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Eigenvalues and Eigenvectors of $AA^T$ vs $ADA^T$
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Assume $D$ is a diagonal matrix. Are the Eigenvalues and Eigenvectors of $AA^T$ and $ADA^T$ related in anyway i.e. are the Eigenvectors same, Eigenvalues same, or is their any spectral relationship between these matrices? Also, the diagonal elements of the matrix $D$ are all greater than 1.
PS: I came across this when trying to prove something related to the operator norms of the two matrices. Any suggestions or references would be very helpful. Thanks.
linear-algebra eigenvalues-eigenvectors norm
asked Jul 20 at 21:35
Ben
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up vote
0
down vote
favorite
Assume $D$ is a diagonal matrix. Are the Eigenvalues and Eigenvectors of $AA^T$ and $ADA^T$ related in anyway i.e. are the Eigenvectors same, Eigenvalues same, or is their any spectral relationship between these matrices? Also, the diagonal elements of the matrix $D$ are all greater than 1.
PS: I came across this when trying to prove something related to the operator norms of the two matrices. Any suggestions or references would be very helpful. Thanks.
linear-algebra eigenvalues-eigenvectors norm
asked Jul 20 at 21:35
Ben
14
Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Assume $D$ is a diagonal matrix. Are the Eigenvalues and Eigenvectors of $AA^T$ and $ADA^T$ related in anyway i.e. are the Eigenvectors same, Eigenvalues same, or is their any spectral relationship between these matrices? Also, the diagonal elements of the matrix $D$ are all greater than 1.
PS: I came across this when trying to prove something related to the operator norms of the two matrices. Any suggestions or references would be very helpful. Thanks.
linear-algebra eigenvalues-eigenvectors norm
asked Jul 20 at 21:35
Ben
14
Assume $D$ is a diagonal matrix. Are the Eigenvalues and Eigenvectors of $AA^T$ and $ADA^T$ related in anyway i.e. are the Eigenvectors same, Eigenvalues same, or is their any spectral relationship between these matrices? Also, the diagonal elements of the matrix $D$ are all greater than 1.
PS: I came across this when trying to prove something related to the operator norms of the two matrices. Any suggestions or references would be very helpful. Thanks.
linear-algebra eigenvalues-eigenvectors norm
asked Jul 20 at 21:35
Ben
14
edited Jul 20 at 21:58
asked Jul 20 at 21:35
Ben
14
asked Jul 20 at 21:35
Ben
14
asked Jul 20 at 21:35
Ben
14
14
Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47
add a comment |Â
Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47
Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47
add a comment |Â
1 Answer
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If $ Ain mathbbC^n times n$
We have that
$$A^TA = (U Sigma V^T)^T(U Sigma V^T) = (V^T)^TSigma^T U^T(USigma V^T) $$
now we know that $U^TU= I $
$$V Sigma^T U^T Sigma V^T = V Sigma^T Sigma V^T $$
and
$ Sigma^T Sigma = Sigma^2 $
$$ A^TA = V Sigma^2 V^T = V Lambda V^T$$
Where $ Lambda $ is the matrix of eigenvalues $ VV^T = V^T V = I$
Now
$$ ADA^T = (U Sigma V^T)D(U Sigma V^T)^T $$
$$ AD A^T = (USigma V^T)D(V Sigma U^T) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps
