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It seems "common knowledge" that the following holds:

Let $T$ be a linear transformation on nxn matrices with complex coefficients that preserves the determinant. Then there exists matrices U and V whose product has determinant 1 such that one of the following holds:

a) For any matrix $A$ we have $T(A)=UAV$
b) For any matrix $A$ we have $T(A)=UA^TV$ where $A^T$ is the transpose of $A$

It seems quite reasonable, but as far as "common knowledge" goes, I have no clue right now on how to prove such a thing?

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  • $\begingroup$ What happened to the answer that was here a few minutes ago? $\endgroup$ – Ohdarkdevil Oct 14 '09 at 22:59
  • $\begingroup$ It was deleted by the person who posted it. $\endgroup$ – Anton Geraschenko Oct 14 '09 at 23:04
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First, some easy observations: $T$ must be injective since for any $A$, there is some $B$ such that $B$ and $A+B$ have different determinants (easy exercise). By multiplying $T$ by $T(1)^{-1}$, it may be assumed that $T(1)=1$.

Now note that $T$ preserves the rank of matrices. Indeed, $T$ must preserve the rank $n$ matrices, and then the rank $n-1$ matrices are just the nonsingular locus in the variety of matrices with determinant $0$. This implies $T$ preserves rank $n-1$ matrices. Rank $n-2$ matrices are then the nonsingular locus in rank $<n-1$ matrices so they are preserved, and so on.

Now rank $k$ projections are exactly those rank $k$ matrices which when subtracted from the identity give you something of rank $n-k$; this is easy to see from Jordan normal form. Thus $T$ sends rank $1$ projections to rank $1$ projections. Two projections have disjoint ranges and commute iff their sum is also a projection. In particular, for $P_i$ the projections onto a basis $e_i$, $T$ sends $P_i$ to projections $Q_i$ onto some other basis $f_i$. Now let $U$ be the change of basis matrix from the $e_i$ to the $f_i$. Conjugating $T$ by $U$ shows that we may assume $T$ fixes each $P_i$. That is, picking the standard basis, $T$ fixes all diagonal matrices.

Now matrices whose only nonzero entries are either all in the first row or all in the first column are characterized by the fact that they are rank $1$ and they remain rank $1$ if their first diagonal entry changes. Similar statements hold for other rows and columns. It follows that $T(e_{ij})$ is a multiple of either $e_{ij}$ or $e_{ji}$ for all $j$ and $i$, where $e_{ij}$ is the matrix with $ij$ entry $1$ and all others $0$. By considering the ranks of matrices with only two nonzero entries, it is now easy to see that we must either always have $T(e_{ij})$ a multiple of $e_{ij}$ or always have $T(e_{ij})$ a multiple of $e_{ji}$. Composing $T$ with the transpose map we may assume we are in the first case.

Now let $a_{ij}$ be the scalars such that $T(e_{ij})=a_{ij}e_{ij}$. We know that $a_{ii}=1$, and by considering permutation matrices, it is easy to see that we must have $a_{ij}a_{jk}=a_{ik}$. It follows that $T$ coincides with conjugation by the diagonal matrix with diagonal entries $a_{1i}$, and in particular $T$ has the form $T(A)=UAV$.

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    $\begingroup$ Well, just to nitpick a bit, there are some minor wrinkles, like if $T\begin{pmatrix} a & b \\ c & d \end{pmatrix}= \begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$. Then we would have to change $T$ one more time by conjugating with $\operatorname{diag}(1,-1)$ or similar. $\endgroup$ – Zach Teitler Jan 24 at 14:48
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    $\begingroup$ You're right, and apparently it took over 10 years for anyone to catch that! I've corrected the argument at the end. $\endgroup$ – Eric Wofsey Jan 24 at 15:46
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The conclusion you indicate is obtained as the main result in the following paper, but with an apparently stronger hypothesis: (EDIT: Not stronger at all, actually - just realized you're assuming the map is linear.)

