Complex factorisation of a psd matrix

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Let $AinmathbbR^ntimes n$ be a symmetric positive definite matrix. Assume that $A=Lcdot L^t$ for $LinmathbbC^ntimes m$. Can we infer that $L$ is intact real, i.e., has only real entries?






matrix-decomposition







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  • Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
    – user382144
    Sep 3 at 9:07











  • Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
    – kimchi lover
    Sep 3 at 12:57










  • Yes.. ätherweise it is clear...
    – user382144
    Sep 3 at 13:36














up vote
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down vote

favorite












Let $AinmathbbR^ntimes n$ be a symmetric positive definite matrix. Assume that $A=Lcdot L^t$ for $LinmathbbC^ntimes m$. Can we infer that $L$ is intact real, i.e., has only real entries?






matrix-decomposition







share|cite|improve this question





















  • Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
    – user382144
    Sep 3 at 9:07











  • Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
    – kimchi lover
    Sep 3 at 12:57










  • Yes.. ätherweise it is clear...
    – user382144
    Sep 3 at 13:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $AinmathbbR^ntimes n$ be a symmetric positive definite matrix. Assume that $A=Lcdot L^t$ for $LinmathbbC^ntimes m$. Can we infer that $L$ is intact real, i.e., has only real entries?






matrix-decomposition







share|cite|improve this question













Let $AinmathbbR^ntimes n$ be a symmetric positive definite matrix. Assume that $A=Lcdot L^t$ for $LinmathbbC^ntimes m$. Can we infer that $L$ is intact real, i.e., has only real entries?






matrix-decomposition




matrix-decomposition






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 3 at 9:06









user382144

111




111











  • Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
    – user382144
    Sep 3 at 9:07











  • Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
    – kimchi lover
    Sep 3 at 12:57










  • Yes.. ätherweise it is clear...
    – user382144
    Sep 3 at 13:36
















  • Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
    – user382144
    Sep 3 at 9:07











  • Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
    – kimchi lover
    Sep 3 at 12:57










  • Yes.. ätherweise it is clear...
    – user382144
    Sep 3 at 13:36















Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
– user382144
Sep 3 at 9:07





Note that we this does not hold one $A$ has not fall rank. For example $L=(2+I& 2-i)^t$ would yield a psd matrix, but the rank is not full.
– user382144
Sep 3 at 9:07













Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
– kimchi lover
Sep 3 at 12:57




Do you really mean to have transpose instead of conjugate transpose in your factorization of $A$?
– kimchi lover
Sep 3 at 12:57












Yes.. ätherweise it is clear...
– user382144
Sep 3 at 13:36




Yes.. ätherweise it is clear...
– user382144
Sep 3 at 13:36










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Let $L=beginpmatrixi&sqrt 2\sqrt 2&-iendpmatrix$.
Then $LL'$ is the $2times2$ identity matrix.






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    Let $L=beginpmatrixi&sqrt 2\sqrt 2&-iendpmatrix$.
    Then $LL'$ is the $2times2$ identity matrix.






    share|cite|improve this answer
























      up vote
      0
      down vote













      Let $L=beginpmatrixi&sqrt 2\sqrt 2&-iendpmatrix$.
      Then $LL'$ is the $2times2$ identity matrix.






      share|cite|improve this answer






















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        up vote
        0
        down vote









        Let $L=beginpmatrixi&sqrt 2\sqrt 2&-iendpmatrix$.
        Then $LL'$ is the $2times2$ identity matrix.






        share|cite|improve this answer












        Let $L=beginpmatrixi&sqrt 2\sqrt 2&-iendpmatrix$.
        Then $LL'$ is the $2times2$ identity matrix.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 3 at 14:09









        kimchi lover

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