$$A^TA = V Lambda V^T $$
$$ |V Lambda V^T | leq | V | Lambda | | V^T |$$
$$ | A^TA| leq | Lambda| =max_1 leq i leq n | | lambda_i|$$
now suppose $D= cI$ where $c$ is a scalar
$$ A^TD A = (VSigma^TU^T cI USigma V^T $$
$$ | A^TDA | leq | V| Sigma^T | |U^T | |cI | | U| | Sigma | |V^T |$$
$$ | A^T D A | leq |Sigma | | cI| | Sigma | $$
Now $ Lambda = Sigma^2$
$$| A^T D A | leq max_1 leq i leq n | |c| max_1 leq i leq n |sigma_i| = |c| max_1 leq i leq n |lambda_i| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
answered Jul 20 at 21:58
RHowe
1,000815
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $ Ain mathbbC^n times n$
We have that
$$A^TA = (U Sigma V^T)^T(U Sigma V^T) = (V^T)^TSigma^T U^T(USigma V^T) $$
now we know that $U^TU= I $
$$V Sigma^T U^T Sigma V^T = V Sigma^T Sigma V^T $$
and
$ Sigma^T Sigma = Sigma^2 $
$$ A^TA = V Sigma^2 V^T = V Lambda V^T$$
Where $ Lambda $ is the matrix of eigenvalues $ VV^T = V^T V = I$
Now
$$ ADA^T = (U Sigma V^T)D(U Sigma V^T)^T $$
$$ AD A^T = (USigma V^T)D(V Sigma U^T) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps
$$A^TA = V Lambda V^T $$
$$ |V Lambda V^T | leq | V | Lambda | | V^T |$$
$$ | A^TA| leq | Lambda| =max_1 leq i leq n | | lambda_i|$$
now suppose $D= cI$ where $c$ is a scalar
$$ A^TD A = (VSigma^TU^T cI USigma V^T $$
$$ | A^TDA | leq | V| Sigma^T | |U^T | |cI | | U| | Sigma | |V^T |$$
$$ | A^T D A | leq |Sigma | | cI| | Sigma | $$
Now $ Lambda = Sigma^2$
$$| A^T D A | leq max_1 leq i leq n | |c| max_1 leq i leq n |sigma_i| = |c| max_1 leq i leq n |lambda_i| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
answered Jul 20 at 21:58
RHowe
1,000815
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
add a comment |Â
up vote
0
down vote
If $ Ain mathbbC^n times n$
We have that
$$A^TA = (U Sigma V^T)^T(U Sigma V^T) = (V^T)^TSigma^T U^T(USigma V^T) $$
now we know that $U^TU= I $
$$V Sigma^T U^T Sigma V^T = V Sigma^T Sigma V^T $$
and
$ Sigma^T Sigma = Sigma^2 $
$$ A^TA = V Sigma^2 V^T = V Lambda V^T$$
Where $ Lambda $ is the matrix of eigenvalues $ VV^T = V^T V = I$
Now
$$ ADA^T = (U Sigma V^T)D(U Sigma V^T)^T $$
$$ AD A^T = (USigma V^T)D(V Sigma U^T) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps
$$A^TA = V Lambda V^T $$
$$ |V Lambda V^T | leq | V | Lambda | | V^T |$$
$$ | A^TA| leq | Lambda| =max_1 leq i leq n | | lambda_i|$$
now suppose $D= cI$ where $c$ is a scalar
$$ A^TD A = (VSigma^TU^T cI USigma V^T $$
$$ | A^TDA | leq | V| Sigma^T | |U^T | |cI | | U| | Sigma | |V^T |$$
$$ | A^T D A | leq |Sigma | | cI| | Sigma | $$
Now $ Lambda = Sigma^2$
$$| A^T D A | leq max_1 leq i leq n | |c| max_1 leq i leq n |sigma_i| = |c| max_1 leq i leq n |lambda_i| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
answered Jul 20 at 21:58
RHowe
1,000815
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $ Ain mathbbC^n times n$
We have that
$$A^TA = (U Sigma V^T)^T(U Sigma V^T) = (V^T)^TSigma^T U^T(USigma V^T) $$
now we know that $U^TU= I $
$$V Sigma^T U^T Sigma V^T = V Sigma^T Sigma V^T $$
and
$ Sigma^T Sigma = Sigma^2 $
$$ A^TA = V Sigma^2 V^T = V Lambda V^T$$
Where $ Lambda $ is the matrix of eigenvalues $ VV^T = V^T V = I$
Now
$$ ADA^T = (U Sigma V^T)D(U Sigma V^T)^T $$
$$ AD A^T = (USigma V^T)D(V Sigma U^T) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps
