If $B$ is a small perturbation of positive-definite matrix $A$, then do we have $B>epsilon A$?
$begingroup$
Suppose $A=(a_{ij})$ is a (symmetric) positive-definite matrix, and $B$ is another symmetric matrix.
Question: If $B$ is in a small neighborhood $U$ of $A$, then it seems that $B$ should also be positive-definite. Moreover for what value of $epsilon>0$ we can find a neighborhood $U$ so that $B>epsilon A$?
If both $A$ and $B$ are diagonal matrices, then this is trivial. But in general since we can only diagonalize one of them and I am afraid there will be some issue.
linear-algebra matrices positive-definite
asked Dec 17 '18 at 1:21
HangHang
492415
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $A=(a_{ij})$ is a (symmetric) positive-definite matrix, and $B$ is another symmetric matrix.
Question: If $B$ is in a small neighborhood $U$ of $A$, then it seems that $B$ should also be positive-definite. Moreover for what value of $epsilon>0$ we can find a neighborhood $U$ so that $B>epsilon A$?
If both $A$ and $B$ are diagonal matrices, then this is trivial. But in general since we can only diagonalize one of them and I am afraid there will be some issue.
linear-algebra matrices positive-definite
asked Dec 17 '18 at 1:21
HangHang
492415
$endgroup$
$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
1
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52
|
show 1 more comment
$begingroup$
Suppose $A=(a_{ij})$ is a (symmetric) positive-definite matrix, and $B$ is another symmetric matrix.
Question: If $B$ is in a small neighborhood $U$ of $A$, then it seems that $B$ should also be positive-definite. Moreover for what value of $epsilon>0$ we can find a neighborhood $U$ so that $B>epsilon A$?
If both $A$ and $B$ are diagonal matrices, then this is trivial. But in general since we can only diagonalize one of them and I am afraid there will be some issue.
linear-algebra matrices positive-definite
asked Dec 17 '18 at 1:21
HangHang
492415
$endgroup$
Suppose $A=(a_{ij})$ is a (symmetric) positive-definite matrix, and $B$ is another symmetric matrix.
Question: If $B$ is in a small neighborhood $U$ of $A$, then it seems that $B$ should also be positive-definite. Moreover for what value of $epsilon>0$ we can find a neighborhood $U$ so that $B>epsilon A$?
If both $A$ and $B$ are diagonal matrices, then this is trivial. But in general since we can only diagonalize one of them and I am afraid there will be some issue.
linear-algebra matrices positive-definite
linear-algebra matrices positive-definite
asked Dec 17 '18 at 1:21
HangHang
492415
asked Dec 17 '18 at 1:21
HangHang
492415
asked Dec 17 '18 at 1:21
HangHang
492415
asked Dec 17 '18 at 1:21
HangHang
492415
asked Dec 17 '18 at 1:21
HangHang
492415
492415
$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
1
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52
|
show 1 more comment
$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
1
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52
$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
1
1
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If you consider $mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)={lambda_1leqcdots leq lambda_n}$ (note that $lambda_1>0$). Let $epsilonin[0,1)$ and $alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=rho(S)$ (the spectral radius of $S$).
We assume that $||H||<alpha$, that implies $x^TBxgeq x^TAx-alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $alpha<lambda_1$. Then $x^TBxgeq lambda_1-alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-epsilon A)xgeq x^TAx(1-epsilon)-alphageq lambda_1(1-epsilon)-alpha$.
We choose $alpha<lambda_1(1-epsilon)$. Then $B>epsilon A$.
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
$endgroup$
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
add a comment |
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$begingroup$
If you consider $mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)={lambda_1leqcdots leq lambda_n}$ (note that $lambda_1>0$). Let $epsilonin[0,1)$ and $alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=rho(S)$ (the spectral radius of $S$).
We assume that $||H||<alpha$, that implies $x^TBxgeq x^TAx-alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $alpha<lambda_1$. Then $x^TBxgeq lambda_1-alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-epsilon A)xgeq x^TAx(1-epsilon)-alphageq lambda_1(1-epsilon)-alpha$.
We choose $alpha<lambda_1(1-epsilon)$. Then $B>epsilon A$.
