Mixing
Patterns for eigenvalues of Vandermonde matrix
$begingroup$
Let $A$ be a Vandermonde type matrix
$A =
begin{bmatrix}
1 & 1 & dots & 1 \
x_1 & x_2 &dots & x_n \
dots& dots & dots& dots\
x_1^{n-1} &x_2^{n-1} &dots & x_n^{n-1}
end{bmatrix}$
When I was testing some such real matrices with positive entries for their eigenvalues I have noticed that eigenvalues are also real and positive.
- How such property (if true in general case) can be explained ?
Moreover in examples which I tested (when it was assumed for $i<j$ that $x_i <x_j$) the greatest eigenvalue was always close to the greatest value in this matrix and the smallest one always less than $1$.
- How to explain also these facts ?
Examples of $3 times 3$ matrices generated from natural numbers
$A =
begin{bmatrix}
1 & 1 & 1 \
2 & 3 & 5 \
4 & 9 & 25
end{bmatrix}$
Eigenvalues: ${ 27.09 , 0.12 , 1.79 } $
$A =
begin{bmatrix}
1 & 1 & 1 \
3 & 7 & 11 \
9 & 49 & 121
end{bmatrix}$
Eigenvalues: ${ 125.63, 0.37 , 3.03 } $
Of course if two eigenvalues in the examples above are rather small then the third must be rather great from the trace of the matrix, but why these two must be small ?
- If the answer for general case is hard to find let dimension of a
matrix be specific i.e. $ 3 times 3$ and entries only natural.
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
$endgroup$
|
show 1 more comment
$begingroup$
Let $A$ be a Vandermonde type matrix
$A =
begin{bmatrix}
1 & 1 & dots & 1 \
x_1 & x_2 &dots & x_n \
dots& dots & dots& dots\
x_1^{n-1} &x_2^{n-1} &dots & x_n^{n-1}
end{bmatrix}$
When I was testing some such real matrices with positive entries for their eigenvalues I have noticed that eigenvalues are also real and positive.
- How such property (if true in general case) can be explained ?
Moreover in examples which I tested (when it was assumed for $i<j$ that $x_i <x_j$) the greatest eigenvalue was always close to the greatest value in this matrix and the smallest one always less than $1$.
- How to explain also these facts ?
Examples of $3 times 3$ matrices generated from natural numbers
$A =
begin{bmatrix}
1 & 1 & 1 \
2 & 3 & 5 \
4 & 9 & 25
end{bmatrix}$
Eigenvalues: ${ 27.09 , 0.12 , 1.79 } $
$A =
begin{bmatrix}
1 & 1 & 1 \
3 & 7 & 11 \
9 & 49 & 121
end{bmatrix}$
Eigenvalues: ${ 125.63, 0.37 , 3.03 } $
Of course if two eigenvalues in the examples above are rather small then the third must be rather great from the trace of the matrix, but why these two must be small ?
- If the answer for general case is hard to find let dimension of a
matrix be specific i.e. $ 3 times 3$ and entries only natural.
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
$endgroup$
1
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25
|
show 1 more comment
$begingroup$
Let $A$ be a Vandermonde type matrix
$A =
begin{bmatrix}
1 & 1 & dots & 1 \
x_1 & x_2 &dots & x_n \
dots& dots & dots& dots\
x_1^{n-1} &x_2^{n-1} &dots & x_n^{n-1}
end{bmatrix}$
When I was testing some such real matrices with positive entries for their eigenvalues I have noticed that eigenvalues are also real and positive.
- How such property (if true in general case) can be explained ?
Moreover in examples which I tested (when it was assumed for $i<j$ that $x_i <x_j$) the greatest eigenvalue was always close to the greatest value in this matrix and the smallest one always less than $1$.
- How to explain also these facts ?
Examples of $3 times 3$ matrices generated from natural numbers
$A =
begin{bmatrix}
1 & 1 & 1 \
2 & 3 & 5 \
4 & 9 & 25
end{bmatrix}$
Eigenvalues: ${ 27.09 , 0.12 , 1.79 } $
$A =
begin{bmatrix}
1 & 1 & 1 \
3 & 7 & 11 \
9 & 49 & 121
end{bmatrix}$
Eigenvalues: ${ 125.63, 0.37 , 3.03 } $
Of course if two eigenvalues in the examples above are rather small then the third must be rather great from the trace of the matrix, but why these two must be small ?
