# Tagged Questions

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### Exterior algebra of a direct sum [duplicate]

Why the exterior algebra of a direct sum of subspaces is isomorphic to a Tensorial product of it's exterior algebras, I mean, Why ⋀(V⊕W)≃⋀V⊗⋀W ?
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### Let $K$ a field, $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor. Prove that $f \wedge f =0$

Let $K$ a field, with $\operatorname{char}(K)=0$. Let $V$ a vector space over $K$, $\dim(V) \geq 1$, and be $f$ a $n$-tensor ($f \in {\mathcal T}_n(V):=\Lambda^{n}(V)$), i.e., $f$ is an multilinear ...
1answer
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### Trace of the multiplication operator

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 ...
1answer
334 views

### Laplace expansion

This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix ...
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### Exterior Algebra of Self-Direct Sum

Suppose $V$ and $W$ are vector spaces and $\bigwedge V$ and $\bigwedge W$ their exterior algebras. Then it is known that $\bigwedge (V \oplus W) \simeq \bigwedge V \otimes \bigwedge W$. Now my ...
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### Basis of the symmetric algebra $S(M)$ given $R$-module basis of $M$ using the diamond lemma?

Over the past week, I read this secret blogging seminar post concerning the diamond lemma, which got me to reading about Bergman's paper on the diamond lemma. Now suppose you have a free $A$-module ...
2answers
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### Why is it that $\det(\phi-x\text{id})=\sum_{i=0}^n (-1)^ic_ix^i$?

I'm trying to understand a certain formula for the determinant in a more general setting. Say you have a free module $M$ of rank $n$ over a (commutative) ring $R$. Let ...
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### Alternating forms tangential to a subspace.

Let $V$ be a finite-dimensional vector space with euclidean product, and let $U$ be a subspace. Now let $P$ be the projection of $V$ onto $U$, and let $\omega$ be any alternating multilinear $k$-form. ...
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### Covectors $\omega^1, …, \omega^k$ are linearly dependent iff their wedge product is zero

How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?
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205 views