Understanding the absolute value of a matrix.
$begingroup$
I believe that the absolute value of a matrix is defined as
$$
|A|=sqrt{A^{dagger}A} .
$$
But the square root of a matrix is not unique wikipedia gives a list of examples to illustrate this.
To understand this, how does one work out the absolute value of:
$$
A=begin{pmatrix}1 & 0\0 & -1end{pmatrix}
$$
Clearly $A^{dagger}=A$ so $|A|=sqrt{A^2}$, but this is not necessarily $A$. I want to pick the identity in this case, since then the eigenvalues of $|A|$ are both 1 (and they were $pm1$ for $A$). But mathematics is not about what I want. So what is $|A|$? Is it well-defined? And how do I do this operation in general, since my application for this is of course far more complex.
linear-algebra matrices absolute-value
asked Jun 23 '18 at 13:25
maormaor
385
$endgroup$
add a comment |
$begingroup$
I believe that the absolute value of a matrix is defined as
$$
|A|=sqrt{A^{dagger}A} .
$$
But the square root of a matrix is not unique wikipedia gives a list of examples to illustrate this.
To understand this, how does one work out the absolute value of:
$$
A=begin{pmatrix}1 & 0\0 & -1end{pmatrix}
$$
Clearly $A^{dagger}=A$ so $|A|=sqrt{A^2}$, but this is not necessarily $A$. I want to pick the identity in this case, since then the eigenvalues of $|A|$ are both 1 (and they were $pm1$ for $A$). But mathematics is not about what I want. So what is $|A|$? Is it well-defined? And how do I do this operation in general, since my application for this is of course far more complex.
linear-algebra matrices absolute-value
asked Jun 23 '18 at 13:25
maormaor
385
$endgroup$
$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
1
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
1
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22
add a comment |
$begingroup$
I believe that the absolute value of a matrix is defined as
$$
|A|=sqrt{A^{dagger}A} .
$$
But the square root of a matrix is not unique wikipedia gives a list of examples to illustrate this.
To understand this, how does one work out the absolute value of:
$$
A=begin{pmatrix}1 & 0\0 & -1end{pmatrix}
$$
Clearly $A^{dagger}=A$ so $|A|=sqrt{A^2}$, but this is not necessarily $A$. I want to pick the identity in this case, since then the eigenvalues of $|A|$ are both 1 (and they were $pm1$ for $A$). But mathematics is not about what I want. So what is $|A|$? Is it well-defined? And how do I do this operation in general, since my application for this is of course far more complex.
linear-algebra matrices absolute-value
asked Jun 23 '18 at 13:25
maormaor
385
$endgroup$
I believe that the absolute value of a matrix is defined as
$$
|A|=sqrt{A^{dagger}A} .
$$
But the square root of a matrix is not unique wikipedia gives a list of examples to illustrate this.
To understand this, how does one work out the absolute value of:
$$
A=begin{pmatrix}1 & 0\0 & -1end{pmatrix}
$$
Clearly $A^{dagger}=A$ so $|A|=sqrt{A^2}$, but this is not necessarily $A$. I want to pick the identity in this case, since then the eigenvalues of $|A|$ are both 1 (and they were $pm1$ for $A$). But mathematics is not about what I want. So what is $|A|$? Is it well-defined? And how do I do this operation in general, since my application for this is of course far more complex.
linear-algebra matrices absolute-value
linear-algebra matrices absolute-value
asked Jun 23 '18 at 13:25
maormaor
385
asked Jun 23 '18 at 13:25
maormaor
385
edited Jun 23 '18 at 13:46
maor
asked Jun 23 '18 at 13:25
maormaor
385
asked Jun 23 '18 at 13:25
maormaor
385
asked Jun 23 '18 at 13:25
maormaor
385
385
$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
1
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
1
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22
add a comment |
$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
1
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
1
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22
$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
1
1
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
1
1
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22
add a comment |
1 Answer
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$begingroup$
If $D$ is a diagonal matrix with positive terms, then the positive square root, $sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $ sqrt A= P^{-1} sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
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add a comment |
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$begingroup$
If $D$ is a diagonal matrix with positive terms, then the positive square root, $sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $ sqrt A= P^{-1} sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
add a comment |
$begingroup$
If $D$ is a diagonal matrix with positive terms, then the positive square root, $sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $ sqrt A= P^{-1} sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
add a comment |
$begingroup$
If $D$ is a diagonal matrix with positive terms, then the positive square root, $sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $ sqrt A= P^{-1} sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
$endgroup$
If $D$ is a diagonal matrix with positive terms, then the positive square root, $sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $ sqrt A= P^{-1} sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
answered Jun 23 '18 at 13:56
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.4k42061
41.4k42061
add a comment |
add a comment |
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$begingroup$
Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root."
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:28
$begingroup$
yes if $|A|$ is "the positive definite square root $sqrt{A^{dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
$endgroup$
– maor
Jun 23 '18 at 13:34
1
$begingroup$
I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come.
$endgroup$
– Ethan Bolker
Jun 23 '18 at 13:36
1
$begingroup$
For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms.
$endgroup$
– Mohammad Riazi-Kermani
Jun 23 '18 at 13:47
$begingroup$
Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
$endgroup$
– maor
Jun 25 '18 at 14:22