Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let ${x_1,x_2,...x_n}$ be positive numbers. Consider the matrix $C$ whose $(i,j)$-th entry is

$$\min\left\{\frac{x_i}{x_j},\frac{x_j}{x_i}\right\}$$

Show that $C$ is non-negative definite (or positive semidefinite, meaning $z^t C z\geq 0$ for all $z\in \mathbb{R}^n$).When is $C$ positive definite?

share|improve this question
2  
Non-negative definite? Is that another way of saying "positive semidefinite"? –  Rod Carvalho Sep 17 '12 at 2:57
    
Where is the problem from? –  Jonas Meyer Sep 17 '12 at 3:42
    
It smells a bit like a homework problem. If so, please tag it with homework –  Arkamis Sep 17 '12 at 3:45
    
Also, I assume that the $x_i$ comprising the matrix are not the elements of the unknown vector $x$. Might want to check your notation a bit. –  Arkamis Sep 17 '12 at 3:47

3 Answers 3

We can suppose that $x_i\leq x_j$ for $i<j$ (otherwise just permute the components). As I'll show below, the determinant of your matrix is $$\det C_n=\frac{x_2^2-x_1^2}{x_2^2}\frac{x_3^2-x_2^2}{x_3^2}\dots\frac{x_n^2-x_{n-1}^2}{x_n^2},\qquad(*)$$ which is $\geq 0$, and it is $>0$ iff all $x_i$'s are different. Notice that if you take the top left corner of $C$ with $k$ rows and columns, you get $C_k$, and $\det C_k\geq0$. Your matrix is therefore positive semidefinite (by Sylvester criterion), and it is positive definite iff all $x_i$'s are different.

Now we need to prove $(*)$. Let $D_n$ be $C_n$ with $i,k$-th element multiplied by $x_i x_j$, so that $\det D_n=\det C_n\times\prod_i x_i^2$. We want to show

$$\det D_n=x_1^2(x_2^2-x_1^2)\dots(x_n^2-x_{n-1}^2).$$

$D_n$ looks like $$ \begin{pmatrix} x_1^2 & x_1^2& x_1^2&x_1^2\\ x_1^2& x_2^2& x_2^2&x_2^2\\ x_1^2& x_2^2&x_3^2&x_3^2\\ x_1^2& x_2^2&x_3^2&x_4^2 \end{pmatrix} $$ (for $n=4$ - I hope the pattern is clear), i.e. $$ \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}. $$ ($a=x_1^2,\dots,d=x_4^2$). If we now subtract the first row from the others and then the first column from the others, we get $$ \begin{pmatrix} a & 0& 0&0\\ 0& b-a& b-a&b-a\\ 0& b-a&c-a&c-a\\ 0& b-a&c-a&d-a \end{pmatrix}. $$ i.e. $$ \det \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}= a\det \begin{pmatrix} b-a& b-a&b-a\\ b-a&c-a&c-a\\ b-a&c-a&d-a \end{pmatrix} $$ Repeating this identity, we get $$\det D_n= \det \begin{pmatrix} a & a& a&a\\ a& b& b&b\\ a& b&c&c\\ a& b&c&d \end{pmatrix}= a\det \begin{pmatrix} b-a& b-a&b-a\\ b-a&c-a&c-a\\ b-a&c-a&d-a \end{pmatrix}= a(b-a)\det \begin{pmatrix} c-b&c-b\\ c-b&d-b \end{pmatrix}= a(b-a)(c-b)(d-c) $$ as we wanted to show.

There must be a more intelligent solution, but for the moment this one should do.

share|improve this answer
    
Just because the determinant is positive, it doesn't mean that the matrix is PSD. –  S.B. Sep 17 '12 at 18:10
    
@S.B. positive determinant for all top left corners; see en.wikipedia.org/wiki/Sylvester%27s_criterion –  user8268 Sep 17 '12 at 20:27
    
I see, I wasn't aware of this theorem. –  S.B. Sep 17 '12 at 22:14
    
Thanks guys.Both the proofs are elegant.the BM motion proof is more revealing...the sort of thing i was looking for.But the determinant proofs has there own charm too. –  dsantanu Sep 18 '12 at 18:05

First, note that the matrix is symmetric: if $x_j > x_i$, then $C_{ij} = \frac{x_i}{x_j}$. Then, $C_{ji} = \min \{ \frac{x_i}{x_j}, \frac{x_j}{x_i} \} = \frac{x_i}{x_j} = C_{ij}$.

Since it is real and symmetric, it is diagonalizable: $C = V^T D V$. Let $y = Vx$. Then, the problem reduces to finding $y^T D y > 0$. This should be easy to finish from here.

share|improve this answer
    
Thanks.But it is not! One must prove that the eigenvalues(the diagonal entries of D) are non-negative. Finding eigen values is much harder problem. –  dsantanu Sep 17 '12 at 3:47
    
There is a relationship between eigenvalues and determinants. –  Arkamis Sep 17 '12 at 3:48
    
product of eigenvalues == determinant.Its not a home work problem.It came up while proving something else. –  dsantanu Sep 17 '12 at 4:04
    
@user41518 Why do you think it is positive semidefinite? –  Alex Becker Sep 17 '12 at 4:05
    
I have made numerical experiments!besides its equivalent to another matrix C' having same property.C' has (i,j)-th entry –  dsantanu Sep 17 '12 at 4:19

Consider positive real numbers $t_i > 0$ and real numbers $\alpha_j \in \mathbb{R}$.

  • The matrix $M_{ij}=\min(t_i, t_j)$ is positive semi-definite since $\sum z_i M_{ij}z_j = \int_0^{\infty} \big( \sum_i z_i 1_{t<t_i}\big)^2 \, dt$. One can also recognize the covariance matrix of $(B_{t_1}, \ldots, B_{t_n})$ where $B$ is a Brownian motion.
  • If $M=(M_{ij})_{ij}$ is positive semi-definite, so is $N_{ij} = \alpha_i \alpha_j M_{ij}$ since $\sum z_i z_j N_{ij} = \sum y_i y_j M_{ij} \geq 0$ where $y_i = \alpha_i z_i$.

The choice $t_i = x_i^2$ and $\alpha_j = \frac{1}{x_j}$ solves the exercise.

share|improve this answer
    
Beautiful; while I was proving that $M$ (in my answer denoted by $D$) is positive semidefinite via determinants, you gave a direct half-line argument –  user8268 Sep 17 '12 at 21:26
    
yes, it is beautiful! –  dsantanu Sep 18 '12 at 18:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.