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I need to know that if the following holds for complex vectors $x=[a\cdot A \mid b\cdot B]u$, and $y= [A \mid B]u$ $$\det(I+d \frac{yy^*}{rI})\leq\det(I+\frac{xx^*}{rI})\leq \det(I+c \frac{yy^*}{rI})$$ where $d= \min \{a,b\}$$, c = \max \{a,b\}$ and $r$ are positive real values. $u$ is a complex valued vector and $^*$ means conjugate transform (Hermitian).

By $[A \mid B]$ I mean the composite matrix made from the two matrices $A, B$. Let $A$ and $B$ be complex valued matrices.

It works in simulations. I think it has something to do with eigenvalues of positive definite matrices. I know that $xx^*$ and $yy^*$ are positive definite and so is $I$. Also I think the division by $I$ is not a problem here. But I don't know how to prove it.

Thank you.

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(Presumably both $A$ and $B$ are nonzero matrices and $a\neq b$, i.e. $0<d<c$.) Contrary to your claim, your inequality never holds universally, i.e. we can always find a vector $u$ that violates the inequality.

For any complex scalar $\lambda$ and any complex vector $v$, we have $\det(I+\lambda vv^\ast)=1+\lambda v^\ast v$. Now, suppose $A$ is $n\times p$, $B$ is $n\times q$ and $M=[A|B]\neq0$. Then $x=M(aI_p\oplus bI_q)u$ and your inequality is equivalent to $$ 1+\frac{u^\ast (dM^\ast M)u}{r} \leq 1+\frac{u^\ast(aI_p\oplus bI_q)M^\ast M(aI_p\oplus bI_q)u}{r} \leq 1+\frac{u^\ast(cM^\ast M)u}{r}.\tag{1} $$ If it holds for all $u$, we must have $$ dM^\ast M \preceq (aI_p\oplus bI_q)M^\ast M(aI_p\oplus bI_q) \preceq cM^\ast M,\tag{2} $$ where $X\preceq Y$ means $Y-X$ is positive semidefinite. In particular, the analogous inequality must hold for the leading principal $p\times p$ and trailing principal $q\times q$ submatrices. That is, \begin{cases} dA^\ast A \preceq a^2A^\ast A \preceq cA^\ast A,\\ dB^\ast B \preceq b^2B^\ast B \preceq cB^\ast B.\tag{3} \end{cases} Since both $A$ and $B$ are nonzero, the above implies that $d\le a^2\le c$ and $d\le b^2\le c$, i.e. $d\le d^2\le c^2\le c$. Yet, by assumption, $0<d<c$. So, the statement that $d\le d^2\le c^2\le c$ is always false and at least one case in $(3)$ is bound to fail. That is, there always exists a vector in either $\mathbb{R}^p\times 0^q$ or $0^p\times\mathbb{R}^q$ such that $(1)$ does not hold.

For an illustrative example, consider $r=1,\ a=d=\frac12,\ b=c=1$ and $$ M=[A|B]=(1,1),\ u=\pmatrix{1\\ 0},\ y=1,\ x=\frac12. $$ Then the first inequality sign in $\det(I+\frac{dyy^*}{r})=\frac32\color{red}{\le}\det(I+\frac{xx^*}{r})=\frac54\le \det(I+\frac{cyy^*}{r})=2$ is violated.

In the uninteresting cases, a similar analysis shows that your inequality holds for all $u$ iff one of the following four conditions is satisfied:

  1. $A=0$ and $B=0$.
  2. $A\ne0,\ B\ne0$ and $a=b=1$,
  3. $A=0,\ B\ne0$ and $d\le b^2\le c$,
  4. $A\ne0,\ B=0$ and $d\le a^2\le c$.
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  • $\begingroup$ Thank you. I have made a mistake in writing the question. I it should be $dydy^*$ not $dyy^*$, same goes with $c$. I want to replace $a,b$ by their min and max. I think now My inequality is correct if I understood your proof? If you wish I can correct my question and then your can change the answer so other readers may know the correct inequality (assuming it is correct when changed). Also a small question about the notation does $(aI_p\oplus bI_q)[A\mid B]$ do same as $[aA\mid bB]$? because $A$ is not a square matrix so I don't know how $aI_pA$ can give $aA$. Once again thank you very much. $\endgroup$ – triomphe Sep 13 '13 at 13:26
  • $\begingroup$ @MLT (1) $aI_p\oplus bI_q$ means $\pmatrix{aI_p\\ &bI_q}$, so the matrix product is $[A|B](aI_p\oplus bI_q)$, not $(aI_p\oplus bI_q)[A|B]$. (2) If the $c$ and $d$ in your question are replaced by $c^2$ and $d^2$, then we would expect the inequality to hold in more cases, but it still does not hold in general. For a counterexample, consider $r=1,\ a=d=1,\ b=c=2,\ [A|B]=(2,1),\ [aA|bB]=(2,2)$. With $u=(1,-1)^T$, we get $y=1,\ x=0$ and the first inequality sign in $1+d^2y^2 = 2 \color{red}{\le} 1+x^2 = 1 \le 1+c^2y^2 = 5$ is still wrong. $\endgroup$ – user1551 Sep 13 '13 at 15:08
  • $\begingroup$ But what about your claim ...implies that $d\le a^2\le c$ and $d\le b^2\le c$,... I thought it will now change to ...implies that $d^2\le a^2\le c^2$ and $d^2\le b^2\le c^2$,... which is true in my case. $\endgroup$ – triomphe Sep 13 '13 at 15:25
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    $\begingroup$ @MLT If a matrix is positive semidefinite, all principal submatrices should be PSD too. It just happens that, in your first version of the problem (using $c,d$ in the inequality), the inspection of two sets of principal submatrices are enough to rule out all interesting cases. Yet in the corrected version (using $c,d$ in the inequality), $(3)$ only constitute two necessary conditions and they are far from being sufficient. $\endgroup$ – user1551 Sep 13 '13 at 16:24
  • $\begingroup$ In the counterexample mentioned my previous comment, we have $(aI_p\oplus bI_q)M^\ast M(aI_p\oplus bI_q)-d^2M^\ast M=\pmatrix{0&2\\ 2&3}$. While the leading and trailing principal $1\times1$ submatrices (i.e. the two diagonal entries) are PSD, the matrix difference is not. $\endgroup$ – user1551 Sep 13 '13 at 16:24

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