Determinant preserving maps on matrix algebras

Gregor Dolinar and Peter Semrl

Linear Algebra and its Applications Volume 348, Issues 1-3, 15 June 2002, Pages 189-192

Let $M_n$ be the algebra of all $n\times n$ complex matrices. If $\phi:M_n→M_n$ is a surjective mapping satisfying $\det(A+\lambda B)=\det(\phi(A)+\lambda\phi(B))$ then either $\phi$ is of the form $\phi(A)=MAN$ or $\phi$ is of the form $\phi(A)=MA^TN$ where $M,N$ are nonsingular matrices with $\det(MN)=1$.

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  • $\begingroup$ It seems to me that the added hypothesis follows from the requirement that phi is linear. Am I missing something? $\endgroup$ – David E Speyer Oct 15 '09 at 0:11
  • $\begingroup$ Of course - just missed that. It may actually make the claim easier to prove, I'm not sure. $\endgroup$ – Alon Amit Oct 15 '09 at 1:14
  • $\begingroup$ link does not work $\endgroup$ – user76479 Oct 17 '15 at 21:12
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    $\begingroup$ @Arul, I updated the link. $\endgroup$ – Alon Amit Oct 17 '15 at 21:16
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    $\begingroup$ Unfortunately, this article does not prove the theorem OP is asking about, i.e., Frobenius's linear determinant preserver theorem. In fact the article uses Frobenius's theorem in its proof. The basic point of the article is to show that the seemingly weaker hypothesis actually implies linearity, and then Frobenius's theorem can be applied. $\endgroup$ – Zach Teitler Jan 24 at 8:21
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Eric Wofsey's answer can be modified a bit to drop the assumption that we're dealing with the field $K=\mathbb{C}$. Let $K$ be an arbitrary field with $|K|>n$ (where $n \in \mathbb{N}$ is our matrix dimension) and suppose $T(I)=CD$ for some matrices $C$ and $D$. The claim is that there exists some invertible matrix $B$ such that $T: A \mapsto CB^{-1}ABD$ or $T: A \mapsto CB^{-1}A^tBD$.

*We can begin by proving that $T$ is injective (and hence bijective) by the same argument as given by Wofsey. We change from $T$ to $T:=T'(.)=C^{-1}T(.)D^{-1}$ such that $T(I)=I$.

*To show that $T$ preserves the rank of matrices, recall that a rank $m$ matrix can be written as $A=Q\Lambda$ where $\Lambda$ is nonsingular and $Q$ has precisely $m$ nonzero linear independent columns (the other columns being zero). With this in mind, we can see that for any matrix $A$ \begin{equation} rank(A)=\max \left\{ m \in \mathbb{N}\left.\right| D \in M_n:\det (\lambda D + A)= c_{n-m}\lambda^{n-m}+...+c_n \lambda^n \text{ with }c_{n-m} \neq 0 \right\}. \end{equation} (If I'm correct, the assumption that $|K|>n$ is important here, in the sense that for a polynomial $P$ of degree $\leq n$ we have $P(\lambda)=0$ for all $\lambda \in K$ only if $P=0$. Hence polynomials $P$ are uniquely fixed by their evaluations $P(\lambda)$)

But from the bijectivity of $T$ it follows that \begin{equation} rank(A)=\max \left\{ m \in \mathbb{N}\left.\right| D \in M_n:\det (\lambda D + A)=\det (T(\lambda D + A))=\det (\lambda T(D) + T(A))= c_{n-m}\lambda^{n-m}+...+c_n \lambda^n \text{ with }c_{n-m} \neq 0 \right\}= \max \left\{ m \in \mathbb{N}\left.\right| D \in M_n:\det (\lambda D +T(A))= c_{n-m}\lambda^{n-m}+...+c_n \lambda^n \text{ with }c_{n-m} \neq 0 \right\}=rank(T(A)) \end{equation}

*The last modification to make is to revise the proof that $T$ maps projectors on projectors of equal rank. It is sufficient to prove that $A$ is a projector of rank $m$ iff $A$ has rank $m$ and $I-A$ has rank $n-m$.

