Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\mathbf{v}_{1},\mathbf{v}_{2},\cdots,\mathbf{v}_{m}$ be $m$ vectors in $n$-dimensional space. Their Gram determinant is defined by:

$$\Gamma=\left|\begin{array}{cccc} \mathbf{v}_{1}^{2} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\ \left(\mathbf{v}_{2}\cdot\mathbf{v}_{1}\right) & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\ \cdots & \cdots & \cdots & \cdots\\ \left(\mathbf{v}_{m}\cdot\mathbf{v}_{1}\right) & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2} \end{array}\right|$$

If $v_{ij}$ is $j$th component of $\mathbf{v}_{i}$, prove that

$$\Gamma=\sum\left|\begin{array}{cccc} v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\ v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\ \cdots & \cdots & \cdots & \cdots\\ v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}} \end{array}\right|^{2}$$

where the summation is extended over all integers $s_{1},s_{2},\cdots,s_{m}$ from 1 to $n$ with $s_{1}<s_{2}<\cdots<s_{m}$.

share|improve this question
    
The answers use the fact that the vectors $v_1,...,v_n$ are ortonormal. What about a case when they are not ortonormal?I suppose that formula is then not true. Is there a similar formula for Gram's determinant using minors and $g_{ij}=<v_i,v_j>$ coefficients? –  L.T Dec 20 '11 at 12:08
    
@L.T: It's not the vectors $\mathbf{v}_i$ that are orthonormal, but rather the (here unnamed) basis vectors. But you are right, the formula assumes that the components are taken w.r.t. an ON basis. –  Hans Lundmark Dec 20 '11 at 12:26
    
Perhaps I should have said it explicitly in the question. To avoid future confussion: I assume that the basis vectors are orthonormal. –  becko Dec 21 '11 at 3:34

2 Answers 2

up vote 6 down vote accepted

Here's how to begin:

Entry $(i,j)$ equals $\mathbf{v}_{i}\cdot\mathbf{v}_{j} = \sum_{s=1}^n v_{is} v_{js}$. Here $s$ is just a dummy variable, so we don't need to call it $s$ in each entry; let's choose to call it $s_j$ in each of the sums in column number $j$. Then, since the determinant is linear in each column separately, we can pull the sums and the factors $v_{j s_{j}}$ outside. Like this for the first column: $$ \Gamma=\left|\begin{array}{cccc} \sum_{s_1} v_{1 s_{1}} v_{1 s_{1}} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\ \sum_{s_1} v_{2 s_{1}} v_{1 s_{1}} & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\ \cdots & \cdots & \cdots & \cdots\\ \sum_{s_1} v_{m s_{1}} v_{1 s_{1}} & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2} \end{array}\right| = \sum_{s_1} v_{1 s_{1}} \left|\begin{array}{cccc} v_{1 s_{1}} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\ v_{2 s_{1}} & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\ \cdots & \cdots & \cdots & \cdots\\ v_{m s_{1}} & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2} \end{array}\right|. $$ And doing the same for each column: $$ \Gamma = \sum_{s_1} \sum_{s_2} \dots \sum_{s_m} v_{1 s_{1}} v_{2 s_{2}} \dots v_{m s_{m}} \left|\begin{array}{cccc} v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\ v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\ \cdots & \cdots & \cdots & \cdots\\ v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}} \end{array}\right|. $$ If two indices $s_j$ coincide, this last determinant has two equal columns and therefore vanishes, so we need only sum over index sets $(s_1,\dots,s_n)$ with all numbers distinct. Can you see how to continue from here?

share|improve this answer
    
Yes, I get it. Thanks. –  becko Dec 20 '11 at 11:47

The most elegant solution is probably to extend the given scalar product on your $m$-dimensional vector space $V$ to the exterior product $\Lambda ^m(V)$.
The recipe is that any orthonormal basis $e_1,...,e_n$ of $V$ yields an orthonormal basis $e_{s_1 }\wedge ...\wedge e_{s_m }$ $(s_{1}<s_{2}<\cdots<s_{m})$ of $\Lambda ^m(V) $ .

