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. >