Proving the rightward implication is easy. For the leftward implication: for any matrix $A$, we have the inclusion $\ker (A) \subset Ran (I-A)$ (easy exercise), which implies $\dim(\ker (A)) \leq \dim(Ran (I-A))$ where equality is attained iff $\ker (A) = Ran (I-A)$. By the dimension theorem, we then have that $n=\dim(Ran(A))+\dim(\ker(A))\leq \dim(Ran(A)) + \dim(Ran(I-A))$ with equality iff $\ker (A) = Ran (I-A)$. We conclude that \begin{equation} Rank(A)+Rank(I-A)=\dim(Ran(A))+\dim(Ran(I-A))=n \Rightarrow \ker (A) = Ran (I-A) \Rightarrow A(I-A)=0 \Rightarrow A\text{ is a projector} \end{equation}

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  • $\begingroup$ For infinite fields, my argument can be adapted more simply as follows. First, my argument applies verbatim to algebraically closed fields. Now note that the identity $\det T(A)=\det(A)$ must also hold over the algebraic closure $\overline{K}$ of $K$ as well, since $M_n(K)$ is Zariski-dense in $M_n(\overline{K})$. Applying my argument in $\overline{K}$, it is easy to see that the $U$ and $V$ produced actually have entries in $K$. (In fact, because $\det(A)$ and $\det T(A)$ are polynomials of degree $\leq n$, you only need $|K|>n$.) $\endgroup$ – Eric Wofsey Oct 18 '15 at 22:10
  • $\begingroup$ Okay, but invoking the Jordan normal form (or whatever related thing) for this purpose feels a bit expensive. You seem to use it in a place where just the dimension theorem suffices. By the way, I suspect that the $|K|>n$-assumption is in fact also unnecessary. $\endgroup$ – Thibaut Demaerel Oct 18 '15 at 22:16
  • $\begingroup$ Sure; your arguments are definitely nice and more elementary than mine. I guess the point of my previous comment is just that it is quite easy to reduce to the algebraically closed case by formal considerations (but your answer is still useful as providing a more elementary argument even in that case). $\endgroup$ – Eric Wofsey Oct 18 '15 at 22:29
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Here is a more geometric proof. Put $V=\mathbb{C}^n$. Consider the embedding $i:\mathbb{P}(V)\times \mathbb{P}(V^*)\hookrightarrow \mathbb{P}(\mathrm{End}(V))$ which associates to $(x,x^*)$ the endomorphism $z\mapsto \langle x^*,z\rangle x\ $; its image is the locus of rank 1 endomorphisms. Since $T$ preserves the rank, it induces an automorphism of $\mathbb{P}(V)\times \mathbb{P}(V^*)$. Now $i$ is the embedding defined by the global sections of the line bundle $\mathcal{O}(1)\boxtimes \mathcal{O}(1)$; any automorphism of $\mathbb{P}(V)\times \mathbb{P}(V^*)$ preserves this line bundle, hence is induced by a unique automorphism of $\mathbb{P}(\mathrm{End}(V))$. These automorphisms are of the form $(x,x^*)\mapsto (u(x),{}^{t}\!v(x^*))$, with $u,v\in \mathrm{Aut}(V)$, or $(x,x^*)\mapsto ({}^{t}\!v(x^*),u(x))$, where $u$ (resp. $v$) is an isomorphism of $V$ onto $V^*$ (resp. $V^*$ onto $V$). It follows that $T(f)=ufv\ $ or $\ u{}^{t}\!fv$.

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  • $\begingroup$ What can be said about trace preserving maps? $\endgroup$ – Ali Taghavi Jun 9 '19 at 14:04
  • $\begingroup$ Then you are just asking for linear endomorphisms of a vector space preserving a given linear form. This is an easy and rather uninteresting exercise. $\endgroup$ – abx Jun 9 '19 at 14:37
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    $\begingroup$ yes you are right. But what about if we give a bundle flavour to our question: we can consider a fiber bundle over the unit sphere of $V^*$. Each fober over a form $ \alpha \in V^*$ is all element of$ Gl(V)$ preserving $\alpha$ $\endgroup$ – Ali Taghavi Jun 9 '19 at 16:18
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    $\begingroup$ one can consider the similar construction to ontain a vector bundle and corresponding principal bundle. $\endgroup$ – Ali Taghavi Jun 9 '19 at 16:20
  • $\begingroup$ the same construction on $PV^*$ seems more interesting.what do you think? $\endgroup$ – Ali Taghavi Jun 9 '19 at 16:52

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