Relation between different presentations of a positive semidefinite matrix












1












$begingroup$


Let $P$ be a positive semidefinite $ntimes n$ matrix of rank $r$. Then we may write $$ P=sum_{j=1}^r v_jv_j^*, text{where} v_1,ldots,v_rinmathbb C^n.$$



If we also have $P=sum_{j=1}^r w_jw_j^*$ for some $w_1,ldots,w_rinmathbb C^n$,




What's the relation between $v_1,ldots,v_r$ and $w_1,ldots,w_r$?







linear-algebra matrices functional-analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $P$ be a positive semidefinite $ntimes n$ matrix of rank $r$. Then we may write $$ P=sum_{j=1}^r v_jv_j^*, text{where} v_1,ldots,v_rinmathbb C^n.$$



    If we also have $P=sum_{j=1}^r w_jw_j^*$ for some $w_1,ldots,w_rinmathbb C^n$,




    What's the relation between $v_1,ldots,v_r$ and $w_1,ldots,w_r$?







    linear-algebra matrices functional-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Let $P$ be a positive semidefinite $ntimes n$ matrix of rank $r$. Then we may write $$ P=sum_{j=1}^r v_jv_j^*, text{where} v_1,ldots,v_rinmathbb C^n.$$



      If we also have $P=sum_{j=1}^r w_jw_j^*$ for some $w_1,ldots,w_rinmathbb C^n$,




      What's the relation between $v_1,ldots,v_r$ and $w_1,ldots,w_r$?







      linear-algebra matrices functional-analysis







      share|cite|improve this question











      $endgroup$




      Let $P$ be a positive semidefinite $ntimes n$ matrix of rank $r$. Then we may write $$ P=sum_{j=1}^r v_jv_j^*, text{where} v_1,ldots,v_rinmathbb C^n.$$



      If we also have $P=sum_{j=1}^r w_jw_j^*$ for some $w_1,ldots,w_rinmathbb C^n$,




      What's the relation between $v_1,ldots,v_r$ and $w_1,ldots,w_r$?







      linear-algebra matrices functional-analysis




      linear-algebra matrices functional-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 22 '18 at 22:16







      Martin Argerami

















      asked Nov 22 '18 at 4:20









      Martin ArgeramiMartin Argerami

      126k1182180




      126k1182180






















          2 Answers
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          active

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          2












          $begingroup$

          The answer can be more succinct. Put together the vectors $v_i$s to form a matrix $V$ and define $W$ analogously. Then both $V$ and $W$ have full column ranks and $VV^ast=WW^ast$. Hence $V^ast x=0$ if and only if $W^ast x=0$, i.e. the column spaces of $V$ and $W$ have a common orthogonal complement $mathcal X$. Let $X$ be a matrix whose columns form a basis of $mathcal X$. Then both augmented matrices $[V|X]$ and $[W|X]$ are invertible. Since $VV^ast=WW^ast$, we get $[V|X][V|X]^ast = [W|X][W|X]^ast$. Thus $U=[W|X]^{-1}[V|X]$ is a unitary matrix and $V=WU$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very neat. $ $
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 5:27



















          1












          $begingroup$

          The relation is that there exists a unitary $Uin M_r(mathbb C)$ such that $$w_j=sum_{k=1}^r U_{jk}v_k.$$



