1
$\begingroup$

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.

  • 1
    $\det I=1$, so they are inverses (multiplicatively speaking).2012-03-26
  • 1
    Why do you necessarily have $AA' = I$, though?2012-03-26
  • 5
    The determinant of a map between two different vector spaces is not well-defined.2012-03-26
  • 0
    1 - I know they're inverses. I need them to have the same determinant too to get my result. Daniel Montealegre 2 - I'm assuming AA' = I. 3 - Switch it to R^n -> R^n then..2012-03-26

3 Answers 3

2

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 $A$.

  • 0
    The determinant has the $\operatorname{sgn}(\sigma)$ incorporated as in Robert Israel's answer.2012-03-26
  • 0
    @robjohn Yes. I forgot that.2012-03-26
  • 0
    very cool. I've never seen determinants defined that way.2012-03-27
  • 0
    @HenriqueTyrrell The definition on Wikipedia contains this formula.2012-03-27
3

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.

  • 0
    Suppose we characterize the determinant as the unique $n$-linear, alternating function of the columns, with value $1$ on the identity matrix. $\det(A')$ is then an $n$-linear function on the *rows* of $A$. Is there an easy way to see that this is also an $n$-linear function on the columns of $A$? Furthermore, is it easy to see that the $n$-linear function on the rows is the same as the $n$-linear function on the columns?2012-03-26
2

(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 $\det A=\det D$.