. And subtracts the following on both sides:

$f \circ \alpha(R) - \alpha(R)\circ \lambda_i id_V = \alpha(R) \circ f - \alpha(R)\circ \lambda_i id_V$

This changes to:

$(f - \lambda_i \times id_V) \circ \alpha(R) = \alpha(R) \circ (f - \lambda_i \times id_V)$

Now we apply the Schur's Lemma from last time and find out that $f - \lambda_i \times id_V$ has to be either invertible or the 0 function. But it cannot be invertible because λi is an eigenvalue (actually the only eigenvalue). And we have what we wanted.