Pllllllllleeeeeeeeaaaaaaasssssseeeeee help!!! Proof involving moore-penrose pseudo inverse!!!?
If A^+ denotes the moore penrose pseudoinverse of A could someone please prove (A*A)^+=(A^+)(A*)^+ . This is really important so i would really appreciate any help. Naturally 10 points will be awarded to the best answer!
- 1 decade agoFavorite Answer
It's been a while since I've done matrix work, and I never got to pseudoinverses, so the following, in all probability, (a) is longer than it needs to be; and (b) should be checked by you for correctness. Still, even if it's not satisfactory, it may be enough for you to get where you want to go.
I make use of identities whose short proofs are given at the Wikipedia link below. Note also that conjugate transpose (*) commutes with pseudoinverse (+).
For any A, A+ = A+ (A*)+ A* and (A+)* = A A+ (A*)+. We can multiply these together to get
A+ (A*)+ = A (A+)* = A+ (A*)+ A* A A+ (A*)+
We also have A* = A* A A+ and A = (A*)+ A* A. And again, multiplying, we have
A* A = A* A A+ (A*)+ A* A
A* A A+ (A*)+ = A* (A*)+ = A+ A A* (A*)+ = A+ A = A+ (A+)* A* A = (A* A A+ (A*)+)*
A+ (A*)+ A* A = A+ A = A+ A A* (A*)+ = A* (A*)+ = A* A A+ (A*)+ = (A+ (A*)+ A* A)*
Now, let U = (A* A) and V = A+ (A*)+. We have just shown that
V = V U V
U = U V U
U V = (U V)*
V U = (V U)*
Since for any U, U+ exists and is unique, and V satisfies the
requirements of U+, V = U+.