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| ``Re: invariants of the tensor product''
by bwebste on 2003-10-06 03:57:29 |
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| | this is possible, if not all that illuminating. recall that the $k$th coefficient of the characteristic polynomial of A is $(-1)^k {\rm tr}(\wedge^k A)$. Thus, for $A\otimes B$ we get ${\rm tr}(\wedge^n A\otimes B)={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A\otimes B)+\cdots={\rm tr}(\wedge^n A)+{\rm tr}(\wedge^{n-1} A){\rm tr}(B)+\cdots$. |
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