I have written a demonstration of the existence of SVD based on the existence of QR, but I am not sure if it's correct or if I am missing something.
QR and L'Q'
- We know that B = QR , with Q orthogonal and R upper triangular
- We know that Bt = Q'R', with Q' orthogonal and R' upper triangular
- B = (Bt)t = (Q'R')t = R'tQ't
- Since R't is lower triangular, we can rename it as L'.
- The transpose operation perserves orthogonality, so Q't is orthogonal.
QDQ'
- A = QR
- R = L'Q't.
- Since R is upper triangular, L' is both upper triangular and lower triangular, which means that L' is actually a diagonal matrix. We rename it as D
- A = QR = Q(DQ't) = QDQ't
Conclusion:
Since Q and Q't are both orthogonal, and D is a diagonal matrix, the existence of QR guarantees the existence of SVD by allowing me to decompose an A matrix into an QDQt' matrix.
It turns out I am making a wrong step, "Since R is upper triangular, L' is both upper triangular and lower triangular, which means that L' is actually a diagonal matrix. We rename it as D" is NOT necessarily true.
L' can be lower triangular and still return a valid QL'Q't = A, which means it's not equal to a diagonal matrix. Without that, we cannot compose a SVD.
There is no such obvious relationship between QR and SVD.