# Formula for the Ratio of Gram Determinants

Let $(v_1,v_2,...)$ be linearly independent vectors in an inner product space $(V, \langle\cdot,\cdot\rangle)$.

Let $(e_1, e_2,...)$ be the gram schmidt orthonormalized version of $(v_1,v_2,...)$, so $e_1=\frac{v_1}{||v_1||}$, etc.

Let $G_n$ be the $n \times n$ Gram Matrix, so $(G_n)_{i,j}=\langle v_i,v_j\rangle$.

I want to show that $\frac{det(G_{n+1})}{det(G_{n})}=||v_{n+1}-\sum_{i=1}^{n}\langle v_{n+1},e_i\rangle e_i||^2$

I'm trying to use induction on The $n=1$ case is pretty easy to prove. The inductive hypothesis is that $\exists n \in \mathbb{N}$ such that

$\frac{det(G_{n+1})}{det(G_{n})}=||v_{n+1}-\sum_{i=1}^{n}\langle v_{n+1},e_i\rangle e_i||^2$, and from this I need to show that $$\frac{det(G_{n+2})}{det(G_{n+1})}=||v_{n+2}-\sum_{i=1}^{n+1}\langle v_{n+2},e_i\rangle e_i||^2$$

This is where I get stuck. Expanding the determinant seems to be way too messy and I wasn't able to get anywhere for it, even the $n=2$ case was hard to manage. I did manage to show that for $j<i$ we have

$$\langle v_i, v_j\rangle = \langle v_i, e_j\rangle||v_j - \sum_{k=1}^{j-1}\langle v_j,e_k\rangle e_k|| + \sum_{k=1}^{i-1}\langle v_i, e_k \rangle \langle e_k, v_i \rangle$$ And also that $$\langle v_i, v_i\rangle = ||v_i - \sum_{k=1}^{i-1}\langle v_i,e_k\rangle e_k||^2 + \sum_{k=1}^{i-1}\langle v_i, e_k \rangle \langle e_k, v_i \rangle$$

Finally after messing around a lot I think that $$det(G_n)=\prod_{i=1}^{n}(||v_i||^2-\sum_{j=1}^{i-1}\langle v_i, e_j \rangle \langle e_j, v_i \rangle)$$

If this is true, then the result I'm after becomes trivial, but I also haven't managed to get very far on proving this identity either.

Any help on this would be much appreciated, it doesn't seem like it should be this complicated! Also I can provide more details if needed. Thanks!

EDIT: I found a similar question here: Special Gram's inequality Which makes me think that this has something to do with the generalized distance, and suggests there might be a less brute force method for my problem.

## 1 Answer

HINT: You can check by induction that for each $m$, the span of the vectors $(v_1, v_2, \ldots, v_m)$ equals to the span of the vectors $(e_1, e_2, \ldots, e_m)$.

Since the vector $v'_{n+1} =v_{n+1}-\sum_{i=1}^{n}\langle v_{n+1},e_i\rangle e_i$ is perpendicular to all of the vectors $e_1, \ldots, e_n$, it is perpendicular to all of the vectors $v_1, \ldots, v_n$ .

Consider the Gram determinant $G(v_1,v_2, \ldots, v_n, v'_{n+1})$. It is easy to see that it equals the Gram determinant $G(v_1, \ldots, v_n, v_{n+1})= G_{n+1}$, since we have an equation of form $$v'_{n+1} = v_{n+1} + \sum_{m=1}^n a_m v_m$$

But using the orthogonality of $v'_{n+1}$ to $v_1$, $\ldots$, $v_n$ we get $$G(v_1,v_2, \ldots, v_n, v'_{n+1}) = G_n \cdot \|v'_{n+1}\|^2$$

• Thanks for your help! I understand much better now, but I still can't see why $G(v_1,v_2,...,v_n,v'_{n+1})=G(v_1,v_2,...,v_n,v_{n+1})$. If you could explain why this id I'd be very grateful. – J.doe Oct 7 '17 at 3:21
• Sorry, I think I figured it out! You use the linearity of the determinant and reduce $G(v_1,v_2,...,v'_{n+1})$ to $G(v_1,v_2,...,v_{n+1})$ and a sum of other determinants, but because $v'_{n+1}$ is $v_{n+1}$ plus a linear combination of $v_m$ for $m \leq n$, each determinant has a duplicate row or column and so is zero! Thank you so much for all your help! – J.doe Oct 7 '17 at 5:25
• @J.doe: I am glad you got it on your own. It would be fun to see how the gram determinant changes when we modify the system by a matrix. It is in fact more convenient to consider $G(v_1, \ldots v_n; w_1, \ldots w_n)$. Now the dependence is bilinear. You can write it formally as $(\langle v_1, \ldots , \langle v_n)^t \cdot (w_1 \rangle, \ldots , w_n\rangle)$. What if we change $v_i$ by a matrix $A$ and $w_i$ by a matrix $B$. It streamlines some things. There are more fun things involving Gram dets. – orangeskid Oct 7 '17 at 11:44