          Indeed, fix an orthonormal basis $e_1,ldots,e_n$ of $mathbb C^n$, and we may write
          $$
          v_k=sum_{k=1}^ns_{jk}e_j, w_k=sum_{j=1}^rt_{jk}e_j.
          $$
          Now consider $S,Tin M_r(mathbb C)$ with $S_{kj}=s_{kj}$, $T_{kj}=t_{kj}$. Then
          $$
          P=sum_{k=1}^rv_kv_k^*=sum_{k=1}^rsum_{j,ell=1}^rs_{jk}overline{s_{ell k}}e_je_ell^*=sum_{k=1}^rsum_{j,ell=1}^rS_{jk}{(S^*)_{k ell}}e_je_ell^*=sum_{j,ell=1}^r(SS^*)_{jell}e_je_ell^*.
          $$
          Similarly,
          $$
          P=sum_{j,ell=1}^r(TT^*)_{jell}e_je_ell^*.
          $$
          As $e_je_ell^*$ are the matrix units in $M_r(mathbb C)$, we get that $TT^*=SS^*$. Now we write the polar decompositions
          $$
          S^*=W(SS^*)^{1/2}, T^*=V(TT^*)^{1/2},
          $$
          where $W,V$ are unitaries. Then $S=(SS^*)^{1/2}W^*$, and
          $$
          T=(TT^*)^{1/2}V^*=(SS^*)^{1/2}V^*=SWV^*=SZ,
          $$
          where $Z=WV^*$. Now
          begin{align}
          w_j&=sum_{k=1}^r t_{kj}e_k=sum_{k=1}^r(SZ)_{kj}e_k=sum_{k,m=1}^rS_{km}Z_{mj}e_k\ \
          &=sum_{m=1}^r Z_{mj}sum_{k=1}^rS_{km}e_k=sum_{m=1}^rZ_{mj}v_m\ \
          &=sum_{m=1}^rU_{jm}v_m,
          end{align}
          where $U=Z^T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
            $endgroup$
            – mathrookie
            Nov 22 '18 at 21:06












          • $begingroup$
            If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 22:15











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          2 Answers
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          2 Answers
          2






          active

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          active

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          votes






          active

          oldest

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          2












          $begingroup$

          The answer can be more succinct. Put together the vectors $v_i$s to form a matrix $V$ and define $W$ analogously. Then both $V$ and $W$ have full column ranks and $VV^ast=WW^ast$. Hence $V^ast x=0$ if and only if $W^ast x=0$, i.e. the column spaces of $V$ and $W$ have a common orthogonal complement $mathcal X$. Let $X$ be a matrix whose columns form a basis of $mathcal X$. Then both augmented matrices $[V|X]$ and $[W|X]$ are invertible. Since $VV^ast=WW^ast$, we get $[V|X][V|X]^ast = [W|X][W|X]^ast$. Thus $U=[W|X]^{-1}[V|X]$ is a unitary matrix and $V=WU$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very neat. $ $
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 5:27
















          2












          $begingroup$

          The answer can be more succinct. Put together the vectors $v_i$s to form a matrix $V$ and define $W$ analogously. Then both $V$ and $W$ have full column ranks and $VV^ast=WW^ast$. Hence $V^ast x=0$ if and only if $W^ast x=0$, i.e. the column spaces of $V$ and $W$ have a common orthogonal complement $mathcal X$. Let $X$ be a matrix whose columns form a basis of $mathcal X$. Then both augmented matrices $[V|X]$ and $[W|X]$ are invertible. Since $VV^ast=WW^ast$, we get $[V|X][V|X]^ast = [W|X][W|X]^ast$. Thus $U=[W|X]^{-1}[V|X]$ is a unitary matrix and $V=WU$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Very neat. $ $
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 5:27














          2












          2








          2





          $begingroup$

          The answer can be more succinct. Put together the vectors $v_i$s to form a matrix $V$ and define $W$ analogously. Then both $V$ and $W$ have full column ranks and $VV^ast=WW^ast$. Hence $V^ast x=0$ if and only if $W^ast x=0$, i.e. the column spaces of $V$ and $W$ have a common orthogonal complement $mathcal X$. Let $X$ be a matrix whose columns form a basis of $mathcal X$. Then both augmented matrices $[V|X]$ and $[W|X]$ are invertible. Since $VV^ast=WW^ast$, we get $[V|X][V|X]^ast = [W|X][W|X]^ast$. Thus $U=[W|X]^{-1}[V|X]$ is a unitary matrix and $V=WU$.






