Due to R. Penrose. Any matrix can be partitioned in the form
[
A B
C CA⁻¹B
]
(using a suitable arrangement of rows and columns), A being any non-singular submatrix whose rank is equal to that of the whole matrix. Please explain proof (verify) that AA∗ + BB∗ and A∗A + B∗B are positive definite.
1) We recall that a positive definite matrix A is characterized by AX=0⇔X=0 for any vector X. Also, all eigenvalues are >0.
2) As A is non-singular, then, AA* is positive definite.
3) BB* is positive subdefinite, so all eigenvalues are ≥0.
4) Now, for any vector X, (AA* +BB* )X=0, then,
(AA* )X+(BB* )X=0 ⇔(BB* )X=-(AA* )X.
From 2) and 3), the equality in 4) holds if and only if X=0. Consequently, from 1), AA* +BB* is positive definite.
( Clarification: there exist λ ≥0 eigenvalue of BB* such that BB* X=λX=-AA* X⇔AA* X=-λX ).
In your question, as indicated, You use partitioned matrix with Shur complement, I wish if someone find a solution by that way.
Dear Prof. Dr.Hanifa Zekraoui,
Thank you so much for your kindly answers and advice.
Let me know, why BB* is positive subdefinite?
Regards,
Wiwat Wanicharpichat
BB* is self -adjoint and there is no condition on B to be non-singular. If B is non-singular, then BB* is like AA* . It is just a special case of my answer.
In general B need not a square.
I would like to know that, why AA* + BB* is positive definite? If BB* is positive definite then we get AA* + BB* is also positive definite. But in this problem we have no any information about B.
Yes, every matrix M of the form: M=NN* (N* N) is positive semi definite ( also called positive sub-definite ), in French: semi-définie positive.
There are 4 equivalent conditions for a symmetric matrix to be positive sub-definite, including this one:
There exists a real (complex ) matrix N such that M=NtN resp (N* N)
There is no mention on N to be square. You can see this link:
https://fr.wikipedia.org/wiki/Matrice_autoadjointe_positive
Thank you very much for your useful informations.
We see that, for y = B∗x, for all nonzero x∈ℂr
x*BB*x = (B∗x)*(B∗x) = y*y = ∑rk=1(yk)2 ≥ 0.
Therefore BB* is positive semi-definite.
Hi Wiwat,
On top of what you just mentioned, that x*BB*x is non-negative. This proves that BB* (and AA*) are positive semi-definite. Hence AA* + BB* is positive semi-definite. You still need to prove why it is positive definite. Turns out that because A is non-singular then you could say AA* is positive definite and hehnce AA* + BB* is also positive definite.
Regards
Dear Ahmad,
Due to R. Penrose.
R. Penrose, On best approximate solutions of linear matrix equations, Proc. Cambridge Philos. Soc., 52 (1955) 17-19. (p.18).
Regards
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