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Can a positive-definite diagonal matrix have square roots that are positive-definite but not diagonal?
Consider a a diagonal $ntimes n$ matrix $M$ with real diagonal elements $m_{11},m_{22},dots,m_{nn}>0$. I am interested in finding positive-definite $n times n$ matrices $X$ that solve
begin{equation}M=X^2,.tag*{(1)}end{equation}
An obvious solution $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.
matrices matrix-equations positive-definite
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
add a comment |
Consider a a diagonal $ntimes n$ matrix $M$ with real diagonal elements $m_{11},m_{22},dots,m_{nn}>0$. I am interested in finding positive-definite $n times n$ matrices $X$ that solve
begin{equation}M=X^2,.tag*{(1)}end{equation}
An obvious solution $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.
matrices matrix-equations positive-definite
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
1
On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07
add a comment |
Consider a a diagonal $ntimes n$ matrix $M$ with real diagonal elements $m_{11},m_{22},dots,m_{nn}>0$. I am interested in finding positive-definite $n times n$ matrices $X$ that solve
begin{equation}M=X^2,.tag*{(1)}end{equation}
An obvious solution $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.
matrices matrix-equations positive-definite
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
Consider a a diagonal $ntimes n$ matrix $M$ with real diagonal elements $m_{11},m_{22},dots,m_{nn}>0$. I am interested in finding positive-definite $n times n$ matrices $X$ that solve
begin{equation}M=X^2,.tag*{(1)}end{equation}
An obvious solution $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$. Is that the only positive-definite solution?
If I do not restrict my search to positive-definite $M$ and $X$, then I can find $2times 2$ $M$ for which non-diagonal $X$ solve equation (1). There are plenty of examples on wikipedia. All examples listed there are not positive-definite. I haven't been able to come up with a positive-definite example, thus my hypothesis that $X=diag(sqrt{m_{11}},sqrt{m_{22}},dots,sqrt{m_{nn}})$ is the only positive-definite solution for positive-definite $M$.
matrices matrix-equations positive-definite
matrices matrix-equations positive-definite
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
asked Nov 21 '18 at 1:02
Alice Schwarze
9114
9114
1
On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07
add a comment |
1
On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07
1
1
On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07
On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07
add a comment |
1 Answer
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Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $sqrt{M}$.
In your case, since $M=operatorname{diag}(m_{11},ldots,m_{nn})$ is positive semidefinite and $X=operatorname{diag}(sqrt{m_{11}},ldots,sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.
answered Nov 21 '18 at 3:22
user1551
71.6k566125
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
active
oldest
votes
Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $sqrt{M}$.
In your case, since $M=operatorname{diag}(m_{11},ldots,m_{nn})$ is positive semidefinite and $X=operatorname{diag}(sqrt{m_{11}},ldots,sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.
answered Nov 21 '18 at 3:22
user1551
71.6k566125
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
add a comment |
Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $sqrt{M}$.
In your case, since $M=operatorname{diag}(m_{11},ldots,m_{nn})$ is positive semidefinite and $X=operatorname{diag}(sqrt{m_{11}},ldots,sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.
answered Nov 21 '18 at 3:22
user1551
71.6k566125
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
add a comment |
Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $sqrt{M}$.
In your case, since $M=operatorname{diag}(m_{11},ldots,m_{nn})$ is positive semidefinite and $X=operatorname{diag}(sqrt{m_{11}},ldots,sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.
answered Nov 21 '18 at 3:22
user1551
71.6k566125
Among all square roots of a positive semidefinite matrix $M$, exactly one of them is positive semidefinite. This fact is usually summarised as "every positive semidefinite matrix has a unique positive semidefinite square root". This positive semidefinite square root is usually denoted by $M^{1/2}$ or $sqrt{M}$.
In your case, since $M=operatorname{diag}(m_{11},ldots,m_{nn})$ is positive semidefinite and $X=operatorname{diag}(sqrt{m_{11}},ldots,sqrt{m_{nn}})$ is already a positive semidefinite square root of $M$, the only positive semidefinite square root of $M$ is $X$. So, the answer to the question in the title is no, $M$ has not any non-diagonal positive definite square root.
answered Nov 21 '18 at 3:22
user1551
71.6k566125
answered Nov 21 '18 at 3:22
user1551
71.6k566125
answered Nov 21 '18 at 3:22
user1551
71.6k566125
answered Nov 21 '18 at 3:22
user1551
71.6k566125
71.6k566125
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
add a comment |
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
Thank you. I didn't know that every positive semidefinite matrix has a unique positive semidefinite square root, so I looked for a proof and found one here
– Alice Schwarze
Nov 21 '18 at 6:15
add a comment |
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On the wikipedia page you cite, it says: "a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root".
– Sambo
Nov 21 '18 at 1:07