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 John Armstrong | Algebra, Linear Algebra

7 Comments »

I’ve always liked the definition of determinants via taking alternating products- can we prove this fact using that? For instance, do we know that taking adjoints commutes with exterior products?

Comment by Akhil Mathew | July 31, 2009 | Reply
I’m not sure offhand what you mean by “by taking alternating products”. You mean the way I defined it originally? It should be possible to prove that in a basis-free manner, but this is the quick and dirty method that I’m going to need before tomorrow’s post.

Comment by John Armstrong | July 31, 2009 | Reply
Yes. I was thus curious if $\bigwedge^n T^* = (\bigwedge^n T)^*$ or something like that?

The problem is that although we can get a canonical basis of $\bigwedge^n V$ , given one of $V$ , we don’t necessarily have a canonical bilinear form on $\bigwedge^n V$ (as far as I know, at least). Still, a functorial/basis-free proof would be nice.

Comment by Akhil Mathew | July 31, 2009 | Reply
[…] of matrices over more general fields). So now that we’ve got information about how the determinant and the adjoint interact, we can see what happens when we restrict the determinant homomorphism to these subgroups […]

Pingback by The Determinant of Unitary and Orthogonal Transformations « The Unapologetic Mathematician | July 31, 2009 | Reply
I think it’s enough to know that the sign representation of $S_n$ is isomorphic to its dual representation.

First, for any finite-dimensional space $V$ over a ground field $k$ , we have a natural isomorphism

$\Lambda^n V \cong V^{\otimes n} \otimes_{k S_n} sgn$

(which we could also write as

$\Lambda^n V \cong sgn \otimes_{k S_n} V^{\otimes n}$

by taking advantage of the anti-automorphism on the group algebra $k S_n$ ).

Generally speaking, if $G$ is a finite group and $U, W$ are $G$ -reps with $W$ isomorphic to its contragredient dual representation, then there are canonical isomorphisms

$U^* \otimes_{k G} W \cong \hom_{k G}(U, W) \cong \hom(W^* \otimes_{k G} U, k) \cong \hom(W \otimes_{k G} U, k) = (W \otimes_{k G} U)^*$

natural in $U$ . Now take $G = S_n$ , $W$ equal to the sign representation, and $U = V^{\otimes n}$ as an $S_n$ -representation. This gives the second natural isomorphism in

$(V^*)^{\otimes n} \otimes_{k S_n} sgn \cong (V^{\otimes n})^* \otimes_{k S_n} sgn \cong (sgn \otimes_{k S_n} V^{\otimes n})^*$

where the first isomorphism is easily seen to be natural in $V$ . In other words, we have an isomorphism

$\Lambda^n (V^*) \cong (\Lambda^n V)^*$

natural in $V$ . This naturality is I believe the statement you are after (it applies to $V$ of any dimension, even though the determinant identity refers to the case where dim $(V) = n$ ).

Comment by Todd Trimble | August 1, 2009 | Reply
Yes, that’s the type of functoriality I was looking for. Thanks for the explanation.

Comment by Akhil Mathew | August 1, 2009 | Reply
[…] we know that the determinant of the adjoint of a transformation is the complex conjugate of the determinant of the original transformation (or […]

Pingback by The Determinant of a Positive-Definite Transformation « The Unapologetic Mathematician | August 3, 2009 | Reply

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