What is the intuition behind using eigenvectors of AA^T and A^T A in the SVD?

[daviddoria]

In the SVD, we use the eigenvectors of AA^T and A^T A as the input and output bases for the matrix. Does anyone have any intuition about these matrices? ie. if I multiply a vector x by AA^T, what space (in terms of the column space, etc. of A) will it bring x to?

Thanks,

Dave

 

[HallsofIvy]

If A: U->V, that is, if A maps vector space U to vector space V, then A^T: V-> U so it acts "like" an inverse. The condition is that <Au, v>_V= <u,A^Tv>_V. Note that Au and V are in vector space V and u and A^Tv are in vector space U. That's why I put the "V" and "U" subscripts on the inner products- to indicate which space the inner product is defined in. Of course, an important application of inner product is to define "orthogonal" or perpendicular. If A maps U into a subspace of V, then all vectors, v, in the orthogonal complement of that subspace have the property that <Au,v>= 0. Since <Au,v>= 0= <u,A^Tv>, and u could be any member of U, it follows that A^Tv= 0 for v in the orthogonal complement of AU.