If $A: V\to V$ is a linear map such that $\langle Av, Aw\rangle=\langle v, w\rangle$ for all $v, w\in V$, show that $\det(A)=\pm 1$. Let $V$ be a finite dimensional vector space over the field $K$, with a nondegenerate scalar product$\langle, \rangle$. If $A: V\to V$ is a linear map such that $\langle Av, Aw\rangle=\langle v, w\rangle$ for all $v, w\in V$, show that $\det(A)=\pm 1$.
The solution said:
In the general case we also have $A^*A=AA^*=I$ (where $A^*$ is the transpose of the operator). If $J$ represents the scalar product, then $\langle Av,w\rangle=\langle v,A^*w\rangle$ so that $A^tJ=JA^t$ where $A^t$ is the transpose matrix of $A$. Hence $\det(A^t)=\det(A^*)$, which implies that $\det(A^*)^2=1$.
I don't understand why in solution, $\langle Av,w\rangle=\langle v,A^*w\rangle$ can show that $A^tJ=JA^t$? And why then $\det(A^t)=\det(A^*)$?
(I find a post $\langle Av,Aw\rangle=\langle v,w\rangle$,$\det(A)=1$ proof verification asked about the same problem, but didn't address the same question of this problem.)

 A: Let $B=A^TA$. Then $\langle v, Bw\rangle=\langle v, w\rangle$. Hence $\langle v, Bw-w\rangle=0$ for every $v, w$. Take $v=Bw-w$. You get $||v||=0$, so $v=0$. Thus for every $w, Bw=w$, so $B=A^TA=I$. Since $\det(A)=\det(A^T)$, we have  $\det(A^2)=1$, so
$\det(A)=\pm 1$.
A: $A$ is a linear operator on a coordinate vector space $\mathbb F^n$.  When you say the matrix $J$ 'represents' the non-degerate scalar product, I interpret this to mean
there is some basis chosen and given by the hyper-vector  $\mathbf B$
(this is also a square matrix since we are dealing with a coordinate vector space)  and
$j_{i,k} = \langle \mathbf b_i,\mathbf b_k\rangle$
now consider the effect of $A$ operating on $\mathbf B$
$A\mathbf B = \mathbf B Q$
taking determinants of each side we see $\det\big(A\big) = \det\big(Q\big)$
so it suffices to prove  $\det\big(Q\big)\in\{-1,+1\}$
Consider $v,v' \in V$
$v = \mathbf B\mathbf x$ and $v' =\mathbf B y$;
$\mathbf w = Q\mathbf x$ and $\mathbf z = Q\mathbf y$
$\langle A v, Av'\rangle$
$=\langle A\mathbf B\mathbf x\mathbf , A\mathbf B\mathbf y\rangle$
$=\langle \mathbf B (Q\mathbf x), \mathbf B(Q\mathbf y)\rangle$
$=\langle \mathbf B \mathbf w, \mathbf B\mathbf z\rangle$
$=\langle \sum_{k=1}^n  \mathbf b_k w_k , \sum_{i=1}^n  \mathbf b_i z_i\rangle$
$=\sum_{k=1}^n\sum_{i=1}^n w_kz_i \langle  \mathbf b_k  ,   \mathbf b_i \rangle$
$=\mathbf w^TJ\mathbf z$
$=\mathbf x^T Q^T JQ\mathbf y$
and by nearly identical calculation
$\mathbf x^T J\mathbf y = \langle  v, v'\rangle =  \langle  Av, Av'\rangle=\mathbf x^T Q^T J Q\mathbf y$
the choice of $v, v'$ were arbitrary, so we may run the above argument on $n$ linearly independent vectors in $V$ which gives
$X^TJX = X^T Q^T JQX \implies J = Q^T J Q$
so $Q$ is a stabilizer for $J$  and being non-degenerate $\det\big(J\big) \neq 0$  thus
$\det\big(J\big)=\det\big(Q^T\big)\det\big(J\big) \det\big(Q\big)\implies \det\big(Q\big)^2 =1\implies \det\big(Q\big) \in \{-1,+1\}$
A: The confusing part is that in this solution there is only one notation being used for two different inner products. Let's say that $V = \mathbb{R}^n$. Then an abstract vector $v \in V$ or an abstract linear operator $A \colon V \to V$ may be treated as column vector $[v] \in \mathbb{K}^n$ or a matrix $[A] \in \operatorname{Mat}_n(\mathbb{R})$. Furthermore, when $(-)^T$ denotes the transpose of a vector or matrix, then we have a standard inner product I'll denote by $(v \cdot u) = [v]^T [u]$. For each linear operator $A \colon V \to V$, we define its adjoint with respect to $(- \cdot -)$ to be the operator $A^T \colon V \to V$ satisfying
$$ (A u \cdot v) = (u \cdot A^T v) \text{ for all } u, v \in V.$$
One can prove that the adjoint exists and is unique. Furthermore, applying the definition of $(- \cdot -)$ to the equation above, we get $[u]^T [A]^T [v] = [u] [A^T] [v]$ for all $u, v \in V$, and so it turns out that the matrix $[A^T]$ of the adjoint is equal to the conjugate-transpose of the matrix $[A]$. (This distinction is subtle, but important for what is to come).

