Relation between different presentations of a positive semidefinite matrix











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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







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    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


























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      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















      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













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      edited Nov 22 at 22:16

























      asked Nov 22 at 4:20









      Martin Argerami

      123k1176174




      123k1176174






















          2 Answers
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          I am not sure what purpose does your exercise serve, but 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























          • Very neat. $ $
            – Martin Argerami
            Nov 22 at 5:27


















          up vote
          1
          down vote













          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























          • 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.
            – mathrookie
            Nov 22 at 21:06












          • 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.
            – Martin Argerami
            Nov 22 at 22:15











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          2 Answers
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          2 Answers
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          up vote
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          I am not sure what purpose does your exercise serve, but 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























          • Very neat. $ $
            – Martin Argerami
            Nov 22 at 5:27















          up vote
          2
          down vote













          I am not sure what purpose does your exercise serve, but 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























          • Very neat. $ $
            – Martin Argerami
            Nov 22 at 5:27













          up vote
          2
          down vote










          up vote
          2
          down vote









          I am not sure what purpose does your exercise serve, but 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














          I am not sure what purpose does your exercise serve, but 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








          answered Nov 22 at 5:13


























          community wiki





          user1551













          • Very neat. $ $
            – Martin Argerami
            Nov 22 at 5:27


















          • Very neat. $ $
            – Martin Argerami
            Nov 22 at 5:27
















          Very neat. $ $
          – Martin Argerami
          Nov 22 at 5:27




          Very neat. $ $
          – Martin Argerami
          Nov 22 at 5:27










          up vote
          1
          down vote













          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























          • 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.
            – mathrookie
            Nov 22 at 21:06












          • 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.
            – Martin Argerami
            Nov 22 at 22:15















          up vote
          1
          down vote













          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























          • 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.
            – mathrookie
            Nov 22 at 21:06












          • 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.
            – Martin Argerami
            Nov 22 at 22:15













          up vote
          1
          down vote










          up vote
          1
          down vote









          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














          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 at 13:34

























          answered Nov 22 at 4:20









          Martin Argerami

          123k1176174




          123k1176174












          • 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.
            – mathrookie
            Nov 22 at 21:06












          • 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.
            – Martin Argerami
            Nov 22 at 22:15


















          • 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.
            – mathrookie
            Nov 22 at 21:06












          • 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.
            – Martin Argerami
            Nov 22 at 22:15
















          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.
          – mathrookie
          Nov 22 at 21:06






          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.
          – mathrookie
          Nov 22 at 21:06














          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.
          – Martin Argerami
          Nov 22 at 22:15




          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.
          – Martin Argerami
          Nov 22 at 22:15


















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          Quarter-circle Tiles

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          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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          Mont Emei

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          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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