Mixing
Prove the nxn matrix is positive
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Im trying to solve this exercise:
Let $n$ be a positive integer and A the $nxn$ matrix
$A=begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}$
Prove that A is positive.
Already notice that $A$ is symmetric.
If $X=(x_1,dots,x_n)neq 0$ im trying to prove $X^TAX>0$.
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}
begin{bmatrix}
x_1\
x_2\
vdots\
x_n
end{bmatrix}=$
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
x_1+ x_2/2 + x_3/3 +dots + x_n/n \
x_1/2+x_2/3 + x_3/4 + dots + x_n/(n+1) \
vdots \
x_1/n + x_2/(n+1) + x_3/(n+2) + dots + x_n/(2n-1)
end{bmatrix}=$
$x_1^2+frac{x_1x_2}{2}+frac{x_1x_3}{3}+dots+frac{x_1x_n}{n}+frac{x_2^2}{3}+frac{x_2x_3}{4}+dots+frac{x_2x_n}{n+1}+dots+frac{x_nx_1}{n}+dots+frac{x_n^2}{2n-1}$
Its clear to me that the sum has positive adds ($x_i^2$), but also have negative adds... my problem is how conclude this sum is positive.
linear-algebra
asked yesterday
Malena Manzanares
243
add a comment |
up vote
0
down vote
favorite
Im trying to solve this exercise:
Let $n$ be a positive integer and A the $nxn$ matrix
$A=begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}$
Prove that A is positive.
Already notice that $A$ is symmetric.
If $X=(x_1,dots,x_n)neq 0$ im trying to prove $X^TAX>0$.
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}
begin{bmatrix}
x_1\
x_2\
vdots\
x_n
end{bmatrix}=$
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
x_1+ x_2/2 + x_3/3 +dots + x_n/n \
x_1/2+x_2/3 + x_3/4 + dots + x_n/(n+1) \
vdots \
x_1/n + x_2/(n+1) + x_3/(n+2) + dots + x_n/(2n-1)
end{bmatrix}=$
$x_1^2+frac{x_1x_2}{2}+frac{x_1x_3}{3}+dots+frac{x_1x_n}{n}+frac{x_2^2}{3}+frac{x_2x_3}{4}+dots+frac{x_2x_n}{n+1}+dots+frac{x_nx_1}{n}+dots+frac{x_n^2}{2n-1}$
Its clear to me that the sum has positive adds ($x_i^2$), but also have negative adds... my problem is how conclude this sum is positive.
linear-algebra
asked yesterday
Malena Manzanares
243
I think you should try using determinant.
– Apocalypse
yesterday
2
Possible duplicate as this.
– xbh
yesterday
1
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Im trying to solve this exercise:
Let $n$ be a positive integer and A the $nxn$ matrix
$A=begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}$
Prove that A is positive.
Already notice that $A$ is symmetric.
If $X=(x_1,dots,x_n)neq 0$ im trying to prove $X^TAX>0$.
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}
begin{bmatrix}
x_1\
x_2\
vdots\
x_n
end{bmatrix}=$
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
x_1+ x_2/2 + x_3/3 +dots + x_n/n \
x_1/2+x_2/3 + x_3/4 + dots + x_n/(n+1) \
vdots \
x_1/n + x_2/(n+1) + x_3/(n+2) + dots + x_n/(2n-1)
end{bmatrix}=$
$x_1^2+frac{x_1x_2}{2}+frac{x_1x_3}{3}+dots+frac{x_1x_n}{n}+frac{x_2^2}{3}+frac{x_2x_3}{4}+dots+frac{x_2x_n}{n+1}+dots+frac{x_nx_1}{n}+dots+frac{x_n^2}{2n-1}$
Its clear to me that the sum has positive adds ($x_i^2$), but also have negative adds... my problem is how conclude this sum is positive.
linear-algebra
asked yesterday
Malena Manzanares
243
Im trying to solve this exercise:
Let $n$ be a positive integer and A the $nxn$ matrix
$A=begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}$
Prove that A is positive.
Already notice that $A$ is symmetric.
If $X=(x_1,dots,x_n)neq 0$ im trying to prove $X^TAX>0$.
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
1 & 1/2 & 1/3 &dots & 1/n \
1/2 & 1/3 & 1/4 & dots & 1/(n+1) \
vdots \
1/n & 1/(n+1) & 1/(n+2) & dots & 1/(2n-1)
end{bmatrix}
begin{bmatrix}
x_1\
x_2\
vdots\
x_n
end{bmatrix}=$
$begin{bmatrix}
x_1&x_2&dots&x_n\
end{bmatrix}
begin{bmatrix}
x_1+ x_2/2 + x_3/3 +dots + x_n/n \
x_1/2+x_2/3 + x_3/4 + dots + x_n/(n+1) \
vdots \
x_1/n + x_2/(n+1) + x_3/(n+2) + dots + x_n/(2n-1)
end{bmatrix}=$
$x_1^2+frac{x_1x_2}{2}+frac{x_1x_3}{3}+dots+frac{x_1x_n}{n}+frac{x_2^2}{3}+frac{x_2x_3}{4}+dots+frac{x_2x_n}{n+1}+dots+frac{x_nx_1}{n}+dots+frac{x_n^2}{2n-1}$
Its clear to me that the sum has positive adds ($x_i^2$), but also have negative adds... my problem is how conclude this sum is positive.
linear-algebra
linear-algebra
asked yesterday
Malena Manzanares
243
asked yesterday
Malena Manzanares
243
asked yesterday
Malena Manzanares
243
asked yesterday
Malena Manzanares
243
asked yesterday
Malena Manzanares
243
243
I think you should try using determinant.
– Apocalypse
yesterday
2
Possible duplicate as this.
– xbh
yesterday
1
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday
add a comment |
I think you should try using determinant.
– Apocalypse
yesterday
2
Possible duplicate as this.
– xbh
yesterday
1
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday
I think you should try using determinant.
– Apocalypse
yesterday
I think you should try using determinant.
– Apocalypse
yesterday
2
2
Possible duplicate as this.
– xbh
yesterday
Possible duplicate as this.
– xbh
yesterday
1
1
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday
add a comment |
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I think you should try using determinant.
– Apocalypse
yesterday
2
Possible duplicate as this.
– xbh
yesterday
1
en.wikipedia.org/wiki/Hilbert_matrix
– Will Jagy
yesterday
Thank !!! Actually is a duplicate, so sorry. Thanks again!!
– Malena Manzanares
yesterday