          share|cite|improve this answer











          $endgroup$



          The answer can be more succinct. Put together the vectors $v_i$s to form a matrix $V$ and define $W$ analogously. Then both $V$ and $W$ have full column ranks and $VV^ast=WW^ast$. Hence $V^ast x=0$ if and only if $W^ast x=0$, i.e. the column spaces of $V$ and $W$ have a common orthogonal complement $mathcal X$. Let $X$ be a matrix whose columns form a basis of $mathcal X$. Then both augmented matrices $[V|X]$ and $[W|X]$ are invertible. Since $VV^ast=WW^ast$, we get $[V|X][V|X]^ast = [W|X][W|X]^ast$. Thus $U=[W|X]^{-1}[V|X]$ is a unitary matrix and $V=WU$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 12 at 12:21


























          community wiki





          2 revs, 2 users 67%
          user1551













          • $begingroup$
            Very neat. $ $
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 5:27


















          • $begingroup$
            Very neat. $ $
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 5:27
















          $begingroup$
          Very neat. $ $
          $endgroup$
          – Martin Argerami
          Nov 22 '18 at 5:27




          $begingroup$
          Very neat. $ $
          $endgroup$
          – Martin Argerami
          Nov 22 '18 at 5:27











          1












          $begingroup$

          The relation is that there exists a unitary $Uin M_r(mathbb C)$ such that $$w_j=sum_{k=1}^r U_{jk}v_k.$$



          Indeed, fix an orthonormal basis $e_1,ldots,e_n$ of $mathbb C^n$, and we may write
          $$
          v_k=sum_{k=1}^ns_{jk}e_j, w_k=sum_{j=1}^rt_{jk}e_j.
          $$
          Now consider $S,Tin M_r(mathbb C)$ with $S_{kj}=s_{kj}$, $T_{kj}=t_{kj}$. Then
          $$
          P=sum_{k=1}^rv_kv_k^*=sum_{k=1}^rsum_{j,ell=1}^rs_{jk}overline{s_{ell k}}e_je_ell^*=sum_{k=1}^rsum_{j,ell=1}^rS_{jk}{(S^*)_{k ell}}e_je_ell^*=sum_{j,ell=1}^r(SS^*)_{jell}e_je_ell^*.
          $$
          Similarly,
          $$
          P=sum_{j,ell=1}^r(TT^*)_{jell}e_je_ell^*.
          $$
          As $e_je_ell^*$ are the matrix units in $M_r(mathbb C)$, we get that $TT^*=SS^*$. Now we write the polar decompositions
          $$
          S^*=W(SS^*)^{1/2}, T^*=V(TT^*)^{1/2},
          $$
          where $W,V$ are unitaries. Then $S=(SS^*)^{1/2}W^*$, and
          $$
          T=(TT^*)^{1/2}V^*=(SS^*)^{1/2}V^*=SWV^*=SZ,
          $$
          where $Z=WV^*$. Now
          begin{align}
          w_j&=sum_{k=1}^r t_{kj}e_k=sum_{k=1}^r(SZ)_{kj}e_k=sum_{k,m=1}^rS_{km}Z_{mj}e_k\ \
          &=sum_{m=1}^r Z_{mj}sum_{k=1}^rS_{km}e_k=sum_{m=1}^rZ_{mj}v_m\ \
          &=sum_{m=1}^rU_{jm}v_m,
          end{align}
          where $U=Z^T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
            $endgroup$
            – mathrookie
            Nov 22 '18 at 21:06












          • $begingroup$
            If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 22:15
















          1












          $begingroup$

          The relation is that there exists a unitary $Uin M_r(mathbb C)$ such that $$w_j=sum_{k=1}^r U_{jk}v_k.$$