As a consequence, if you write $v_i=\Sigma v_{ij}e_j$ you get $$v_1\wedge...\wedge v_m=\Sigma c_{s_1...s_m} e_{s_1 }\wedge ...\wedge e_{s_m } \quad (+)$$ with

$$c_{s_1...s_m}= \left|\begin{array}{cccc} v_{1s_{1}} & v_{1s_{2}} & \cdots & v_{1s_{m}}\\ v_{2s_{1}} & v_{2s_{2}} & \cdots & v_{2s_{m}}\\ \cdots & \cdots & \cdots & \cdots\\ v_{ms_{1}} & v_{ms_{2}} & \cdots & v_{ms_{m}} \end{array}\right| \quad \quad (*)$$

Then taking the square of the lengths of both sides of this equality $(+) $ you obtain
$$||v_1\wedge...\wedge v_m||^2=\Sigma |c_{s_1...s_m}|^2\quad (**)$$

Now remember that in a $m$- dimensional vector space spanned by the vectors $v_1,..., v_m$ (assumed independent: everything is trivial if these vectors are dependent ) we have the equality for squared length $$||v_1\wedge...\wedge v_m||^2=\Gamma =\left|\begin{array}{cccc} \mathbf{v}_{1}^{2} & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{2}\right) & \cdots & \left(\mathbf{v}_{1}\cdot\mathbf{v}_{m}\right)\\ \left(\mathbf{v}_{2}\cdot\mathbf{v}_{1}\right) & \mathbf{v}_{2}^{2} & \cdots & \left(\mathbf{v}_{2}\cdot\mathbf{v}_{m}\right)\\ \cdots & \cdots & \cdots & \cdots\\ \left(\mathbf{v}_{m}\cdot\mathbf{v}_{1}\right) & \left(\mathbf{v}_{m}\cdot\mathbf{v}_{2}\right) & \cdots & \mathbf{v}_{m}^{2} \end{array}\right|\quad \quad (***)$$ ( determinant of Gram matrix).

Replacing both sides of $(**)$ by their values given in $(*)$ and $(***)$ proves the required formula.

Edit
1) I forgot to mention that formula $(*)$ can be thought of as a generalization of Pythagoras' theorem, with $||v_1\wedge...\wedge v_m||^2$ playing the role of the squared length of the hypotenuse $v_1\wedge...\wedge v_m$.

2) I feel that the euclidean structure inherited by $\Lambda ^mV \;\;$ from a euclidean structure on $V$ is not as well-known as it deserves.
One of the rare treatments of this theme in the textbook literature is given in MacLane-Birkhoff's Algebra (cf. especially Theorem 18, page 557).

share|improve this answer
    
What happens when $v_1,...,v_m$ are not ortonormal? –  L.T Dec 20 '11 at 12:09
    
Dear L.T., the $v_i$'s are not supposed orthonormal anywhere in the above. –  Georges Elencwajg Dec 20 '11 at 12:28
    
Sorry, obviously you have right. I read not carefully. I was interested to a question similar to this one given by becko but in more general situation. Assume that we have arbitrary (not necessary ON) basis $e_1,...,e_n$ and we have coordinates $v_{ij}$ of vectors $v_i$, $i=1,...,m$. Is it a formula on Gram determinants of vectors $v_1,...,v_m$ using minors of matrice $[v_{ij}]$ and coefficients $g_{ij}:=<e_i,e_j>?$ ? –  L.T Dec 20 '11 at 14:02
1  
@L.T. You should use $(+)$ and bilinearity to calculate $(v_1\wedge...\wedge v_m|v_1\wedge...\wedge v_m)$. We know that $(e_{s_1 }\wedge ...\wedge e_{s_m }|e_{t_1 }\wedge ...\wedge e_{t_m })$ is the determinant of $(g_{s_i t_j})$. The calculation then reduces to calculating the $ c_{s_1...s_m}$ ' s . I don't know the formula but it has no longer anything to do with exterior algebra. The problem is reduced to : in a euclidean space , calculate the coefficients of a vector in a fixed basis , given the mutual scalar products of all pairs of basis vectors. –  Georges Elencwajg Dec 20 '11 at 18:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.