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Let $V$ be vector space, $\dim V=N$. Define the multiplication operator $L_{\mathbf{b}}$ as $L_{\mathbf{b}}:\omega\to \mathbf{b}\wedge\omega$, where $\omega\in\wedge V$ ($\wedge V$ is the entire exterior algebra) and $\mathbf{b}\in V$. I want to compute the trace of $L_{\mathbf{b}}$.

If $N=1,2$ choosing a basis in $V$ we can obtain a basis of $\wedge V$ and with some calculations we obtain $tr(L_{\mathbf{b}})=0$. However this method will be very complicated for larger value of $N$ because $\dim(\wedge V)=2^N$. So I search for a coordinate-free (basis-free) calculations applying the following definition of trace.

If $A=\sum_{k=1}^N \mathbf{v}_k\otimes\mathbf{f}_k^{*}\in V\otimes V^{*}$ then $tr(A)=\sum_{k=1}^N \mathbf{f}_k^{*}(\mathbf{v}_k)$.

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up vote 5 down vote accepted

You don't need to choose any particular basis; all you need is the fact that $\wedge V$ is graded and $L_b$ raises the degree by $1$, so if you choose any basis that respects the gradation, then $L_b$ sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish.

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Thanks. So not only the trace is zero, but every element in diagonal is zero. –  vesszabo Aug 8 '12 at 9:13
    
@vesszabo: In such a basis, yes. Of course you can transform to some weird basis that doesn't respect the gradation and get non-zero diagonal elements. –  joriki Aug 8 '12 at 9:14

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