# Matrix eigenvalues inequality (1)

Assume that $$A$$ is a $$n\times n$$ positive matrix, whose eigenvalues are $$a_1\ge a_2\ldots \ge a_n>0;$$ $$B$$ is a $$p \times p$$ positive matrix, whose eigenvalues are $$b_1\ge b_2\ldots \ge b_p>0.$$ For $$n\times p$$ full rank matrix $$X$$ with $$n\ge p,$$

How to prove $$\det(X'AX+B)\ge c(X)\prod_{i=1}^p(a_{n-p+i}+b_{i})，$$

where $$c(X)$$ is a postive constant that depends only on $$X$$.

• Maybe you have something more precise in mind. For instance, $c(X)=0$ always works. – Suvrit Jan 28 '19 at 19:18
• @Suvrit Thanks. $C(X)>0.$ – Xiaopai Song Jan 28 '19 at 20:41
• What do we get if $X=0$? – Fedor Petrov Jan 30 '19 at 21:28
• @Fedor Petrov Thank you again. It's full rank. – Xiaopai Song Jan 30 '19 at 21:57
• @Denis Serre Thanks for your edition. Are you interesting in another inequality mathoverflow.net/q/322262/134602? – Xiaopai Song Feb 2 '19 at 15:29

Your both conjectural inequalities are equivalent to each other and false even for $$n=p$$, $$A=B$$, $$X=I$$ (and $$a_i=\lambda^i$$ for large $$\lambda$$, for example).

What is true that $$|X'AX+B|\geqslant c(X)\prod_{i=1}^p (a_{n-p+i}+b_i).$$

Proof. Let $$\lambda_1\geqslant \lambda_{2}\geqslant \ldots \geqslant \lambda_p$$ be eigenvalues of $$X'AX+B$$. It suffices to prove that $$\lambda_{i}\geqslant \rho(b_i+a_{n-p+i})\quad (*)$$ for all $$i=1,\dots,p$$, where $$\rho=\rho(X)$$ depends only on $$X$$. Consider two cases.

1) $$b_i\geqslant a_{n-p+i}$$. Then $$(*)$$ holds with $$\rho=1/2$$, since $$X'AX+B\geqslant B$$, thus $$\lambda_i\geqslant b_i$$.

2) $$a_{n-p+i}\geqslant b_i$$. Let $$L$$ be the image of the operator (identified with the matrix) $$X$$, $$\dim L=p$$ and $$X$$ is a linear isomorphism onto $$L$$. Denote by $$4\rho^2$$, $$\rho>0$$, the norm of the inverse map $$X^{-1}:L\mapsto X$$. The intersection $$L_i$$ of $$L$$ and the space generated by $$u_1,\dots,u_{n-p+i}$$, where $$Au_i=a_i u_i$$ and $$u_i$$ are orthogonal eigenvectors of $$A$$, has dimension at least $$i$$. We have $$((X'AX+B) x,x)\geqslant (X'AX x,x)=(AXx,Xx)\geqslant a_{n-p+i}(Xx,Xx)\geqslant 2\rho a_{n-p+i}(x,x)$$ whenever $$x\in X^{-1}L_i.$$ Therefore by the variational (Courant) principle $$\lambda_i\geqslant 2\rho(X)a_{n-p+i}\geqslant \rho(a_{n-p+i}+b_i)$$.

• Thank you again. Your answer is right. – Xiaopai Song Jan 31 '19 at 17:12
• Are you interested in other similar inequalities? mathoverflow.net/q/322262/134602 and mathoverflow.net/q/322265/134602. For now, I can't prove it. – Xiaopai Song Feb 2 '19 at 3:36
• Again, estimate consecutive eigenvalues of all involved matrices using variational principle – Fedor Petrov Feb 2 '19 at 9:24
• Thanks for your constructive comment. I try to handle the inequality (2) mathoverflow.net/q/322262/134602 and makes comments in it. I am not familiar with the variational principle. So I can't follow you overall. If you have time, could you give some comments or answer it? Your comments are very important to me. – Xiaopai Song Feb 2 '19 at 15:17
• I mean en.wikipedia.org/wiki/Min-max_theorem In other words, in order to estimate the $i$-th eigenvalue from below you should find a subspace of dimension $i$ onto which you have a lower bound for the quadratic form – Fedor Petrov Feb 2 '19 at 19:23