Mixing
Is it possible to compute the eigenvalues and eigenvectors if I know the svd?
0
$begingroup$
I have a matrix $A$
A =
2.00000 3.00000 1.00000 0.50000 4.00000
4.00000 5.00000 7.00000 0.10000 1.00000
5.00000 3.00000 6.00000 19.20000 9.00000
1.00000 4.00000 1.00000 4.00000 7.00000
3.00000 1.00000 6.00000 2.00000 6.00000
I did the SVD on matrix $A$.
[U,S,V] = svd(A)
U =
-0.149403 -0.278046 -0.422317 -0.209000 0.823612
-0.198548 -0.731899 0.473681 -0.422812 -0.147509
-0.872587 0.409042 0.244648 -0.055987 0.091043
-0.304238 -0.128976 -0.728923 -0.262419 -0.539085
-0.290325 -0.450625 -0.078065 0.839971 -0.031671
S =
Diagonal Matrix
25.63602 0 0 0 0
0 10.08558 0 0 0
0 0 6.01867 0 0
0 0 0 3.71319 0
0 0 0 0 0.98157
V =
-0.2586652 -0.2894557 0.2176915 -0.0354640 0.8947963
-0.2171159 -0.4197117 -0.1924603 -0.8399028 -0.1850004
-0.3440842 -0.5730769 0.5257015 0.3427795 -0.3991605
-0.7273295 0.6171432 0.2428507 -0.1592856 -0.0760113
-0.4884167 -0.1754271 -0.7617289 0.3878607 0.0027517
But then when I checked the eigenvalues and eigenvectors, they where not the same.
[v, e] = eig(A)
v =
-0.18903 + 0.00000i 0.15906 - 0.21827i 0.15906 + 0.21827i -0.87710 + 0.00000i -0.20055 + 0.00000i
-0.39291 + 0.00000i 0.25894 + 0.33885i 0.25894 - 0.33885i 0.15307 + 0.00000i -0.83881 + 0.00000i
-0.74945 + 0.00000i -0.69885 + 0.00000i -0.69885 - 0.00000i 0.42831 + 0.00000i 0.12425 + 0.00000i
-0.30382 + 0.00000i 0.12855 - 0.36818i 0.12855 + 0.36818i 0.13432 + 0.00000i -0.05592 + 0.00000i
-0.39484 + 0.00000i 0.17079 + 0.27486i 0.17079 - 0.27486i -0.07597 + 0.00000i 0.48746 + 0.00000i
e =
Diagonal Matrix
21.3589 + 0.0000i 0 0 0 0
0 -1.9810 + 6.6826i 0 0 0
0 0 -1.9810 - 6.6826i 0 0
0 0 0 1.2580 + 0.0000i 0
0 0 0 0 4.3450 + 0.0000i
As you can see, the matrix $A$ have complex eigenvalues, but not in the $S$ matrix of SVD. Why? Is it possible for me to compute the eigenvalues and eigenvectors from svd if I know $A,U,S,V$?
matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
$endgroup$
|
show 5 more comments
0
$begingroup$
I have a matrix $A$
A =
2.00000 3.00000 1.00000 0.50000 4.00000
4.00000 5.00000 7.00000 0.10000 1.00000
5.00000 3.00000 6.00000 19.20000 9.00000
1.00000 4.00000 1.00000 4.00000 7.00000
3.00000 1.00000 6.00000 2.00000 6.00000
I did the SVD on matrix $A$.
[U,S,V] = svd(A)
U =
-0.149403 -0.278046 -0.422317 -0.209000 0.823612
-0.198548 -0.731899 0.473681 -0.422812 -0.147509
-0.872587 0.409042 0.244648 -0.055987 0.091043
-0.304238 -0.128976 -0.728923 -0.262419 -0.539085
-0.290325 -0.450625 -0.078065 0.839971 -0.031671
S =
Diagonal Matrix
25.63602 0 0 0 0
0 10.08558 0 0 0
0 0 6.01867 0 0
0 0 0 3.71319 0
0 0 0 0 0.98157
V =
-0.2586652 -0.2894557 0.2176915 -0.0354640 0.8947963
-0.2171159 -0.4197117 -0.1924603 -0.8399028 -0.1850004
-0.3440842 -0.5730769 0.5257015 0.3427795 -0.3991605
-0.7273295 0.6171432 0.2428507 -0.1592856 -0.0760113
-0.4884167 -0.1754271 -0.7617289 0.3878607 0.0027517
But then when I checked the eigenvalues and eigenvectors, they where not the same.
