Positive definite matrix implies the **infimum** of eigenvalues are positive?
$begingroup$
Suppose $P(x): mathbb{R} to mathbb{R}^{n times n}$ is always a positive definite matrix, does it imply that the infimum (over $mathbb{R}$) of the minimum eigenvalue of $P(x)$ is always positive?, that is, $inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0$ In a mathematical way:
Is the following conclusion correct?
begin{equation}
forall x in mathbb{R},;P(x) succ 0 implies inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0
end{equation}
Please be careful with the infimum.
Please go to the other similar question here, which is exactly the real (and more difficult in my opinion) question that I want to ask.
real-analysis linear-algebra matrices supremum-and-infimum positive-definite
asked Dec 7 '18 at 10:53
winstonwinston
522318
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add a comment |
$begingroup$
Suppose $P(x): mathbb{R} to mathbb{R}^{n times n}$ is always a positive definite matrix, does it imply that the infimum (over $mathbb{R}$) of the minimum eigenvalue of $P(x)$ is always positive?, that is, $inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0$ In a mathematical way:
Is the following conclusion correct?
begin{equation}
forall x in mathbb{R},;P(x) succ 0 implies inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0
end{equation}
Please be careful with the infimum.
Please go to the other similar question here, which is exactly the real (and more difficult in my opinion) question that I want to ask.
real-analysis linear-algebra matrices supremum-and-infimum positive-definite
asked Dec 7 '18 at 10:53
winstonwinston
522318
$endgroup$
add a comment |
$begingroup$
Suppose $P(x): mathbb{R} to mathbb{R}^{n times n}$ is always a positive definite matrix, does it imply that the infimum (over $mathbb{R}$) of the minimum eigenvalue of $P(x)$ is always positive?, that is, $inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0$ In a mathematical way:
Is the following conclusion correct?
begin{equation}
forall x in mathbb{R},;P(x) succ 0 implies inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0
end{equation}
Please be careful with the infimum.
Please go to the other similar question here, which is exactly the real (and more difficult in my opinion) question that I want to ask.
real-analysis linear-algebra matrices supremum-and-infimum positive-definite
asked Dec 7 '18 at 10:53
winstonwinston
522318
$endgroup$
Suppose $P(x): mathbb{R} to mathbb{R}^{n times n}$ is always a positive definite matrix, does it imply that the infimum (over $mathbb{R}$) of the minimum eigenvalue of $P(x)$ is always positive?, that is, $inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0$ In a mathematical way:
Is the following conclusion correct?
begin{equation}
forall x in mathbb{R},;P(x) succ 0 implies inf_{x in mathbb{R}} {lambda_{{rm min}}(P(x)) } > 0
end{equation}
Please be careful with the infimum.
Please go to the other similar question here, which is exactly the real (and more difficult in my opinion) question that I want to ask.
real-analysis linear-algebra matrices supremum-and-infimum positive-definite
real-analysis linear-algebra matrices supremum-and-infimum positive-definite
asked Dec 7 '18 at 10:53
winstonwinston
522318
asked Dec 7 '18 at 10:53
winstonwinston
522318
edited Dec 7 '18 at 11:09
winston
asked Dec 7 '18 at 10:53
winstonwinston
522318
asked Dec 7 '18 at 10:53
winstonwinston
522318
asked Dec 7 '18 at 10:53
winstonwinston
522318
522318
add a comment |
add a comment |
1 Answer
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No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
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$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
$endgroup$
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
add a comment |
$begingroup$
No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
$endgroup$
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
add a comment |
$begingroup$
No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
$endgroup$
No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
answered Dec 7 '18 at 10:55
user1551user1551
73.5k566129
73.5k566129
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
add a comment |
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
I think it should be $P(x)=e^{-x}$.
$endgroup$
– winston
Dec 7 '18 at 10:59
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
@winston $e^x$ is OK. It is always positive and $e^xto0$ when $xto-infty$.
$endgroup$
– user1551
Dec 7 '18 at 11:00
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Yes,you are right.
$endgroup$
– winston
Dec 7 '18 at 11:01
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
$begingroup$
Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/…
$endgroup$
– winston
Dec 7 '18 at 11:07
add a comment |
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