Prove the matrix is positive

Question

Consider the matrix $A=\begin{bmatrix} 1 & 1/2 & 1/3 &\dots &1/n \\ 1/2 & 1/3 & 1/4 &\dots &1/(n+1) \\ \vdots & \vdots & \vdots & \vdots & \vdots\\ 1/n & 1/(n+1) & 1/(n+2) & \dots& 1/(2n-1) \end{bmatrix}$

Prove that $A$ is positive.

My work: $A$ is diagonalisable, symmetric but I can't seem to put these facts togheter to help me. I tried to prove by induction (a naive attempt) that the determinant of its minors is always positive but knowing $det(A^{k,k})>0$ there is no information of $det(A^{k+1,k+1}).

A pretty quick way is to see $A$ as a Gram matrix of an inner product under some basis. — xbh, Nov 13 '18 at 01:44
That was fast, thank you! Can you please post it as answer so I can accept it? — 2ndYearFreshman, Nov 13 '18 at 01:50
@2ndYearFreshman if you wish someone to be notified of your reply to a comment, begin your reply with an @ sign and the beginning of that username. — Will Jagy, Nov 13 '18 at 02:02
@SangchulLee I think the OP here is addressing you, requesting you to post a full answer — Will Jagy, Nov 13 '18 at 02:02

score 2 · Answer 1 · answered Nov 13 '18 at 01:55

tried two , Sylvester Inertia

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ \frac{ 1 }{ 3 } & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 60 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & \frac{ 1 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 60 & 30 & 20 \\ 30 & 20 & 15 \\ 20 & 15 & 12 \\ \end{array} \right) $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 1 & 0 \\ \frac{ 1 }{ 4 } & \frac{ 9 }{ 10 } & \frac{ 3 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 420 & 0 & 0 & 0 \\ 0 & 35 & 0 & 0 \\ 0 & 0 & \frac{ 7 }{ 3 } & 0 \\ 0 & 0 & 0 & \frac{ 3 }{ 20 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 1 & \frac{ 9 }{ 10 } \\ 0 & 0 & 1 & \frac{ 3 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 420 & 210 & 140 & 105 \\ 210 & 140 & 105 & 84 \\ 140 & 105 & 84 & 70 \\ 105 & 84 & 70 & 60 \\ \end{array} \right) $$

Upvoted because it taught me the Sylvester Inertia. – 2ndYearFreshman Nov 13 '18 at 20:26 — 2ndYearFreshman, Nov 13 '18 at 20:26

score 2 · Accepted Answer · answered Nov 13 '18 at 02:05

(Migrated from comment)

The matrix in question is called the Hilbert matrix. To see that $A$ is positive-definite, let $x \in \mathbb{R}^n$. Then

$$ x^T A x = \sum_{i, j = 1}^{n} \frac{x_i x_j}{i+j-1} = \sum_{i, j = 1}^{n} \int_{0}^{1} t^{i+j-2} x_i x_j \, dt = \int_{0}^{1} \left( \sum_{i=1}^{n} x_i t^{i-1} \right)^2 \, dt \geq 0 $$

Moreover, the equality in the last step holds if and only if $\sum_{i=1}^{n} x_i t^{i-1} \equiv 0$ on $t \in [0, 1]$, which is equivalent to $x = 0$. Therefore $A$ is positive-definite as required.

score 1 · Answer 3 · answered Nov 13 '18 at 02:12

This is not a new answer. It is essentially the same ideas as already presented by Sangchool Lee. I'm giving it just as yet another application of the identity

$$ \frac{1}{x} = \int_0^\infty e^{-sx} ds.$$

Observe:

$A_{i,j} = \frac{1}{i+j-1}$.

\begin{align*} (Av,v) &= \sum_{i,j} \frac{v_j v_i}{i+j-1}\\ & = \sum_{i,j} \int_0^\infty (e^{-(j+1/2)s} v_j) (e^{-(i+1/2)s}v_i) ds \\ & = \int_0^\infty \sum_{i,j} (e^{-(j+1/2)s} v_j) (e^{-(i+1/2)s}v_i)ds \\ & = \int_0^\infty \sum_{j} (e^{-(j+1/2)s} v_j)^2 ds\\ & \ge 0 \end{align*}

Note that the proof works whenever $A_{i,j} = 1/(f(i) + f(j))$ where $f$ is a positive function.

Prove the matrix is positive

3 Answers3