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Let $V$ be a finite dimensional vector space over $\mathbb{R}$ of dimension $n$.

The vector space $\underbrace{V \oplus V \oplus \cdots \oplus V}_k$ has dimension $kn$, and the vector space $\underbrace{V \otimes V \otimes \cdots \otimes V}_k$ has dimension $n^k$.

Is there a similar construction (that does not depend on a basis of $V$), that gives a vector space of dimension $2^n$?

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    Is there any particular reason (apart from curiousity) why you ask this question? (as far as I know the answer to your question is no)2011-03-25
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    Similar construction to what? The one for $k=2$ where your vectors are $(\vec{a},\vec{b})$, addition is componentwise and scalar multiplication is $c(\vec{a},\vec{b})=(c\vec{a},c\vec{b})$?2011-03-25

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You might be looking for the exterior algebra of $V$. Its definition does not depend on choosing a basis, and its dimension is $2^n$.

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    Which is a special case of the [Clifford algebra](http://en.wikipedia.org/wiki/Clifford_algebra) .2011-03-25
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    This is exactly what I am looking for. I can't believe I forgot about it in the first place.2011-03-25
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    @Jonas: Doesn't the exterior algebra of an $n$-dimensional space have dimension $\binom{n}{2}$?2011-03-25
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    the full algebra has dimension $\sum{n\choose k}=2^n$2011-03-25
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    @Arturo: What yoyo said; $\binom{n}{2}$ is the dimension of the exterior square.2011-03-25
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    @Jonas, yoyo: Damn; I was halfway through writing when I realized what my mistake was (I was thinking "exterior square", not "exterior algebra"; and of course, since the $k$-th exterior product has dimension $\binom{n}{k}$, once you add them all together you get $2^n$), and I *thought* I had canceled the comment. Apparently, I posted it by mistake. Sorry!2011-03-25