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It seems "common knowledge" that the following holds:


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


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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 $0$. It is now easy to check that we must either always have $T(e_{ij})=e_{ij}$ or always have $T(e_{ij})=e_{ji}$, i.e. $T(A)=A$ or $T(A)=A^t$.

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Determinant preserving maps on matrix algebras

Gregor Dolinar and Peter Semrl


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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link does not work – Arul Oct 17 '15 at 21:12
@Arul, I updated the link. – Alon Amit Oct 17 '15 at 21:16

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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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 $|K|>n$.) – Eric Wofsey Oct 18 '15 at 22:10
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. – Thibaut Demaerel Oct 18 '15 at 22:16
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). – Eric Wofsey Oct 18 '15 at 22:29

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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