Cfrgtkky

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}).

(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.

tried two , Sylvester Inertia

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

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

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.