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In the Euclidean space $\mathbb{R}^n$ consider two (ordered) sets of vectors $a_1 \ldots a_k$ and $b_1 \ldots b_k$ with $k \le n$.

Question

  1. What is the geometrical interpretation of $\det(a_i \cdot b_j)?$
  2. Is it true that $\det(a_i\cdot b_j)=\det(a'_p\cdot b'_q)$ if $a_1\wedge \ldots \wedge a_k=a'_1\wedge \ldots \wedge a'_k$ and $b_1\wedge \ldots \wedge b_k=b'_1\wedge \ldots \wedge b'_k?$

Since $\det(a_i\cdot a_j)$ equals the squared $k$-volume spanned by $a_1\ldots a_k$, I guess that $\det(a_i\cdot b_j)$ may be interpreted as the $k$-volume spanned by some kind of projection of $b_1\wedge \ldots \wedge b_k$ (thought of as an oriented $k$-parallelogram) onto $a_1\wedge \ldots \wedge a_k$. For the same reason I would answer affirmatively to the second question.

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By $\det(a_i \cdot b_j)$, do you mean the determinant of the matrix $(a_i \cdot b_j)_{ij}$? Isn't this matrix the same as $AB$, where $A = (a_1, \dotsc, a_k)$ and $B = (b_1, \dotsc, b_k)^t$? –  André Caldas Jun 1 '12 at 0:51
    
@AndréCaldas: Yes to both questions. –  Giuseppe Negro Jun 1 '12 at 1:06
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+1: A very interesting question on a rather neglected subject: euclidean structures on exterior products. –  Georges Elencwajg Jun 1 '12 at 9:54

1 Answer 1

up vote 1 down vote accepted
  1. Given a Euclidean structure on $\mathbb R^n$, you deduce a canonical euclidean structure on each $\Lambda ^k\mathbb R^n$.
    On pairs of decomposable elements $a_1\wedge \ldots \wedge a_k, b_1\wedge \ldots \wedge b_k\in\Lambda ^k\mathbb R^n$ it is given by the formula $$ (a_1\wedge \ldots \wedge a_k\mid b_1\wedge \ldots \wedge b_k)= \det(a_i\cdot b_j) $$
    So your determinant is the scalar product of two vectors, but in a new vector space.
    In particular you have the pleasant interpretation that the volume of the parallelipiped spanned by the vectors $a_1, \ldots ,a_k\in V$ is the length of the vector $a_1\wedge \ldots \wedge a_k\in \Lambda ^k\mathbb R^n$.

  2. The above interpretation makes it now obvious that given $$\omega =a_1\wedge \ldots \wedge a_k, \:\omega '=a'_1\wedge \ldots \wedge a'_k, \eta=b_1\wedge \ldots \wedge b_k, \eta'=b'_1\wedge \ldots \wedge b'_k$$ the equalities $\omega=\omega'$ and $\eta=\eta'$ imply that $$(\omega\mid\eta)=\det(a_i\cdot b_j)=(\omega'\mid\eta')=\det(a'_p\cdot b'_q)$$

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Very nice, thank you. I'm linking this to another closely related question: math.stackexchange.com/questions/140180/… –  Giuseppe Negro Jun 1 '12 at 18:44
    
This is another related answer, more algebraic in nature, that I found useful: math.stackexchange.com/a/44212/8157 –  Giuseppe Negro Jun 2 '12 at 9:18

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