Closed just12321 closed 2 months ago
π is not a mathematically orthogonal matrix, and it is not even a square matrix, so πTπ = πΌ is not valid in this situation.
The βfinal_matrixβ in the code is generated using multiple orthogonal matrix blocks (i.e., each block is an orthogonal matrix), but the fact that the final_matrix is stitched together and scaled does not guarantee that the entire final_matrix is still orthogonal.
The βfinal_matrixβ in the code is generated using multiple orthogonal matrix blocks (i.e., each block is an orthogonal matrix), but the fact that the final_matrix is stitched together and scaled does not guarantee that the entire final_matrix is still orthogonal.
Oh, I see, I review the code and notice that I've mistakenly thought nb_column and nb_row are H and W(=H) of the feature, thanks! π
I am confused by the following: Given that πβ=πππ and πβ=ππT , then: π*(πβ)T=(ππT)(πππ)T=πππ(πππ). Since Ο is an orthogonal matrix, we have πTπ=πΌ. Therefore: πβ(πβ)π=πππ.
What is the issue here? Or the π just servers for 1/C(x) (namely the D_inv in the code)? Could someone please help to clarify!