- If the answer for general case is hard to find let dimension of a
matrix be specific i.e. $ 3 times 3$ and entries only natural.
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
$endgroup$
Let $A$ be a Vandermonde type matrix
$A =
begin{bmatrix}
1 & 1 & dots & 1 \
x_1 & x_2 &dots & x_n \
dots& dots & dots& dots\
x_1^{n-1} &x_2^{n-1} &dots & x_n^{n-1}
end{bmatrix}$
When I was testing some such real matrices with positive entries for their eigenvalues I have noticed that eigenvalues are also real and positive.
- How such property (if true in general case) can be explained ?
Moreover in examples which I tested (when it was assumed for $i<j$ that $x_i <x_j$) the greatest eigenvalue was always close to the greatest value in this matrix and the smallest one always less than $1$.
- How to explain also these facts ?
Examples of $3 times 3$ matrices generated from natural numbers
$A =
begin{bmatrix}
1 & 1 & 1 \
2 & 3 & 5 \
4 & 9 & 25
end{bmatrix}$
Eigenvalues: ${ 27.09 , 0.12 , 1.79 } $
$A =
begin{bmatrix}
1 & 1 & 1 \
3 & 7 & 11 \
9 & 49 & 121
end{bmatrix}$
Eigenvalues: ${ 125.63, 0.37 , 3.03 } $
Of course if two eigenvalues in the examples above are rather small then the third must be rather great from the trace of the matrix, but why these two must be small ?
- If the answer for general case is hard to find let dimension of a
matrix be specific i.e. $ 3 times 3$ and entries only natural.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
edited Mar 9 '17 at 8:57
Widawensen
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
asked Mar 8 '17 at 14:49
WidawensenWidawensen
4,50321446
4,50321446
1
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25
|
show 1 more comment
1
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25
1
1
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix
answered Mar 8 '17 at 14:55
shdpshdp
32114
$endgroup$
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
add a comment |
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$begingroup$
If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix
answered Mar 8 '17 at 14:55
shdpshdp
32114
$endgroup$
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
add a comment |
$begingroup$
If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix
answered Mar 8 '17 at 14:55
shdpshdp
32114
$endgroup$
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
add a comment |
$begingroup$
If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix
answered Mar 8 '17 at 14:55
shdpshdp
32114
$endgroup$
If you assume $x_{i}<x_{j}$ for $i<j$, it will always be positive, https://en.wikipedia.org/wiki/Vandermonde_matrix
answered Mar 8 '17 at 14:55
shdpshdp
32114
answered Mar 8 '17 at 14:55
shdpshdp
32114
answered Mar 8 '17 at 14:55
shdpshdp
32114
answered Mar 8 '17 at 14:55
shdpshdp
32114
32114
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
add a comment |
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
$begingroup$
Can you tell us what the "it" is?
$endgroup$
– ancientmathematician
Mar 8 '17 at 14:58
1
1
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
$begingroup$
@shdp Why do you think so? Because the determinant is positive?
$endgroup$
– Widawensen
Mar 8 '17 at 14:59
add a comment |
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1
$begingroup$
Is this helpful? - mathoverflow.net/questions/155845/…
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:07
$begingroup$
@ancientmathematician Yes, it provides some knowledge although not for everything I mentioned here. It seems from mathoverflow that the problem is rather difficult, I'm afraid that explanation there is rather hard for me. What does it mean "strictly totally positive" for the problem( this meaning of minors) I was thinking that the problem is easier..
$endgroup$
– Widawensen
Mar 8 '17 at 15:14
$begingroup$
it means all the minors are positive. If one wants to understand the eigenvalues of the Vandermonde matrix, then I think you'd need to grapple with this stuff. (I may be wrong).
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:20
$begingroup$
@ancientmathematician so the answer lies somewhere in 35-pages paper, hmm ... it can be rather a long way to understanding ..
$endgroup$
– Widawensen
Mar 8 '17 at 15:22
$begingroup$
No royal road to geometry.
$endgroup$
– ancientmathematician
Mar 8 '17 at 15:25