Determinant in calculation of orientation in Riemannian Geometry

Question

Assuming not all of $y_1,..., y_n$ are zero. When $i\ne j$, define $$ c_{ij}= \frac{2y_iy_j}{\left(\sum_{k=1}^n y_k^2\right)^2} $$ and $$ c_{ii}=\frac{-\left(\sum_{k=1}^n y_k^2\right)+2y_i^2}{\left(\sum_{k=1}^n y_k^2\right)^2}. $$ How to show that $$ \det(c_{ij})>0 $$ when $n$ is odd and $$ \det(c_{ij})<0 $$ when $n$ is even?

Answer 1

First use the fact that for any scalar $\alpha$ and $n\times n$ matrix $A$ we have $\det(\alpha A) = \alpha^n \det(A)$ to note that the sign of $\det(c_{ij})$ is the same as the sign of $\det(b_{ij})$ where

$$ b_{ij}= 2y_iy_j $$

for $i \ne j$ and

$$ b_{ii}=2y_i^2-\sum_{k=1}^n y_k^2. $$

Next, define $d = \sum_{k=1}^n y_k^2 = y^Ty \gt 0$ and note that

$$ \det(b_{ij}) = \det(2yy^T - d I) = (-d)^n \det\left(I - \frac{2yy^T}{d}\right). $$

Now, by Weinstein–Aronszajn identity we have

$$ \det(b_{ij}) = (-d)^n \left(1 - \frac{2y^Ty}{d}\right) = (-1)^{n+1} d^n. $$

Therefore,

$$ \det(b_{ij})>0 $$ when $n$ is odd and $$ \det(b_{ij})<0 $$ when $n$ is even. Conclusion follows from the fact that $\det(c_{ij})$ has the same sign as $\det(b_{ij})$.