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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          The Binding of Isaac: Rebirth/Afterbirth

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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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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