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Suppose $G$ is a Hermitian $n \times n$ matrix and $A$ is some $n \times n$ matrix over complex numbers. such that $G-A^H G A$ is positive-definite.

Then can we show that $G$ is invertible?

Also, can we conclude anything about the eigenvalues of the matrix $A$ in terms of the eigenvalues of $G$ (for instance, relate the number of positive eigenvalues of $G$ with the eigenvalues of $A$ of norm less than 1)?

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4 Answers 4

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Since $G-A^*GA$ is Hermitian and positive definite, we can find $m>0$ such that $$\forall x\in\Bbb C^n,\quad x^*Gx-(Ax)^*G(Ax)\geq m\lVert x\rVert^2.$$ This gives, by induction, that for all $x\in\Bbb C^n$, $$(A^px)^*G(A^px)\leq x^*Gx-m\sum_{j=0}^{p-1}\lVert A^jx\rVert^2.$$ The initial assumption gives that $G-(A^p)^*GA^p\geq 0$ and that each eigenvalue of $A$ is of modulus $<1$. Therefore, using Jordan form of $A$, we get that $G\geq 0$. We get $$m\sum_{j=0}^{p-1}\lVert A^jx\rVert^2\leq x^*Gx,$$ and the wanted result.

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I am not clear about the conclusion that each eigenvalue of $A$ is less than $1$, and the statement about the Jordan form. Could you please clarify these? –  BharatRam Sep 15 '12 at 12:27
    
It's unfortunately not necessarily true for example when $G=-I$ and $A=2I$. So my argument works if $G\geq 0$, but it can be deduced of the others. I conclude I need to think more about the problem. –  Davide Giraudo Sep 15 '12 at 17:01

Let $G$ be non invertible matrix. Then singular value decomposition of $G$, $G=USV^T$ will have at least one singular value $\sigma_n=0$.

Now consider the singluar value decomposition of $G-A^HGA=U^{'}S^{'}V^{'T}$ where we have $\sigma^{'}_i\neq 0$ for all $i$.

As $G$ has singularity, $A^HGA$ shouldnt such that $G-A^HGA$ is positive definite. From this, we can write the SVD of $A^HGA$ as

$$A^HGA=(U^*S^*V^{*})^HUSV^TU^*S^*V^{*}=usv^T$$

where

$s=S^*S^*S$

which has $\sigma_n=0$. Therefore $G$ can not be non-invertible.

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So you have decomposed $G$, $A$, and $A^H G A$ into their singular value decompositions.I couldn't follow your conclusions. If possible, could you please clarify it more? –  BharatRam Sep 2 '12 at 4:11
    
Yes I can. If $G$ is invertible it shouldnt have any singular value equals to $0$. Likewise if it is noninvertible it should have at least one singular value (that is the smallest one) equals to zero. Now I assume that $G$ is non-invertible and check if this provides a contradiction with the restrcition that $G-A^HGA>0$. If it does then I can conclude that $G$ should be invertible. Now I check $G-A^HGA$ with $G$ has singular value decomposition with at least one singular value equals to $0$. Since $G-A^HGA$ is positive definite I also know that it shouldnt have any singular value equals to $0$ –  Seyhmus Güngören Sep 2 '12 at 9:32
    
Now I must force $G-A^HGA$ to have no singular value equals to zero, given $G$ has at least one singular value equals to zero. As I showed that this is impossible, I can conclude that if $G$ is non-invertible then $G-A^HGA$ can not be positive definite as positive definite matix doesnt allow any zero singular value. Therefore I finally conclude that $G$ should be positive definite such that $G-A^HGA>0$ –  Seyhmus Güngören Sep 2 '12 at 9:38

$G-A^TGA$ is positive definite, so $$x^T(G-A^TGA)x>0$$ or $$x^TGx>x^TA^TGAx$$ for any $x \in \mathbb{C}^n$.

Then suppose $G$ is not invertible, then there exists $y \in \mathbb{C}^n$ such that $Gy=0$, then we have $$y^TA^TGAy<y^TGy=0$$

Since $A$ is arbitrary, the left side of the above inequality means $G$ is negative definite, which is contradict to the assumption that $G$ is not invertible.

So $G$ is invertible.

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$A$ is some fixed matrix such that $G-A^H G A$ is positive definite. I did not follow how you concluded that $G$ is negative definite. –  BharatRam Sep 2 '12 at 1:34
    
In your question, it says $A$ is any n∗n matrix over complex numbers –  chaohuang Sep 2 '12 at 1:37
    
Well, it still has to satisfy the positive definiteness condition of $G-A^H G A$. It cannot be just any matrix. I shall edit it if thats ambiguous. –  BharatRam Sep 2 '12 at 1:41

Another solution would be to show that $G$ does not admit the eigenvalue $0$. (Then G $is$ invertible because it is hermitien). This can be shown by contradiction: Let $x$ be the eigenvector to the eigenvalue $0$ of $G$. Then $x^t(G-A^tGA)x^t>0$ gives (since $Gx=0$):

$-(Ax)^tG(Ax)>0$.

Now you can argue in the same way as chaohuang.

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How do you proceed with that? chaohuang's proof is incorrect. If G has negative eigenvalues, your hypothesis will certainly be true anyway. How does it prove that G is invertible? –  BharatRam Sep 9 '12 at 11:25
    
Yes, chaohuang is of course false, I did not read it properly. So in fact this does not work properly how I tried. –  Lucien Sep 10 '12 at 12:11

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