Is it true that $\det (A) = \det (A')$? Let $A: \mathbb{R}^n \rightarrow \mathbb{R}^n$ a linear transformation, and let $A'$ be its hilbert adjoint.
Is it true that $\det(A) = \det(A')$?
Trying to prove:
$A A' = I$ in $\mathbb{R}^n \ \Rightarrow \ \det(A) = \pm 1$.
Since $AA' = I$ we have $\det(I) = \det(AA') = \det(A)\det(A')$, and if they're equal, then I have the result. 
Thanks in advance.
 A: How you will want to prove $\det(A) = \det(A')$ may depend on how you define the determinant. If you use $\sum_\sigma \text{sgn}(\sigma) a_{1, \sigma(1)} \ldots a_{n, \sigma(n)}$, notice that $a_{\sigma(1),1} \ldots a_{\sigma(n),1} = a_{1,\rho(1)} \ldots a_{n,\rho(n)}$ where $\rho$ is the inverse of the permutation $\sigma$.
Or if you characterize the determinant as the unique $n$-linear alternating function of the columns with value $1$ on the identity matrix, note that $A \to \det (A')$ is also an $n$-linear alternating function of the columns of $A$ and is $1$ on the identity.
A: (you need $n=m$ for the determinant to make sense)
It is true. Basically doing the determinant of $A$ by the first row is the same as doing the determinant of $A'$ by the first column. Not sure how easy it is to write a proof, though. 
This is the way I would do your problem:
Since $AA'=I$, one can show that $A'A=I$. Then $A$ is normal and thus diagonalizable through an orthogonal matrix: $A=SDS'$, with $D$ diagonal and $SS'=S'S=I$. Then
$$
D^2=D'D=(S'AS)'(S'AS)=S'A'SS'AS=S'A'AS=S'IS=S'S=I.
$$
So each diagonal entry of $D$ satisfies $D_{kk}^2=1$, so $D_{kk}=\pm1$. 
The determinant of $D$ is then $\pm1$, and $\det A=\det D$.
A: If $A=(a_{ij})_{1 \leq i,j \leq n}$ then then the definition of the determinant is
$$ \det(A)= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\cdots a_{n \sigma(n)} $$
where $\sigma \in S_n$ means $\sigma :\{1,\ldots,n\} \to \{1,\ldots,n\}$ bijective. Now it is easy to see that writing the sum with the inverse permutation of $\sigma$ yields
$$ \det(A)=\sum_{\tau \in S_n} \operatorname{sgn}(\tau)a_{\tau(1)1}a_{\tau(2)2}\cdots a_{ \tau(n)n}$$
and this is the determinant of the transpose of $A$.
