If $B$ is a small perturbation of positive-definite matrix $A$, then do we have $B>epsilon A$?












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 '18 at 1:21









HangHang

482315




482315












  • $begingroup$
    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










  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $B>epsilon A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

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 $B>epsilon A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 $B>epsilon A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 $B>epsilon A$.






share|cite|improve this answer









$endgroup$



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 $B>epsilon A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 18 '18 at 17:27









loup blancloup blanc

23.2k21851




23.2k21851












  • $begingroup$
    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










  • $begingroup$
    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


















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