Linear transformation that preserves the determinant. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:39:34Z http://mathoverflow.net/feeds/question/522 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/522/linear-transformation-that-preserves-the-determinant Linear transformation that preserves the determinant. Ohdarkdevil 2009-10-14T22:27:52Z 2009-10-15T01:13:22Z <p>It seems "common knowledge" that the following holds:</p> <p>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:</p> <p>a) For any matrix A we have T(A)=UAV<br> b) For any matrix A we have T(A)=UBV where B is the transpose of A</p> <p>It seems quite reasonable, but as far as "common knowledge" goes, I have no clue right now on how to prove such a thing?</p> http://mathoverflow.net/questions/522/linear-transformation-that-preserves-the-determinant/528#528 Answer by Alon Amit for Linear transformation that preserves the determinant. Alon Amit 2009-10-14T23:19:05Z 2009-10-15T01:13:22Z <p>The conclusion you indicate is obtained as the main result in the <a href="http://www.sciencedirect.com/science?%5Fob=ArticleURL&amp;%5Fudi=B6V0R-44RNPBX-3&amp;%5Fuser=10&amp;%5Frdoc=1&amp;%5Ffmt=&amp;%5Forig=search&amp;%5Fsort=d&amp;%5Fdocanchor=&amp;view=c&amp;%5FsearchStrId=1048734509&amp;%5FrerunOrigin=google&amp;%5Facct=C000050221&amp;%5Fversion=1&amp;%5FurlVersion=0&amp;%5Fuserid=10&amp;md5=a45e0ac01af91c0b7f9d22d73102e1ec" rel="nofollow">following paper</a>, but with an apparently stronger hypothesis: (EDIT: Not stronger at all, actually - just realized you're assuming the map is linear.)</p> <p>Determinant preserving maps on matrix algebras</p> <p>Gregor Dolinar and Peter Semrl</p> <p>Linear Algebra and its Applications Volume 348, Issues 1-3, 15 June 2002, Pages 189-192</p> <p>Let Mn be the algebra of all n×n complex matrices. If φ:Mn→Mn is a surjective mapping satisfying det(A+λB)=det(φ(A)+λφ(B)) then either φ is of the form φ(A)=MAN or φ is of the form φ(A)=MAtN where M,N are nonsingular matrices with det(MN)=1.</p> http://mathoverflow.net/questions/522/linear-transformation-that-preserves-the-determinant/534#534 Answer by Eric Wofsey for Linear transformation that preserves the determinant. Eric Wofsey 2009-10-15T00:30:03Z 2009-10-15T00:57:12Z <p>First, some easy observations: T is injective since for any A, there is some B such that the matrices tA+B have different determinants for different scalars t (easy exercise). By multiplying T by T(1)^{-1}, it may be assumed that T(1)=1.</p> <p>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 &lt; n-1 matrices so they are preserved, and so on.</p> <p>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 Pi the projections onto a basis ei, T sends Pi to projections Qi onto some other basis fi. Now let U be the change of basis matrix from the ei to the fi. Conjugating T by U shows that we may assume T fixes each Pi. That is, picking the standard basis, T fixes all diagonal matrices.</p> <p>Now matrices whose only nonzero entries are all either in the first row or first column are characterized by the fact that they are rank 1 and they remain rank 1 if their first diagonal entry changes but they become rank 2 if any other diagonal entry changes. Similar statements hold for other rows and columns. It follows that T(eij) is a multiple of either eij or eji for all j and i, for eij the matrix with ij entry 1 and all others 0. It is now easy to check that we must either always have T(eij)=eij or always have T(eij)=eji, i.e. T(A)=A or T(A)=A^t.</p>