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Let $V$ be an $n$-dimensional vector space and let $(v_1, \dots, v_n)$ denote any oriented basis for $V$. Also, let $g$ be an inner product on $V$ and let $G$ denote the Gram matrix of inner products $G = [g(v_i, v_j)]$. I am trying to show that if $v_j = A^k_je_k$, where $e_k$ denotes a basis that is orthonormal with respect to g, then $\det{(A^i_j)} = \sqrt{G}$.

I believe I have found a useful intermediate result, but I'm not really sure how to close the deal. For vectors $v_i$ and $v_j$ we have:

$$ g(v_i, v_j) = g(A^k_i e_k, A^r_j e_r) = A^k_iA^r_j \delta_{kr} = \sum\limits_{m=1}^n A^m_iA^m_j = \langle A_i | A_j\rangle $$

where $A_k$ denotes the $k^{th}$ column of $A$ and $\langle\cdot | \cdot\rangle$ denotes the standard Euclidean inner product. Therefore, the matrix $G$ is given by

$$ G = [\langle A_i|A_j \rangle] $$

At this point, I'm not sure what to do. I'm thinking there's some essential fact I need to know in order to continue.

So, my question is, am I on the right track and if so what should my next step be?

Edit: I updated this question to change the assumption that the $e_i$ are the standard basis vectors to the assumption that the $e_i$ are orthonormal with respect to $g$

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Two nitpicks: First, it's the Gram determinant (after Jørgen Pedersen Gram), second: it is much better to use \langle and \rangle for inner products: compare $\langle x,y \rangle = < x, y >$ (since < and > are relations, this results in awkward spacing). –  t.b. Jun 23 '11 at 21:31
@Theo Buehler Thanks for fixing those things –  ItsNotObvious Jun 23 '11 at 21:40
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1 Answer

up vote 7 down vote accepted

In fact, you're almost done.

You have calculated that $G = A^{T}A$ (since you're working with the standard inner product!). Now $\det{G} = \det{(A^TA)} = \det{(A)}^2$ and taking square roots gives what you want.

The Gram determinant is uniquely characterized by the following properties.

Let $v_{k} : V^{k} = V \times \cdots \times V \to \mathbb{R}_{\geq 0 }$ be a map satisfying

  1. $v_k (a_{1},\ldots, a_{i-1}, \lambda a_{i}, a_{i+1}, \ldots, a_{k}) = |\lambda| v_k(a_{1},\ldots, a_{k})$ for all $(a_{1},\ldots,a_k) \in V^{k}$, all $i$ and all $\lambda \in \mathbb{R}$.
  2. $v_{k}(a_{1}, \ldots, a_{i-1}, a_{i} + a_{j}, a_{i+1}, \ldots, a_k)$ for all $i \neq j$.
  3. $v_{k}(a_{1},\ldots,a_{k}) = 1$ if the $a_{i}$ are orthonormal.

Then $v_{k}(A) = \sqrt{\det{A^{T}A}}$, where $A$ is the $n \times k$ matrix with columns $A = (a_1,\ldots, a_k)$. Clearly, the expression $A \mapsto \sqrt{\det{A^TA}}$ has the desired properties.

Added: On 3Sphere's request, I'm sketching an argument.

First of all, note that a function satisfying the three properties above allows us to perform the following things while keeping track of the value of $v_{k}$:

  • Multiply a column by a scalar.
  • Adding one column to another.

These are the two things one needs to do to perform Gaussian elimination and Gram-Schmidt. If the vectors $a_1, \ldots, a_k$ are linearly dependent, then we can express one column as linear combination of the others using properties 1. and 2. of $v_k$, hence $v_k (a_1,\ldots, a_k) = 0$ (by property 1). Now Gram-Schmidt tells us that there is a way of transforming $(a_1,\ldots,a_k)$ into an orthonormal system spanning the same $k$-dimensional subspace, so $v_k (a_1, \ldots, a_k)$ is determined uniquely by property 3.

Since the expression $v_k(A) = \sqrt{\det{(A^TA)}}$ has the three desired properties, we see that $v_k$ exists and is uniquely determined by these properties.

Of course, $v_k(A)$ is nothing but the $k$-dimensional volume of the parallelepiped spanned by $(a_1,\ldots,a_k)$, and the proof is essentially the same as one of the proofs of the existence and uniqueness of the determinant.

The proof appears this way in the very nice German textbook Analysis 2 by K. Königsberger, but I don't think that there is a translation into other languages.

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@Theo Buehler I guess what I missed (am missing) is that the matrix $G$ I have computed is actually $A^TA$. I'll have to ponder that a bit. –  ItsNotObvious Jun 23 '11 at 22:04
Well, $\langle x | y \rangle = x^T y$, no? Unfortunately I can't elaborate my answer further because the rendering problems make this virtually impossible for me at the moment. By the way, feel free to call me Theo (and @Theo suffices for pings). –  t.b. Jun 23 '11 at 22:09
@Theo No need for elaboration on this point; I see this now. Thanks for your help! –  ItsNotObvious Jun 23 '11 at 22:24
@Theo Umm...well, now that I read over my entire argument again, I realized that I assume that $g(e_i, e_k) = \delta^i_j$ and I'm not sure this is justified; Even though the $e_i$'s are orthonormal $g$ is an abribrary inner product. How can I justify this step? –  ItsNotObvious Jun 23 '11 at 22:37
Wait a minute. I read over this. I assumed that the $e_i$'s are orthonormal with respect to $g$ and that $A$ is given with respect to $e_i$. –  t.b. Jun 23 '11 at 22:39
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