$$A^TA = V Lambda V^T $$
$$ |V Lambda V^T | leq | V | Lambda | | V^T |$$
$$ | A^TA| leq | Lambda| =max_1 leq i leq n | | lambda_i|$$
now suppose $D= cI$ where $c$ is a scalar
$$ A^TD A = (VSigma^TU^T cI USigma V^T $$
$$ | A^TDA | leq | V| Sigma^T | |U^T | |cI | | U| | Sigma | |V^T |$$
$$ | A^T D A | leq |Sigma | | cI| | Sigma | $$
Now $ Lambda = Sigma^2$
$$| A^T D A | leq max_1 leq i leq n | |c| max_1 leq i leq n |sigma_i| = |c| max_1 leq i leq n |lambda_i| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
answered Jul 20 at 21:58
RHowe
1,000815
If $ Ain mathbbC^n times n$
We have that
$$A^TA = (U Sigma V^T)^T(U Sigma V^T) = (V^T)^TSigma^T U^T(USigma V^T) $$
now we know that $U^TU= I $
$$V Sigma^T U^T Sigma V^T = V Sigma^T Sigma V^T $$
and
$ Sigma^T Sigma = Sigma^2 $
$$ A^TA = V Sigma^2 V^T = V Lambda V^T$$
Where $ Lambda $ is the matrix of eigenvalues $ VV^T = V^T V = I$
Now
$$ ADA^T = (U Sigma V^T)D(U Sigma V^T)^T $$
$$ AD A^T = (USigma V^T)D(V Sigma U^T) $$
I can't come up any obvious connections other then $D=I$ then the eigenvalues are the same
I can only think of a relation perhaps
$$A^TA = V Lambda V^T $$
$$ |V Lambda V^T | leq | V | Lambda | | V^T |$$
$$ | A^TA| leq | Lambda| =max_1 leq i leq n | | lambda_i|$$
now suppose $D= cI$ where $c$ is a scalar
$$ A^TD A = (VSigma^TU^T cI USigma V^T $$
$$ | A^TDA | leq | V| Sigma^T | |U^T | |cI | | U| | Sigma | |V^T |$$
$$ | A^T D A | leq |Sigma | | cI| | Sigma | $$
Now $ Lambda = Sigma^2$
$$| A^T D A | leq max_1 leq i leq n | |c| max_1 leq i leq n |sigma_i| = |c| max_1 leq i leq n |lambda_i| $$
The eigenvalues are scaled up in this case by the c. Or you can observe at the least the maximum eigenvalue will be?
answered Jul 20 at 21:58
RHowe
1,000815
edited Jul 20 at 22:41
answered Jul 20 at 21:58
RHowe
1,000815
answered Jul 20 at 21:58
RHowe
1,000815
answered Jul 20 at 21:58
RHowe
1,000815
1,000815
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
add a comment |Â
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
Thanks for the answer Geronimo. I added a small point to the question: diagonal elements of the matrix $D$ are greater than 1. Would this change anything?
– Ben
Jul 20 at 22:00
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
The diagonal elements of D are the eigenvalues of D. I imagine this would have small scaling factor on the other matrices and you may be able to show that through some norm comparisons. I think when I wrote this down I wrote $A^TA$ however the eigenvalues of $A^TA $ are equal to $AA^T$ ...
– RHowe
Jul 20 at 22:04
add a comment |Â
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Have you tried simple 2x2 examples, before asking?
– Cosmas Zachos
Jul 20 at 21:47
Yes. I tried small matrices and it seems that the Eigenvalues of $ADA^T$ are greater than eigenvalues of $AA^T$. But I can't prove this analytically.
– Ben
Jul 20 at 22:01
Note that $ADA^T = (AsqrtD)(AsqrtD)^T$, so you're trying to compare singular values of $A$ and $AsqrtD$, where $sqrtD$ is an essentially arbitrary diagonal matrix consisting of values greater than $1$. I haven't parsed it properly, but maybe this answer might help? math.stackexchange.com/a/1793562/248286
– Theo Bendit
Jul 20 at 22:47