Prove that $A: (v_1)^\perp \to (v_1)^\perp ,\ e_j\mapsto Pv_j$'s determinant stays the same for all ONBs $e_j$, where $P$ is an orthogonal projection. I am sorry for the scuffed title, but there are not enough characters. Here is the problem in full:
Take a basis $(v_1,\dots,v_n)$ of $\mathbb{R}^n$ and $W=(v_1)^\perp$. Let $P$ be the orthogonal projection from $\mathbb{R}^n$ onto $W$. Take an orthonormal basis (ONB) $(e_2,\dots,e_n)$ of $W$ and define an endomorphism $A: e_j\mapsto Pv_j,\ j\geq 2$.
Show that $|\det A|$ is independent of the choice of ONBs for $W$ (that is that it stays the same for different bases). Furthermore, show that $|\det(v_1,\dots,v_n)|=|| v_1||\cdot  |\det A|$.
My attempt: At first let us assume there is a $v_i, \ i\neq 1$ s.t. $(v_1,v_i)\neq 0$. Thus, $v_i \not\in W$ and $Pv_i=0$. However, as we know that $Ae_j=Pv_j$, it follows that $Ae_i=Pv_i=0$ and thus, a whole column of the matrix $A$ will be zero. Therefore, $\det A=0$ in this case and it doesn't depend on the choice of the ONB.
If we exclude the existence of such a $v_i$ i.e. $(v_1,v_i)=0,\ j\geq 2$, then it follows that $Pv_j=v_j$ and, thus, $Av=v$ and we get that $A$ has the eigenvalue 1. As $(v_2,\dots,v_n)$ form a basis for $W$, there can't be another eigenvalue and thus, we get $\det A=1$ which would also prove the independence on the choice of the ONB.
My problem: The problem with my "solution" is that the value for $\det A$ is too nice and I think I have gone wrong somewhere in the second part of my proof. Could anyone point out the mistake and suggest a better way? Thank you very much! Your help is greatly appreciated!
 A: For the first statement, let $(e_2,\dots e_n)$ and $(f_2,\dots ,f_n)$ denote two ONB of $W$, and let $O$ be the linear transformation such that $Oe_i=f_i$. We observe that $O$ is orthogonal (i.e. $O^tO=1$) so that $|\det(O)|=1$.
We observe that if $A_e$ denotes the endomorphism with respect to $(e_2,\dots e_n)$ and $A_f$ denotes the endomorphism with respect to $(f_2,\dots f_n)$ then $A_fO=A_e$. Taking determinants and absolute values, we obtain $|\det(A_f)|=|\det(A_e)|$.
The second statement will follow from a convient choice of $(e_2,\dots e_n)$.
Let us first assume that $(v_1,\dots v_n)$ are linearly independant.
Let $(w_1,w_2,\dots w_n)$ denote the Graham-Schmidt orthogonalization of $(v_1,\dots v_n)$, and let $G$ denote the lower-triangular matrix such that $Gw_i=v_i$. Recall from classical Linear algebra that $|mathrm{det}(G)|=|\mathrm{det}(v_1,\dots v_n)|$, and that $G_{11}=\|v_1\|$.
The key observation is that, with respect to the ONB $(w_2,\dots w_n)$ of $W$, we have that the matrix corresponding to $A$ coincides with the principal minor of $G$ (i.e. the matrix obtained from omitting the first row and column). Let us call this matrix $G_1$.
Now, observe that as $G$ is lower-triangular, we have that $\det(G)=\|v_1\|\det(G_1)$. Combining this with the above, we obtain that $|\det(G)|=\|v_1\||\det(A)|$ for this, and thus all choices of ONB.
Now in the case where $(v_1,\dots v_n)$ are linearly dependant, it suffices to show that $\|v_1\||\det(A)|=0$. This is true when $v_1=0$ so assume $v_1\neq 0$.
Now take a nontrivial a linear relation $\sum_{i=1}^n a_iv_i=0$. We may assume that some $a_i\neq 0$ for $i>1$ since $v_1\neq 0$. Furthermore, note that by orthogonality $\sum_{i=2}^{n}a_ie_i\neq 0$.
On the otherhand $A(\sum_{i=2}^{n}a_ie_i)=\sum_{i=2}^na_iP(v_i)=P(\sum_{i=1}^{n}a_iv_i)=0$. Thus $\ker(A)\neq 0$ and so $\det(A)=0$.
