Confusion about the uniqueness of the square root of a positive definite matrix
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There is a lot of material showing that a positive definite matrix $A$ has a unique positive definite square root, $B$ such that $B^2=A$. During a self study session, I needed to use this fact for a proof sketch, but I was confused about whether $B$ is the only square root of $A$ and it happens to be positive definite or is the uniqueness only valid for the positive definiteness; such that one can come up with a matrix $C$, not positive definite and $C^2=A$. I am confused about which is the correct one.
linear-algebra matrices positive-definite
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
add a comment |
$begingroup$
There is a lot of material showing that a positive definite matrix $A$ has a unique positive definite square root, $B$ such that $B^2=A$. During a self study session, I needed to use this fact for a proof sketch, but I was confused about whether $B$ is the only square root of $A$ and it happens to be positive definite or is the uniqueness only valid for the positive definiteness; such that one can come up with a matrix $C$, not positive definite and $C^2=A$. I am confused about which is the correct one.
linear-algebra matrices positive-definite
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
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What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23
add a comment |
$begingroup$
There is a lot of material showing that a positive definite matrix $A$ has a unique positive definite square root, $B$ such that $B^2=A$. During a self study session, I needed to use this fact for a proof sketch, but I was confused about whether $B$ is the only square root of $A$ and it happens to be positive definite or is the uniqueness only valid for the positive definiteness; such that one can come up with a matrix $C$, not positive definite and $C^2=A$. I am confused about which is the correct one.
linear-algebra matrices positive-definite
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
$endgroup$
There is a lot of material showing that a positive definite matrix $A$ has a unique positive definite square root, $B$ such that $B^2=A$. During a self study session, I needed to use this fact for a proof sketch, but I was confused about whether $B$ is the only square root of $A$ and it happens to be positive definite or is the uniqueness only valid for the positive definiteness; such that one can come up with a matrix $C$, not positive definite and $C^2=A$. I am confused about which is the correct one.
linear-algebra matrices positive-definite
linear-algebra matrices positive-definite
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
asked Dec 18 '18 at 0:09
Ufuk Can BiciciUfuk Can Bicici
1,24711127
1,24711127
$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
$begingroup$
What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23
add a comment |
$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
$begingroup$
What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23
$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
$begingroup$
What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23
$begingroup$
What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I think in the case $A=Iin mathbb{R}^{ntimes n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ left[ {begin{array}{cc}
1 & 0 \
0 & 1 \
end{array} } right]$ $left[ {begin{array}{cc}
1 & 0 \
0 & -1 \
end{array} } right]$ $left[ {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right]$ or $
left[ {begin{array}{cc}
-1 & 0 \
0 & -1 \
end{array} } right]
$
but only one of them is positive definite. It is used for the uniqueness.
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
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add a comment |
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There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2times 2$ square roots of the identity include every reflection matrix of the form $begin{bmatrix}a&b\b&-aend{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $ntimes n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
$endgroup$
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I think in the case $A=Iin mathbb{R}^{ntimes n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ left[ {begin{array}{cc}
1 & 0 \
0 & 1 \
end{array} } right]$ $left[ {begin{array}{cc}
1 & 0 \
0 & -1 \
end{array} } right]$ $left[ {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right]$ or $
left[ {begin{array}{cc}
-1 & 0 \
0 & -1 \
end{array} } right]
$
but only one of them is positive definite. It is used for the uniqueness.
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
If I think in the case $A=Iin mathbb{R}^{ntimes n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ left[ {begin{array}{cc}
1 & 0 \
0 & 1 \
end{array} } right]$ $left[ {begin{array}{cc}
1 & 0 \
0 & -1 \
end{array} } right]$ $left[ {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right]$ or $
left[ {begin{array}{cc}
-1 & 0 \
0 & -1 \
end{array} } right]
$
but only one of them is positive definite. It is used for the uniqueness.
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
$endgroup$
add a comment |
$begingroup$
If I think in the case $A=Iin mathbb{R}^{ntimes n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ left[ {begin{array}{cc}
1 & 0 \
0 & 1 \
end{array} } right]$ $left[ {begin{array}{cc}
1 & 0 \
0 & -1 \
end{array} } right]$ $left[ {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right]$ or $
left[ {begin{array}{cc}
-1 & 0 \
0 & -1 \
end{array} } right]
$
but only one of them is positive definite. It is used for the uniqueness.
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
$endgroup$
If I think in the case $A=Iin mathbb{R}^{ntimes n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ left[ {begin{array}{cc}
1 & 0 \
0 & 1 \
end{array} } right]$ $left[ {begin{array}{cc}
1 & 0 \
0 & -1 \
end{array} } right]$ $left[ {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right]$ or $
left[ {begin{array}{cc}
-1 & 0 \
0 & -1 \
end{array} } right]
$
but only one of them is positive definite. It is used for the uniqueness.
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
answered Dec 18 '18 at 0:16
Gabriel PalauGabriel Palau
1126
1126
add a comment |
add a comment |
$begingroup$
There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2times 2$ square roots of the identity include every reflection matrix of the form $begin{bmatrix}a&b\b&-aend{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $ntimes n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
$endgroup$
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
add a comment |
$begingroup$
There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2times 2$ square roots of the identity include every reflection matrix of the form $begin{bmatrix}a&b\b&-aend{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $ntimes n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
$endgroup$
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
add a comment |
$begingroup$
There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2times 2$ square roots of the identity include every reflection matrix of the form $begin{bmatrix}a&b\b&-aend{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $ntimes n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
$endgroup$
There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2times 2$ square roots of the identity include every reflection matrix of the form $begin{bmatrix}a&b\b&-aend{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $ntimes n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
answered Dec 18 '18 at 0:36
jmerryjmerry
15.8k1632
15.8k1632
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
add a comment |
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
"When a $n times n$ matrix has repeated eigenvalues, then it has infinitely many square roots; is this true?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:52
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
$begingroup$
The repeated eigenvalue needs at least two eigenvectors for that to happen; the infinite possibilities come from freely choosing a basis for the space of eigenvectors.
$endgroup$
– jmerry
Dec 18 '18 at 21:05
add a comment |
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$begingroup$
You also have to understand that uniqueness in this statement means that the positive semidefinite square root of a positive semidefinite matrix is unique up to unitary similarity.
$endgroup$
– chhro
Dec 18 '18 at 0:41
$begingroup$
What does this mean?
$endgroup$
– Ufuk Can Bicici
Dec 18 '18 at 7:23