[v, e] = eig(A)
v =
-0.18903 + 0.00000i 0.15906 - 0.21827i 0.15906 + 0.21827i -0.87710 + 0.00000i -0.20055 + 0.00000i
-0.39291 + 0.00000i 0.25894 + 0.33885i 0.25894 - 0.33885i 0.15307 + 0.00000i -0.83881 + 0.00000i
-0.74945 + 0.00000i -0.69885 + 0.00000i -0.69885 - 0.00000i 0.42831 + 0.00000i 0.12425 + 0.00000i
-0.30382 + 0.00000i 0.12855 - 0.36818i 0.12855 + 0.36818i 0.13432 + 0.00000i -0.05592 + 0.00000i
-0.39484 + 0.00000i 0.17079 + 0.27486i 0.17079 - 0.27486i -0.07597 + 0.00000i 0.48746 + 0.00000i
e =
Diagonal Matrix
21.3589 + 0.0000i 0 0 0 0
0 -1.9810 + 6.6826i 0 0 0
0 0 -1.9810 - 6.6826i 0 0
0 0 0 1.2580 + 0.0000i 0
0 0 0 0 4.3450 + 0.0000i
As you can see, the matrix $A$ have complex eigenvalues, but not in the $S$ matrix of SVD. Why? Is it possible for me to compute the eigenvalues and eigenvectors from svd if I know $A,U,S,V$?
matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
$endgroup$
$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
2
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14
|
show 5 more comments
0
0
0
$begingroup$
I have a matrix $A$
A =
2.00000 3.00000 1.00000 0.50000 4.00000
4.00000 5.00000 7.00000 0.10000 1.00000
5.00000 3.00000 6.00000 19.20000 9.00000
1.00000 4.00000 1.00000 4.00000 7.00000
3.00000 1.00000 6.00000 2.00000 6.00000
I did the SVD on matrix $A$.
[U,S,V] = svd(A)
U =
-0.149403 -0.278046 -0.422317 -0.209000 0.823612
-0.198548 -0.731899 0.473681 -0.422812 -0.147509
-0.872587 0.409042 0.244648 -0.055987 0.091043
-0.304238 -0.128976 -0.728923 -0.262419 -0.539085
-0.290325 -0.450625 -0.078065 0.839971 -0.031671
S =
Diagonal Matrix
25.63602 0 0 0 0
0 10.08558 0 0 0
0 0 6.01867 0 0
0 0 0 3.71319 0
0 0 0 0 0.98157
V =
-0.2586652 -0.2894557 0.2176915 -0.0354640 0.8947963
-0.2171159 -0.4197117 -0.1924603 -0.8399028 -0.1850004
-0.3440842 -0.5730769 0.5257015 0.3427795 -0.3991605
-0.7273295 0.6171432 0.2428507 -0.1592856 -0.0760113
-0.4884167 -0.1754271 -0.7617289 0.3878607 0.0027517
But then when I checked the eigenvalues and eigenvectors, they where not the same.
[v, e] = eig(A)
v =
-0.18903 + 0.00000i 0.15906 - 0.21827i 0.15906 + 0.21827i -0.87710 + 0.00000i -0.20055 + 0.00000i
-0.39291 + 0.00000i 0.25894 + 0.33885i 0.25894 - 0.33885i 0.15307 + 0.00000i -0.83881 + 0.00000i
-0.74945 + 0.00000i -0.69885 + 0.00000i -0.69885 - 0.00000i 0.42831 + 0.00000i 0.12425 + 0.00000i
-0.30382 + 0.00000i 0.12855 - 0.36818i 0.12855 + 0.36818i 0.13432 + 0.00000i -0.05592 + 0.00000i
-0.39484 + 0.00000i 0.17079 + 0.27486i 0.17079 - 0.27486i -0.07597 + 0.00000i 0.48746 + 0.00000i
e =
Diagonal Matrix
21.3589 + 0.0000i 0 0 0 0
0 -1.9810 + 6.6826i 0 0 0
0 0 -1.9810 - 6.6826i 0 0
0 0 0 1.2580 + 0.0000i 0
0 0 0 0 4.3450 + 0.0000i
As you can see, the matrix $A$ have complex eigenvalues, but not in the $S$ matrix of SVD. Why? Is it possible for me to compute the eigenvalues and eigenvectors from svd if I know $A,U,S,V$?
matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
$endgroup$
I have a matrix $A$
A =
2.00000 3.00000 1.00000 0.50000 4.00000
4.00000 5.00000 7.00000 0.10000 1.00000
5.00000 3.00000 6.00000 19.20000 9.00000
1.00000 4.00000 1.00000 4.00000 7.00000
3.00000 1.00000 6.00000 2.00000 6.00000
I did the SVD on matrix $A$.