Now, suppose we have any other inner product $\langle -, - \rangle \colon V \times V \to \mathbb{R}$ defined on $V$. We can again define the adjoint $A^\dagger$ of a linear operator $A \colon V \to V$, this time with respect to the new inner product, to be the unique linear operator satisfying
$$ \langle A u, v \rangle = \langle u, A^\dagger v \rangle \text{ for all } u, v \in V. $$
The crucial point is that $A^T \neq A^\dagger$ in general: adjoints with repsect to different inner products are different. How can we relate them to each other? The standard way to do this is to express the new inner product $\langle -, - \rangle$ in terms of the old inner product $(- \cdot -)$: there is a unique linear operator $J \colon V \to V$ such that
$$ \langle u, v \rangle = (u \cdot J v) \text{ for all } u, v \in V.$$
If all we're starting with is the standard basis of $V = \mathbb{R}^n$ then, we can calculate any new inner product $\langle u, v \rangle$ by working out the column vectors $[u]$, $[v]$, and the matrix $[J]$, and then doing
$$ \langle u, v \rangle = (u \cdot J v) = [u]^T [J] [v]. $$
What is the relationship between $A^\dagger$ and $A^T$? Just by manipulating the defining equations, we have
$$ \begin{aligned}
\langle Au, v \rangle = (Au \cdot Jv) = (u \cdot A^T J v) &&&\text{ for all } u, v \in V, \text{ and } \\
\langle u, A^\dagger v \rangle = (u \cdot J A^\dagger v) &&&\text{ for all } u, v \in V.
\end{aligned}$$
Since these are equal, we have the relationship $J A^\dagger = A^T J$.

Having said all of that, the proof is confusing when it should be simple. We can work with any inner product $\langle -, - \rangle \colon V \times V \to \mathbb{R}$, and with any operator $A \colon V \to V$ which is unitary with respect to that inner product, meaning $\langle Au, Av \rangle = \langle u, v \rangle$ for all $u, v \in V$. The key point is realising that being unitary is equivalent to the condition $A^\dagger = A^{-1}$. The proof should proceed in simple steps:

*

*Determine that for a unitary operator $A$, we have $A A^\dagger = I$.

*Determine that for any operator $A$, we have $\det(A^\dagger) = \det(A)$.

*Conclude that for a unitary operator, $\det(A)^2 = 1$.

Parts 1 and 3 should be easy, straight from the definition. Part 2 is more tricky if your definition of determinants uses matrices, and this is where we have to go via the standard inner product $(- \cdot -)$, by writing $A^\dagger = J^{-1} A^T J$, applying $\det$ to get $\det(A^\dagger) = \det(A^T)$, and then using the fact that $[A^T] = [A]^T$ to get
$$ \det(A^\dagger) = \det(J^{-1} A^T J) =\det(A^T) = \det([A^T]) = \det([A]^T) = \det([A]) = \det(A), $$
where in the second-last equals we have used the fact that the determinant of the transpose of a matrix is equal to the determinant of the matrix.
There are other ways to prove part 2 if you have alternative definitions of the determinant that don't require going via a matrix, but I suspect this is the definition of the determinant used in the book.
