I'm reaching out to you because of theorem 1.17 in the textbook "Multivariable calculus with applications". The theorem states the following:
Theorem Every positively oriented ordered list of n linearly independent vectors $\vec{V_{1}},...,\vec{V_{n}}\in \mathbb{R}^n$ can be deformed into a list of unit vectors $\vec{E_1},\dots,\vec{E_n}$.
A proof outframe is given but it is kind of vague. So the problem I'm having is with the statement that there is a rotation that takes the vector $V_n$ into $pE_n$. yes that may be true for some matrix $M_{n\times m}$ but later on, it says that the mentioned rotation "carries the vectors $V_j$ in the vectors $W_j,j<n$" which I assume is true because $V_{n-k}M_{n-k\times n-k}=pE_{n-k}=W_{n-k}$, however it later says that we "shrink" the n-th component of $W_j$ to zero by adding $W_n=pE_n$, which is something of the form $W_j(0,\dots, p, \dots, 0)+W_n(0, \dots, p)=W'(0,\dots, p, \dots, p)$ I suppose? Then it says that the resulting vectors in the hyperplane $x_n=0$ are linearly independent which is true, yes but I don't see the relevance of shrinking and doing everything else that I've mentioned.
You can find the whole thing here page 41, Theorem 1.17.
(the author) after careful consideration, I've reached several conclusions. The first one is the following:
We choose one vector namely the $\vec{V_n}$ which we multiply by the rotation matrix in such a way to reach the form $\vec{V_n}M_{n\times n}=p\vec{E_n}$, we choose the rotation matrix in such a way as to bring the vector back to the basis of $\mathbb{R}^{n}$ for example:
The vector $\vec{V_2}(1,1)$ makes a 45 degrees angle with the x-axis, which means that it will take him 45 degrees to get back to the y-axis and be normalized. Essentially we have: \begin{equation} \begin{pmatrix} 1\\ 1 \end{pmatrix} \begin{pmatrix} \cos(-\frac{\pi}{4}) & \sin(-\frac{\pi}{4}) \\ -\sin(-\frac{\pi}{4}) & \cos(-\frac{\pi}{4}) \end{pmatrix} = \begin{pmatrix} 1\\ 1 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{2}}{2} &-\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} &\frac{\sqrt{2}}{2} \end{pmatrix}= \begin{pmatrix} \frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2} \end{pmatrix}= \begin{pmatrix} 0 \\ \sqrt{2} \end{pmatrix} \end{equation} The resulted vector is of the form $\sqrt{2}E_2$ which is what we were aiming for. From here we can write $V_1$ as $V_1(a,b)$, to obtain the vector mentioned in the theorem (namely $W_1$) we multiply it by the same rotational matrix: \begin{equation} \begin{pmatrix} a\\ b \end{pmatrix} \begin{pmatrix} \cos(-\frac{\pi}{4}) & \sin(-\frac{\pi}{4}) \\ -\sin(-\frac{\pi}{4}) & \cos(-\frac{\pi}{4}) \end{pmatrix}= \begin{pmatrix} \frac{\sqrt{2}}{2}(a -b) \\ \frac{\sqrt{2}}{2}(a +b) \end{pmatrix} \end{equation} We now create the determinant: \begin{equation} \begin{vmatrix} \frac{\sqrt{2}}{2}(a-b)&\frac{\sqrt{2}}{2}(a +b)\\ 0 & \sqrt{2} \end{vmatrix} =(-1)^{2+2}\sqrt{2}\left|\frac{\sqrt{2}}{2}(a -b)\right| \end{equation}
As far as I understand the determinant $\left|\frac{\sqrt{2}}{2}(a -b)\right|$ should be positive due to the positive orientation of the vectors taken.
The same is with the theorem. We don't aim for all vectors but we are deforming only the n-th one ($V_n$) which bring us the following determinant: \begin{equation} \begin{vmatrix} a_{11} & a_{12} &\dots &a_{1n}\\ \dots &\dots & \dots&\dots\\ 0 & 0 &\dots &p\\ \end{vmatrix} \end{equation} When we multiply the n-th row by $-\frac{a_{jn}}{p}$ and add it to the j-th row we get the n-th column of the j-th row to be zero, so we obtain something of the form: \begin{equation} \begin{vmatrix} a_{11} & a_{12} &\dots &0\\ \dots &\dots & \dots&0\\ 0 & 0 &\dots &p\\ \end{vmatrix} \end{equation} Factor out p so to get: \begin{equation} (-1)^{n+n}p\begin{vmatrix} a_{11} & a_{12} &\dots &a_{1,n-1}\\ \dots &\dots & \dots&\dots\\ a_{n-1,1} & a_{n-1,2} &\dots &a_{n-1,n-1}\\ \end{vmatrix} \end{equation} And you do this inductively for all the rest of the $n-1$ cases to obtain $det(Z)>0$. The final question is to determine whether $n+n$ is odd or even. If n-even then $2k+2+2k+2=2(k+k)+4$ which is even, if odd $2k+1+2k+1=2(k+k)+2$ which is even as well. Either way our determinant has a positive sign.