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Line 88:
G^TG = I
</math>
such that <math>\operatorname{rank}(G) = k</math>. In addition, there are identity
:<math>
F\cdot \mathbb{I} = \mathbb{I}
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No edit summary |
No edit summary |
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Line 88:
G^TG = I
</math>
such that <math>\operatorname{rank}(G) = k</math>. In addition, there are identity
:<math>
F\cdot \mathbb{I} = \mathbb{I}
|