ok, i’ve published everything i planned to put out there for the SVD. not to say i won’t need to add things, if anyone ever has any questions about what i meant….
yes, i need to put out a bibliography, too. but i’ve been at this all day, and it’s time to stop. incidentally, the SVD stuff is a category of its own – not a page as i originally set up the first (last!) installment.
i should probably emphasize that the SVD category is posted in the order i intend it to be read.
it looks like quite a pile of stuff, and it’s scary to think that it’s all only background for principal components analysis!
one advantage to doing as much work as i did today is that i’ve managed to streamline the process. take a piece of the latex output from mathematica®, drop it into TextWrangler, edit it massively there, then copy a preliminary version into the wordpress editor, and finally edit there until it works.
well, i’m thrilled to have found a way to play with the SVD, after all the years i’ve meant to. i need to actually write down the list of all the other things i’ve been meaning to look at.
Rip's Applied Mathematics Blog 
December 17, 2007 at 6:02 pm
a friend asks:
1) In some places it looked like you were interchanging inverse and transpose. They
are obviously different operations, under what circumstances do they yield the same
result? What part of “Linear Algebra” should I reread?
my answer:
. that implies that
a real square matrix is called orthogonal if (i have to try this)
the columns of A are a set of orthonormal vectors
the matrix is invertible
A^-1 = A^T
A A^T = I
the matrix is a rotation
to answer your specific question, the transpose = the inverse iff the matrix is orthogonal.
NOTE my remarks about an orthonormal matrix being one that doesn’t have enough columns.
the corresponding property for a complex matrix is “unitary”, and requires that the conjugate transpose = the inverse.
look up “orthogonal matrix” and “unitary matrix”.
December 17, 2007 at 6:09 pm
the same friend also asked:
2) If a map from 4-space to 3-space has rank 2, the result set is a plane in 3 space.
If the rank is 2.15, does this imply that the result set is somehow “thin”, i.e. close to a
plane?
my response:
good question! the matrix is of (integer) rank 3, so its range really is 3D. the matrix itself is close to one of rank 2, and the image of that is only 2D. but the image space of a matrix is a vector space, so rank 3 means that the image is all of a 3D space.
added here: strang’s term for what i called real-valued rank is “effective rank”.
February 21, 2008 at 5:17 pm
Great stuff, I wish I had found it sooner. I just wrote up a little summary of SVD with link to Matlab code for visualizing the SVD for 3×3 matrices, which can be found here.