is is possible to reconstruct MDS generator matrix from partial columns of it?
367788
I dun think so buddy..
953148
For found an answer watch here:\nhttp://en.wikipedia.org/wiki/Transition_matrix
613106
You can decode it with a supercomputer.  Ask the lab at Caltech for assistance.
1072880
You probably need to download RAR, a type of archive software.\nPlease go to this link to download:\nhttp://www.rarlab.com/download.htm
643617
Try Irfanfiew. It converts many different formats and is free.
199628
Assuming, of course, that each of the vectors has 3 entries...\n\nDo you know how to put a matrix formed from those four vectors in row-echelon form?\n\nI'm trying to decide the best way to teach this to you, which is why I need to know if you know how to get the matrices in REF, but the final "Answer" is to see if three of those vectors are linearly independent. If three of them are linearly independent, then they must span R^3.\n\nFor your second question, no. Remember that a singular matrix is a square matrix that is not invertible. Therefore, a singular matrix will not have a pivot in every column, which means that the columns are linearly dependent, which means that the rank is less than the dimension. Any square matrix with a column that is all 0's will will be singular.
842942
Here's a discussion that might interest you\nhttp://forums.techguy.org/business-applications/491904-convert-wdb-xls.html\n\n///
458991
No, but if you go to List mode you can enter your matrix as a list and manipulate it.  Check out the owner's manual for help on how to do it.  I've included a link to it.  Hope this helps.
190014
http://photostore.nba.com/perl/gal?gallery_id=1241&process=gallery&provider_id=202&ptp_photo_id=nba%3A210136&sequencenum=0
372589
How about making yourself a model, finding its point group then working out the reducible representation?
122666
http://www.seedbiology.de/structure.asp
202093
You can't.  You need the entire rar file collection in order to decompress a split rar file.
789766
You can do the conversion using\nAny DWG DXF Converter, downloadable at\n\nhttp://www.topshareware.com/Any-DWG-DXF-Converter-download-17667.htm
871507
Assuming that there are enough eigenvectors to give a basis, yes. This will happen if all the eigenvalues are distinct. Here's how:\n\nLet the eigenvectors be v_1, v_2, v_3 with eigenvalues c_1, c_2 and c_3, respectively.\n\nThen, in the basis {v_1, v_2, v_3}, the matrix is diagonal(c_1, c_2, c_3). To get the matrix in the standard basis, you let A be the matrix obtained by putting the components of {v_1, v_2, v_3} down the columns. Find the inverse matrix A^(-1). Then the matrix you want is\n\nA*diag(c_1,c_2,c_3)*A^(-1).\n\nThis genralizes to higher dimensions, by the way.\n\nIf there is not a basis of eigenvectors, the matrix is not uniquely specified.
777528
just extract the 1st one and the rest will follow automatically. It's in-built.
515796
go to: http://softwareclub.ws/v_split.html  i hope this will help a li'l bit
573927
single replacement, double replacement, decomposition, synthesis