18 авг. 2020 г. · Let matrix A∈Rn×n have k diagonal elements, where k<n, and rest of the elements are zero. I am trying to find the pseudoinverse of A+λI when λ ...

1 февр. 2011 г. · For a matrix A∈Km×n with rank(A)=r and a singular value decomposition A=UΣV∗, the Moore-Penrose pseudo-inverse of A is the matrix A†:=VΣ†U∗∈Kn×m ...

21 мая 2019 г. · Let A∈Rn×p, let D be a diagonal matrix with positive entries. † denotes the Moore-Penrose pseudoinverse. Is it true in general that: (A⊤DA)†A ...

29 окт. 2013 г. · If m≥n, write Σ=(Sk×k0k×(n−k)0(n−k)×k0(n−k)×(n−k)0(m−n)×k0(m−n)×(n−k)) where S is an invertible diagonal matrix. So, you may directly verify ...

25 мая 2020 г. · Yes it is. Just check the Moore-Penrose conditions by doing matrix multiplication.

12 янв. 2015 г. · It turns out the condition A†BA=I together with the non-negativity implies that you cannot have more than one non-zero entry in any column ...

3 авг. 2013 г. · In all other cases, even if A is invertible, the solution if it exist will be in the form x=A+b+(I−A+A)w, where A+is called pseudoinvers of A, ...

7 авг. 2021 г. · A block matrix ( A 0 0 B ) with A and B square is invertible if and only if A and B are invertible -- otherwise, the rank is less than full.

28 нояб. 2015 г. · Pseudo inverse of a singular value decomposition (SVD) is equal to it's "real" inverse for a square matrix. It said it is quite clear that they are equal.

18 янв. 2014 г. · Then, the pseudo inverse becomes: A+=VΣ+UT. The problem comes to: when I was checking the SVD decomposition, I found A≠UΣ ...