What shown belove is a reference from the text Analysis on Manifolds by James Munkres.
So with this notions I ask to prove the following result. If $\vec v_1,..,\vec v_k$ are $k\le n$ linearly independent vectors of $\Bbb R^n$ and so if $\vec v_{k+1}\cdots,v_n$ are mutually orthonormal vectors of $\langle\vec v_1,\cdots,\vec v_k\rangle^\bot$ such that the base $\{\vec v_1,\cdots,\vec v_k,\vec v_{k+1},\cdots,\vec v_n\}$ is right-handed then
$$
\text{vol}(\vec v_1,..,\vec v_k)=\det\begin{bmatrix}v_{1,1}&&\cdots&&v_{1,n}\\\vdots&&\ddots&&\vdots\\v_{k,1}&&\cdots&&v_{k,n}\\v_{k+1,1}&&\cdots&&v_{k+1,n}\\\vdots&&\ddots&&\vdots\\v_{n,1}&&\cdots&&v_{n,n}\end{bmatrix}
$$
where the elements of first $i$-th row are the coordinates of $\vec v_i$ for $i=1,…,n$ with respect a orthonormal right-handed frame $\hat e_1,\cdots,\hat e_n$. In particular I tried to prove the statement using the second point of the theorem $21.3$: indeed it seem that if I define an orthogonal transformation $f$ such that
$$
f(\vec v_i)=\hat e_i
$$
for any $i=(k+1),\cdots,n$ then the statement follows rememberig that any orthogonal matrix has unit determinant.
Best Answer
So putting $$ X:=\begin{bmatrix} v_{1,1}&&\cdots&&v_{1,n}\\ \vdots&&\ddots&&\vdots \\v_{k,1}&&\cdots&&v_{k,n}\\ v_{k+1,1}&&\cdots&&v_{k+1,n}\\ \vdots&&\ddots&&\vdots\\ v_{n,1}&&\cdots&&v_{n,n} \end{bmatrix} $$ first of all we observe that $$ \det\Big(X^{tr}\cdot X\Big)=\det\begin{bmatrix} \vec v_1\cdot\vec v_1&&\cdots&&\vec v_1\cdot\vec v_k&&\vec v_1\cdot\vec v_{k+1}&&\cdots&&\vec v_1\cdot\vec v_n\\ \vdots&&\ddots&&\vdots&&\vdots&&\ddots&&\vdots\\ \vec v_k\cdot\vec v_1&&\cdots&&\vec v_k\cdot\vec v_k&&\vec v_k\cdot\vec v_{k+1}&&\cdots&&\vec v_k\cdot\vec v_n\\ \vec v_{k+1}\cdot\vec v_1&&\cdots&&\vec v_{k+1}\cdot\vec v_k&&\vec v_{k+1}\cdot\vec v_{k+1}&&\cdots&&\vec v_{k+1}\cdot\vec v_n\\ \vdots&&\ddots&&\vdots&&\vdots&&\ddots&&\vdots\\ \vec v_n\cdot\vec v_1&&\cdots&&\vec v_n\cdot\vec v_k&&\vec v_n\cdot\vec v_{k+1}&&\cdots&&\vec v_n\cdot\vec v_n \end{bmatrix}=\\ \det\begin{bmatrix} \vec v_1\cdot\vec v_1&&\cdots&&\vec v_1\cdot\vec v_k&&0&&\cdots&&0\\ \vdots&&\ddots&&\vdots&&\vdots&&\ddots&&\vdots\\ \vec v_k\cdot\vec v_1&&\cdots&&\vec v_k\cdot\vec v_k&&0&&\cdots&&0\\ 0&&\cdots&&0&&1&&\cdots&&0\\ \vdots&&\ddots&&\vdots&&\vdots&&\ddots&&\vdots\\ 0&&\cdots&&0&&0&&\cdots&&1 \end{bmatrix}=\\ \det\begin{bmatrix} \vec v_1\cdot\vec v_1&&\cdots&&\vec v_1\cdot\vec v_k\\ \vdots&&\ddots&&\vdots\\ \vec v_k\cdot\vec v_1&&\cdots&&\vec v_k\cdot\vec v_k \end{bmatrix}\cdot\det\begin{bmatrix} 1&&\cdots&&0\\ \vdots&&\ddots&&\vdots\\ 0&&\cdots&&1 \end{bmatrix}=\\ \det\begin{bmatrix} \vec v_1\cdot\vec v_1&&\cdots&&\vec v_1\cdot\vec v_k\\ \vdots&&\ddots&&\vdots\\ \vec v_k\cdot\vec v_1&&\cdots&&\vec v_k\cdot\vec v_k \end{bmatrix}=\Big(\text{vol}(\vec v_1,\cdots,\vec v_k)\Big)^2 $$ so that remembering that $$ \det X=\det X^{tr} $$ we observe that $$ (\det X)^2=\det X^{tr}\cdot\det X=\det\Big(X^{tr}\cdot X\Big) $$ and thus we conclude that $$ \text{vol}(\vec v_1,\cdots,\vec v_k)=\det X $$ because the bases $\mathcal E:=\{\hat e_1,\dots,\hat e_n\}$ and $\mathcal V:=\{\vec v_1,\dots,\vec v_k,\vec v_{k+1},\dots,\vec v_n\}$ belong to the same orientation and moreover $X$ is the change of base matrix from $\mathcal E$ to $\mathcal V$ so that $\det X$ is positive.