By "solution", I mean a PseudoInverse. By "exact", I mean a formula, not a fit (such as SVD) or a decomposition.
Douglas
You might get a response if you can clarify what you are asking. It is a known fact that every matrix (rectangular or square, real or complex) has at least one pseudoinverse. The set of pseudoinverses always contains a unique pseudoinverse called the Moore-Penrose inverse, which satisfies a list of very desirable properties.. (I am assuming here that you mean "generalized inverse" where you write "pseudoinverse".)
The difficulty with finding the pseudoinverse is much like the difficulty in finding the ordinary inverse of an invertible matrix. In theory it exists, and in theory, it can always be found exactly. There are algorithms for finding the Moore-Penrose inverse that are exact, but they are computationally brutal, and worse, numerically unstable unless you are doing exact arithmetic. In practice, one has to work not with real or complex arithmetic, but rather computationally, using finite, floating point arithmetic. Given that, one opts for computationally efficient and numerically stable algorithms, of which the SVD approach is one common example. Sometimes, when a matrix has a very special format, there are simple, explicit formulae for the entries of its inverse. That suggests that there might be some matrices whose pseudo inverse properties vary in similar way.
If I have not answered your question, I hope that I have at least given you some guidance on how to ask it so that your intent is clearer.
Jeff
Jeff,
This is how I generate and solve this type of matrix.
Thanks,
Doug
(* complete dictionary of words, one per row, number of rows is (alpha^word), using words of length "word" and an alphabet of "alpha" number of characters *)
alpha = 4; word = 3; dict =.;
dict = Partition[Flatten[Tuples[Reverse[IdentityMatrix[alpha]], word]], (alpha*word)];
PseudoInverse[dict] == ((Transpose[dict])*((alpha)^-(word - 1))) - ((word - 1)/(alpha^word)/word)
Output = True
C.C. MacDufee apparently was the first to point out, that a full-rank factorization of a matrix A leads to an explicit formula for its Moore-Penrose inverse, A^{+}.
Let A be any complex matrix of order m by n and let A = FG be a rank factorization. Then it can be easily verified that
F^{+}= (F^{*}F)^{-1}F^{*}, G^{+} = G^{*}(GG^{*})^{-1}
and then
A^{+} = G^{+}F^{+},
where A^{+} is mean its Moore-Penrose inverse, and F^{*} is the conjugate transpose of F.
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