determinant of changing of basis matrix

Question

Let $B=\{f_{1},…,f_{n},g_{1},…g_{n}\}$ the order orthonormal basis of $2n$-dimensional and $B′=\{f_{1}+g_{1},f_{1}−g_{1},…,f_{n}+g_{n},f_{n}−g_{n}\}$ the other order basis. How can I compute determinant of changing of basis matrix from $B$ to $B′$? It suffecies knowing determinant is positive or negetive.

Answer 1

Let $f_1+g_1 = a_{11}f_1+...+a_{1n}f_n+b_{11}g_1+...+b_{1n}g_n$

$f_1-g_1 = a_{21}f_1+...+a_{2n}f_n+b_{21}g_1+...+b_{2n}g_n$

From this we get

$2f_1 = (a_{11}+a_{21})f_1+...+(a_{1n}+a_{2n})f_n+(b_{11}+b_{21})g_1+...+(b_{1n}+b_{2n})g_n$

Similarly

$2g_1 = (a_{11}-a_{21})f_1+...+(a_{1n}-a_{2n})f_n+(b_{11}-b_{21})g_1+...+(b_{1n}-b_{2n})g_n$

Now as above vectors are all basis vectors we get

$(a_{11}+a_{21}) = 2,(a_{12}+a_{22}) = 0,..,(a_{1n}+a_{2n}) = 0,(b_{11}+b_{21}) = 0,..,(b_{1n}+b_{2n}) = 0$

and

$(a_{11}-a_{21}) = 0,(a_{12}-a_{22}) = 0,..,(a_{1n}-a_{2n}) = 0,(b_{11}-b_{21}) = 2,..,(b_{1n}-b_{2n}) = 0$

When we solve the last two sets of equation we get

$a_{11}=a_{21} = 1,a_{12}=a_{22} = 0,...b_{11}=1,b_{21}=-1,....$

The above solution is first two row vectors, if we solve all the pairs of row vectors we will get a matrix, which will look like for n = 1

$\begin{bmatrix} 1& 1\\1 & -1\end{bmatrix}$ for n = 2; $\begin{bmatrix} 1& 0&1 &0\\1& 0&-1 &0 \\ 0& 1&0 &1 \\ 0& 1&0 &-1\end{bmatrix}$ which we notice that can be written as $2\begin{bmatrix} 1& 0 &0\\0& 1&1 \\ 0& 1&-1\end{bmatrix} = 2*-2 = -4$

For any n the determinant will be $(-1)^{\lfloor \dfrac{n+1}{2} \rfloor}.2^n$. (for n = 3, the determinant is 8)