cvMulTransposed and Dimensions of output Matrix?

this might seem as a stupid question but I am having trouble with the output size of the matrix after multiplication.. according to basics, if matrix A has is m*n and matrix B is J*k.. then their product will be of the size m*k..

so according to my code below:

CvMat *meh, *meh2;
    meh = cvCreateMat(2, 9, CV_32FC1);
    meh2 = cvCreateMat(2, 2, CV_32FC1);
    cvMulTransposed(meh, meh2, 1);
    cvSave( "meh.txt", meh2 );

meh is a 2 row, 9 column matrix, so its transpose will be a 9 row 2 column matrix right? and their multiplication will give a 2 row 2 column matrix ( 2*9 multiplied with 9*2).

and that is what I created as my output matrix, meh2.. this is the output is saved to text:

   %YAML:1.0
meh: !!opencv-matrix
   rows: 2
   cols: 2
   dt: f
   data: [ -431602080., -431602080., -431602080., -431602080. ]

as can be seen its filled with junk values..

but when I change the code to make the output 9*9 matrix then I开发者_开发百科 can see actual multiplied values in the saved file..

here, doesnt the output matrix after multiplication have the rows of first matrix and columns of second?

Cheers

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This is the declaration of the function:

void cvMulTransposed(const CvArr* src, CvArr* dst, int order, const CvArr* delta=NULL, double scale=1.0)

You used order = 1, then it means the multiplication is done like this :

dst = scale* transpose[(src-delta)] * (src-delta)

If you use order 0 then the multiplication is :

dst = scale* (src-delta) * transpose[(src-delta)]

In your case as you use order=1, the final matrix is :

  (9,2)     x     (2,9)       =    (9,9)
 transpose     not transpose       result

This is where you can get the documentation of the function