Confusion about the uniqueness of the square root of a positive definite matrix












1












$begingroup$


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







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$endgroup$












  • $begingroup$
    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










  • $begingroup$
    What does this mean?
    $endgroup$
    – Ufuk Can Bicici
    Dec 18 '18 at 7:23
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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










  • $begingroup$
    What does this mean?
    $endgroup$
    – Ufuk Can Bicici
    Dec 18 '18 at 7:23














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      "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











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 0:16









        Gabriel PalauGabriel Palau

        1126




        1126























            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              "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
















            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              "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














            1












            1








            1





            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$



            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $begingroup$
              "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


















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