The Unapologetic Mathematician

Mathematics for the interested outsider

The Determinant of the Adjoint

It will be useful to know what happens to the determinant of a transformation when we pass to its adjoint. Since the determinant doesn’t depend on any particular choice of basis, we can just pick one arbitrarily and do our computations on matrices. And as we saw yesterday, adjoints are rather simple in terms of matrices: over real inner product spaces we take the transpose of our matrix, while over complex inner product spaces we take the conjugate transpose. So it will be convenient for us to just think in terms of matrices over any field for the moment, and see what happens to the determinant of a matrix when we take its transpose.

Okay, so let’s take a linear transformation T:V\rightarrow V and pick a basis \left\{\lvert i\rangle\right\}_{i=1}^n to get the matrix

\displaystyle t_i^j=\langle j\rvert T\lvert i\rangle

and the formula for the determinant reads

\displaystyle\det(T)=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^nt_k^{\pi(k)}=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi(k)\rvert T\lvert k\rangle

and the determinant of the adjoint is

\displaystyle\det(T^*)=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi(k)\rvert T^*\lvert k\rangle=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle k\rvert T\lvert\pi(k)\rangle

where we’ve now taken the transpose.

Now for the term corresponding to the permutation \pi we can rearrange the multiplication. Instead of multiplying from {1} to n, we multiply from \pi^{-1}(1) to \pi^{-1}(n). All we’re doing is rearranging factors, and our field multiplication is commutative, so this doesn’t change the result at all:

\displaystyle\det(T^*)=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi^{-1}(k)\rvert T\lvert\pi(\pi^{-1}(k))\rangle=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi^{-1}(k)\rvert T\lvert k\rangle

But as \pi ranges over the symmetric group S_n, so does its inverse \pi^{-1}. So we relabel to find

\displaystyle\det(T^*)=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi(k)\rvert T\lvert k\rangle

and we’re back to our formula for the determinant of T itself! That is, when we take the transpose of a matrix we don’t change its determinant at all. And since the transpose of a real matrix corresponds to the adjoint of the transformation on a real inner product space, taking the adjoint doesn’t change the determinant of the transformation.

What about over complex inner product spaces, where adjoints correspond to conjugate transposes? Well, all we have to do is take the complex conjugate of each term in our calculation when we take the transpose of our matrix. Then carrying all these through to the end as we juggle indices around we’re left with

\displaystyle\det(T^*)=\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\overline{\langle\pi(k)\rvert T\lvert k\rangle}=\overline{\sum\limits_{\pi\in S_n}\mathrm{sgn}(\pi)\prod\limits_{k=1}^n\langle\pi(k)\rvert T\lvert k\rangle}=\overline{\det(T)}

The determinant of the adjoint is, in this case, the complex conjugate of the determinant.

July 30, 2009 Posted by | Algebra, Linear Algebra | 7 Comments

   

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