          Indeed, fix an orthonormal basis $e_1,ldots,e_n$ of $mathbb C^n$, and we may write
          $$
          v_k=sum_{k=1}^ns_{jk}e_j, w_k=sum_{j=1}^rt_{jk}e_j.
          $$
          Now consider $S,Tin M_r(mathbb C)$ with $S_{kj}=s_{kj}$, $T_{kj}=t_{kj}$. Then
          $$
          P=sum_{k=1}^rv_kv_k^*=sum_{k=1}^rsum_{j,ell=1}^rs_{jk}overline{s_{ell k}}e_je_ell^*=sum_{k=1}^rsum_{j,ell=1}^rS_{jk}{(S^*)_{k ell}}e_je_ell^*=sum_{j,ell=1}^r(SS^*)_{jell}e_je_ell^*.
          $$
          Similarly,
          $$
          P=sum_{j,ell=1}^r(TT^*)_{jell}e_je_ell^*.
          $$
          As $e_je_ell^*$ are the matrix units in $M_r(mathbb C)$, we get that $TT^*=SS^*$. Now we write the polar decompositions
          $$
          S^*=W(SS^*)^{1/2}, T^*=V(TT^*)^{1/2},
          $$
          where $W,V$ are unitaries. Then $S=(SS^*)^{1/2}W^*$, and
          $$
          T=(TT^*)^{1/2}V^*=(SS^*)^{1/2}V^*=SWV^*=SZ,
          $$
          where $Z=WV^*$. Now
          begin{align}
          w_j&=sum_{k=1}^r t_{kj}e_k=sum_{k=1}^r(SZ)_{kj}e_k=sum_{k,m=1}^rS_{km}Z_{mj}e_k\ \
          &=sum_{m=1}^r Z_{mj}sum_{k=1}^rS_{km}e_k=sum_{m=1}^rZ_{mj}v_m\ \
          &=sum_{m=1}^rU_{jm}v_m,
          end{align}
          where $U=Z^T$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
            $endgroup$
            – mathrookie
            Nov 22 '18 at 21:06












          • $begingroup$
            If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 22:15














          1












          1








          1





          $begingroup$

          The relation is that there exists a unitary $Uin M_r(mathbb C)$ such that $$w_j=sum_{k=1}^r U_{jk}v_k.$$



          Indeed, fix an orthonormal basis $e_1,ldots,e_n$ of $mathbb C^n$, and we may write
          $$
          v_k=sum_{k=1}^ns_{jk}e_j, w_k=sum_{j=1}^rt_{jk}e_j.
          $$
          Now consider $S,Tin M_r(mathbb C)$ with $S_{kj}=s_{kj}$, $T_{kj}=t_{kj}$. Then
          $$
          P=sum_{k=1}^rv_kv_k^*=sum_{k=1}^rsum_{j,ell=1}^rs_{jk}overline{s_{ell k}}e_je_ell^*=sum_{k=1}^rsum_{j,ell=1}^rS_{jk}{(S^*)_{k ell}}e_je_ell^*=sum_{j,ell=1}^r(SS^*)_{jell}e_je_ell^*.
          $$
          Similarly,
          $$
          P=sum_{j,ell=1}^r(TT^*)_{jell}e_je_ell^*.
          $$
          As $e_je_ell^*$ are the matrix units in $M_r(mathbb C)$, we get that $TT^*=SS^*$. Now we write the polar decompositions
          $$
          S^*=W(SS^*)^{1/2}, T^*=V(TT^*)^{1/2},
          $$
          where $W,V$ are unitaries. Then $S=(SS^*)^{1/2}W^*$, and
          $$
          T=(TT^*)^{1/2}V^*=(SS^*)^{1/2}V^*=SWV^*=SZ,
          $$
          where $Z=WV^*$. Now
          begin{align}
          w_j&=sum_{k=1}^r t_{kj}e_k=sum_{k=1}^r(SZ)_{kj}e_k=sum_{k,m=1}^rS_{km}Z_{mj}e_k\ \
          &=sum_{m=1}^r Z_{mj}sum_{k=1}^rS_{km}e_k=sum_{m=1}^rZ_{mj}v_m\ \
          &=sum_{m=1}^rU_{jm}v_m,
          end{align}
          where $U=Z^T$.






