User:Mathsci/scratch

Let V be a real finite-dimensional vector space and S an involution on V. Let T be an endomorphism of V such that ST = − TS and ker T2 = ker T. Let V± be the ±1 eigenspaces of S, so that V = V+ ⊕ V−. Then


 * $$\displaystyle{\mathrm{dim}\, V_+ - \mathrm{dim}\,\mathrm{ker}\, T|_{V_+} = \mathrm{dim}\, V_- - \mathrm{dim}\,\mathrm{ker}\, T|_{V_-}.}$$

The assumptions on T imply that TV ⊕ ker T = V. Since S anticommutes with T this gives


 * $$\displaystyle{V_\pm = TV_\mp \oplus \mathrm{ker}\, T|_{V_\pm}.}$$

So the dimension of TV+ is the same as the dimension of TV-.

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