Thank you. (as usual, it kills me how little i know) :)
On Mon, Sep 10, 2012 at 2:14 PM, Ted Dunning <[email protected]> wrote: > You are correct-ish. > > Left and right inverses are still inverses, just not commutative. > > Moreover, since V is full rank and orthonormal, the SVD is trivially: > > svd(V) = V I I > > This means that while V' is the left inverse, it is also the right Penrose > inverse in the sense that the least squares solution of argmin_X norm(V X - > I) is V'. > > On Mon, Sep 10, 2012 at 2:05 PM, Dmitriy Lyubimov <[email protected]> wrote: > >> Yep that's my understanding too. V cannot have inverse since it is not >> square (in common case). >> >> But since it is orthonormal, transpose produces a pseudoinverse which >> we can use just the same. >> >> On Mon, Sep 10, 2012 at 2:03 PM, Sean Owen <[email protected]> wrote: >> > Yes, doesn't the V V' != I bit mean it's not an inverse? I thought >> > only square matrices had a real inverse. This is a one-sided inverse, >> > which is (IIRC?) slightly stronger than being a pseudo-inverse. >> > >> > On Mon, Sep 10, 2012 at 9:59 PM, Ted Dunning <[email protected]> >> wrote: >> >> No. It is the inverse. V' V = I >> >> >> >> On the other hand, V V' != I. We do know that norm(A V V' - A) is >> small. >>
