Consider two positive definite Hermitian matrices $\pmb W_1$ and $\pmb W_2$ such that $\pmb W_1 \succ \pmb W_2$, which means $\pmb W_1-\pmb W_2$ is positive definite. There exist two full cloumn rank matrices $\pmb A_1$ and $\pmb A_2$ with more rows than columns. The number of rows in $\pmb A_1$ and $\pmb A_2$ are the same, while no zero rows are included. Moreover, assume $\pmb A_2=[\pmb A_1,\pmb a]$, where $\pmb a$ is a non-zero vector. Then let $\pmb B_1=\pmb W_1\pmb A_1$ and $\pmb B_2=\pmb W_2\pmb A_2$. I am seeking the proof of the following inequality $$\pmb W_1^H(\pmb I-\pmb B_1\pmb B_1^+)\pmb W_1 \succ \pmb W_2^H(\pmb I-\pmb B_2\pmb B_2^+)\pmb W_2.$$ Here $\pmb B_1^+$ refers to $\pmb B_1^+=(\pmb B_1^H\pmb B_1)^{-1}\pmb B_1^H$, and $\pmb I$ is the identy matrix. Note that, in the context of signal processing, $\pmb A_1$ and $\pmb A_2$ are the so-called steering matrices, whose elements are all complex exponential. $\pmb W_1$ and $\pmb W_2$ can be expressed by $$\pmb W_1=(\pmb R_1^T \otimes \pmb R_1)^{-1/2},\\\pmb W_2=(\pmb R_2^T \otimes \pmb R_2)^{-1/2},$$ where $\otimes$ denotes the Kronecker product. $\pmb R_1$ and $\pmb R_1$ stand for the signal co-variance matrices, and $\pmb R_2 \succ \pmb R_1$.

  • $\begingroup$ It seems that this inequality is determined by $W_1$ and $W_2$. I conducted some simulations and found that there is no evidence for $B_1B_1^+ \succ B_2B_2^+$ or $B_2B_2^+ \succ B_1B_1^+$. But I wonder how to explain the fact that $(I-B_1B_1^+)$ and $(I-B_2B_2^+)$ have little influence on the stated inequality. $\endgroup$ – abao5887 Dec 26 '18 at 10:03
  • $\begingroup$ Note that $P=(I-BB^+)$ is an orthogonal projection matrix, i.e. $P^2=P=P^H$ and consequently $\lVert P \rVert=1$. This alone doesn't prove the inequality but it might give an idea. $\endgroup$ – obareey Dec 26 '18 at 13:25
  • $\begingroup$ This orthogonal projector $I-BB^+$ is indeed contained in the original inequality, and I wrote it explicitly. This norm might be helpful, but I don't know how to proceed afterwards. In addition, I got another point: If we let $W_1=W_2$, then it can be proved that $B_2B_2^+ \succ B_1B_1^+$. However, with different $W$ multiplying $A$, it becomes confusing. $\endgroup$ – abao5887 Dec 27 '18 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.