[U,S,V] = svd(A)
U =
-0.149403 -0.278046 -0.422317 -0.209000 0.823612
-0.198548 -0.731899 0.473681 -0.422812 -0.147509
-0.872587 0.409042 0.244648 -0.055987 0.091043
-0.304238 -0.128976 -0.728923 -0.262419 -0.539085
-0.290325 -0.450625 -0.078065 0.839971 -0.031671
S =
Diagonal Matrix
25.63602 0 0 0 0
0 10.08558 0 0 0
0 0 6.01867 0 0
0 0 0 3.71319 0
0 0 0 0 0.98157
V =
-0.2586652 -0.2894557 0.2176915 -0.0354640 0.8947963
-0.2171159 -0.4197117 -0.1924603 -0.8399028 -0.1850004
-0.3440842 -0.5730769 0.5257015 0.3427795 -0.3991605
-0.7273295 0.6171432 0.2428507 -0.1592856 -0.0760113
-0.4884167 -0.1754271 -0.7617289 0.3878607 0.0027517
But then when I checked the eigenvalues and eigenvectors, they where not the same.
[v, e] = eig(A)
v =
-0.18903 + 0.00000i 0.15906 - 0.21827i 0.15906 + 0.21827i -0.87710 + 0.00000i -0.20055 + 0.00000i
-0.39291 + 0.00000i 0.25894 + 0.33885i 0.25894 - 0.33885i 0.15307 + 0.00000i -0.83881 + 0.00000i
-0.74945 + 0.00000i -0.69885 + 0.00000i -0.69885 - 0.00000i 0.42831 + 0.00000i 0.12425 + 0.00000i
-0.30382 + 0.00000i 0.12855 - 0.36818i 0.12855 + 0.36818i 0.13432 + 0.00000i -0.05592 + 0.00000i
-0.39484 + 0.00000i 0.17079 + 0.27486i 0.17079 - 0.27486i -0.07597 + 0.00000i 0.48746 + 0.00000i
e =
Diagonal Matrix
21.3589 + 0.0000i 0 0 0 0
0 -1.9810 + 6.6826i 0 0 0
0 0 -1.9810 - 6.6826i 0 0
0 0 0 1.2580 + 0.0000i 0
0 0 0 0 4.3450 + 0.0000i
As you can see, the matrix $A$ have complex eigenvalues, but not in the $S$ matrix of SVD. Why? Is it possible for me to compute the eigenvalues and eigenvectors from svd if I know $A,U,S,V$?
matrices eigenvalues-eigenvectors svd singularvalues
matrices eigenvalues-eigenvectors svd singularvalues
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
asked Jan 14 at 20:21
Daniel MårtenssonDaniel Mårtensson
942416
942416
$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
2
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14
|
show 5 more comments
$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
2
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14
$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
2
2
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14
|
show 5 more comments
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$begingroup$
In principle the SVD determines the eigenvalues and eigenvectors because it determines the matrix, but the connection is not a very simple one.
$endgroup$
– Robert Israel
Jan 14 at 20:30
$begingroup$
@RobertIsrael Ok, so due to advanced computations, the values changes a little bit? There is no idea to find the eigenvalues then by using the svd?
$endgroup$
– Daniel Mårtensson
Jan 14 at 20:38
$begingroup$
For example, the eigenvalues are not determined by the singular values.
$endgroup$
– Robert Israel
Jan 14 at 20:40
$begingroup$
See Wikipedia - SVD and spectral decomposition
$endgroup$
– mathcounterexamples.net
Jan 14 at 20:40
2
$begingroup$
The answer is negative. Look at the case $n=1$. You can’t deduce the value of $z in mathbb C$ from $vert zvert^2$ without additional information.
$endgroup$
– mathcounterexamples.net
Jan 14 at 21:14