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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?

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    $\begingroup$ Non-negative definite? Is that another way of saying "positive semidefinite"? $\endgroup$ Commented Sep 17, 2012 at 2:57
  • $\begingroup$ Where is the problem from? $\endgroup$ Commented Sep 17, 2012 at 3:42

3 Answers 3

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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.

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  • $\begingroup$ Just because the determinant is positive, it doesn't mean that the matrix is PSD. $\endgroup$
    – S.B.
    Commented Sep 17, 2012 at 18:10
  • $\begingroup$ @S.B. positive determinant for all top left corners; see en.wikipedia.org/wiki/Sylvester%27s_criterion $\endgroup$
    – user8268
    Commented Sep 17, 2012 at 20:27
  • $\begingroup$ I see, I wasn't aware of this theorem. $\endgroup$
    – S.B.
    Commented Sep 17, 2012 at 22:14
  • $\begingroup$ 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. $\endgroup$
    – dsantanu
    Commented Sep 18, 2012 at 18:05
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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.

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  • $\begingroup$ 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. $\endgroup$
    – dsantanu
    Commented Sep 17, 2012 at 3:47
  • $\begingroup$ There is a relationship between eigenvalues and determinants. $\endgroup$
    – Emily
    Commented Sep 17, 2012 at 3:48
  • $\begingroup$ product of eigenvalues == determinant.Its not a home work problem.It came up while proving something else. $\endgroup$
    – dsantanu
    Commented Sep 17, 2012 at 4:04
  • $\begingroup$ @user41518 Why do you think it is positive semidefinite? $\endgroup$ Commented Sep 17, 2012 at 4:05
  • $\begingroup$ I have made numerical experiments!besides its equivalent to another matrix C' having same property.C' has (i,j)-th entry $\endgroup$
    – dsantanu
    Commented Sep 17, 2012 at 4:19
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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.

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  • $\begingroup$ 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 $\endgroup$
    – user8268
    Commented Sep 17, 2012 at 21:26
  • $\begingroup$ yes, it is beautiful! $\endgroup$
    – dsantanu
    Commented Sep 18, 2012 at 18:06

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