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
$endgroup$
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
add a comment |
$begingroup$
If you consider $mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)={lambda_1leqcdots leq lambda_n}$ (note that $lambda_1>0$). Let $epsilonin[0,1)$ and $alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=rho(S)$ (the spectral radius of $S$).
We assume that $||H||<alpha$, that implies $x^TBxgeq x^TAx-alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $alpha<lambda_1$. Then $x^TBxgeq lambda_1-alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-epsilon A)xgeq x^TAx(1-epsilon)-alphageq lambda_1(1-epsilon)-alpha$.
We choose $alpha<lambda_1(1-epsilon)$. Then $B>epsilon A$.
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
$endgroup$
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
add a comment |
$begingroup$
If you consider $mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)={lambda_1leqcdots leq lambda_n}$ (note that $lambda_1>0$). Let $epsilonin[0,1)$ and $alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=rho(S)$ (the spectral radius of $S$).
We assume that $||H||<alpha$, that implies $x^TBxgeq x^TAx-alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $alpha<lambda_1$. Then $x^TBxgeq lambda_1-alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-epsilon A)xgeq x^TAx(1-epsilon)-alphageq lambda_1(1-epsilon)-alpha$.
We choose $alpha<lambda_1(1-epsilon)$. Then $B>epsilon A$.
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
$endgroup$
If you consider $mathbb{R}^{n^2}$, then that does not work because $B$ may be non-symmetric. Thus we assume that $B=A+H$ where the small matrix $H$ goes through the vector space of symmetric matrices.
Let $spectrum(A)={lambda_1leqcdots leq lambda_n}$ (note that $lambda_1>0$). Let $epsilonin[0,1)$ and $alpha >0$ (to be determined). Since the norms are equivalent, we'll use the spectral norm defined, on the symmetric matrices, by $||S||=rho(S)$ (the spectral radius of $S$).
We assume that $||H||<alpha$, that implies $x^TBxgeq x^TAx-alpha$ when $||x||_2^2=1$ (that we suppose in the sequel).
We choose $alpha<lambda_1$. Then $x^TBxgeq lambda_1-alpha >0$ and $B$ is symmetric$>0$.
Now $x^T(B-epsilon A)xgeq x^TAx(1-epsilon)-alphageq lambda_1(1-epsilon)-alpha$.
We choose $alpha<lambda_1(1-epsilon)$. Then $B>epsilon A$.
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
answered Dec 18 '18 at 17:27
loup blancloup blanc
23.6k21851
23.6k21851
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
add a comment |
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Thank you, but actually I have assumed $B$ is symmetric.
$endgroup$
– Hang
Dec 18 '18 at 17:34
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
Me also my friend. Try to understand my post.
$endgroup$
– loup blanc
Dec 18 '18 at 17:36
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
I do not understand why we cannot simply use the subspace topology in the space of symmetric matrices induced from the topology of $mathbb R^{n^2}$.
$endgroup$
– Hang
Dec 18 '18 at 17:50
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
$begingroup$
Yes, the induced topology is the correct expression
$endgroup$
– loup blanc
Dec 18 '18 at 18:01
add a comment |
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$begingroup$
When you say $B$ is in a small neighborhood of $A$, do you mean that $|B-A| < epsilon$ (operator norm), $|B-A|_F < epsilon$ (Frobenius norm), or some other norm?
$endgroup$
– JimmyK4542
Dec 17 '18 at 1:41
$begingroup$
I just mean the topology of $mathbb R^{n^2}$, but I think I am also interested in the question for different topology.
$endgroup$
– Hang
Dec 17 '18 at 1:49
1
$begingroup$
@Hang You can find some discussion on the topic here math.stackexchange.com/questions/226486/…
$endgroup$
– AnyAD
Dec 17 '18 at 2:07
$begingroup$
You can always find a PDS diagonal matrix as close as you want to the zero matrix. Does this help?
$endgroup$
– Wintermute
Dec 18 '18 at 17:49
$begingroup$
@Wintermute What does PDS stand for?
$endgroup$
– Hang
Dec 18 '18 at 17:52