          share|cite|improve this answer











          $endgroup$



          The relation is that there exists a unitary $Uin M_r(mathbb C)$ such that $$w_j=sum_{k=1}^r U_{jk}v_k.$$



          Indeed, fix an orthonormal basis $e_1,ldots,e_n$ of $mathbb C^n$, and we may write
          $$
          v_k=sum_{k=1}^ns_{jk}e_j, w_k=sum_{j=1}^rt_{jk}e_j.
          $$
          Now consider $S,Tin M_r(mathbb C)$ with $S_{kj}=s_{kj}$, $T_{kj}=t_{kj}$. Then
          $$
          P=sum_{k=1}^rv_kv_k^*=sum_{k=1}^rsum_{j,ell=1}^rs_{jk}overline{s_{ell k}}e_je_ell^*=sum_{k=1}^rsum_{j,ell=1}^rS_{jk}{(S^*)_{k ell}}e_je_ell^*=sum_{j,ell=1}^r(SS^*)_{jell}e_je_ell^*.
          $$
          Similarly,
          $$
          P=sum_{j,ell=1}^r(TT^*)_{jell}e_je_ell^*.
          $$
          As $e_je_ell^*$ are the matrix units in $M_r(mathbb C)$, we get that $TT^*=SS^*$. Now we write the polar decompositions
          $$
          S^*=W(SS^*)^{1/2}, T^*=V(TT^*)^{1/2},
          $$
          where $W,V$ are unitaries. Then $S=(SS^*)^{1/2}W^*$, and
          $$
          T=(TT^*)^{1/2}V^*=(SS^*)^{1/2}V^*=SWV^*=SZ,
          $$
          where $Z=WV^*$. Now
          begin{align}
          w_j&=sum_{k=1}^r t_{kj}e_k=sum_{k=1}^r(SZ)_{kj}e_k=sum_{k,m=1}^rS_{km}Z_{mj}e_k\ \
          &=sum_{m=1}^r Z_{mj}sum_{k=1}^rS_{km}e_k=sum_{m=1}^rZ_{mj}v_m\ \
          &=sum_{m=1}^rU_{jm}v_m,
          end{align}
          where $U=Z^T$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 22 '18 at 13:34

























          answered Nov 22 '18 at 4:20









          Martin ArgeramiMartin Argerami

          126k1182180




          126k1182180












          • $begingroup$
            By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
            $endgroup$
            – mathrookie
            Nov 22 '18 at 21:06












          • $begingroup$
            If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 22:15


















          • $begingroup$
            By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
            $endgroup$
            – mathrookie
            Nov 22 '18 at 21:06












          • $begingroup$
            If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
            $endgroup$
            – Martin Argerami
            Nov 22 '18 at 22:15
















          $begingroup$
          By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
          $endgroup$
          – mathrookie
          Nov 22 '18 at 21:06






          $begingroup$
          By polar decomposition,we have $W,V$ are partial isometries,why are they unitaries?Besides,the other reprentation of $P$ may have $m(m>r)$ sum of rank 1 matrices.
          $endgroup$
          – mathrookie
          Nov 22 '18 at 21:06














          $begingroup$
          If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
          $endgroup$
          – Martin Argerami
          Nov 22 '18 at 22:15




          $begingroup$
          If you do the polar decomposition in $M_r(mathbb C)$, you can always take them to be unitaries. As for $m>r$, enlarge the smallest family with zeroes.
          $endgroup$
          – Martin Argerami
          Nov 22 '18 at 22:15


















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          Mario Kart Wii

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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          Antonio Litta Visconti Arese

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          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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