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
asked Nov 22 at 4:20
Martin Argerami
123k1176174
add a comment |
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
asked Nov 22 at 4:20
Martin Argerami
123k1176174
add a comment |
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
asked Nov 22 at 4:20
Martin Argerami
123k1176174
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
asked Nov 22 at 4:20
Martin Argerami
123k1176174
asked Nov 22 at 4:20
Martin Argerami
123k1176174
edited Nov 22 at 22:16
asked Nov 22 at 4:20
Martin Argerami
123k1176174
asked Nov 22 at 4:20
Martin Argerami
123k1176174
asked Nov 22 at 4:20
Martin Argerami
123k1176174
123k1176174
add a comment |
add a comment |
2 Answers
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2
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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 $V=WU$.
Very neat. $ $
– Martin Argerami
Nov 22 at 5:27
add a comment |
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 $U=Z^T$.
answered Nov 22 at 4:20
Martin Argerami
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
add a comment |
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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 $V=WU$.
Very neat. $ $
– Martin Argerami
Nov 22 at 5:27
add a comment |
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 $V=WU$.
Very neat. $ $
– Martin Argerami
Nov 22 at 5:27
add a comment |
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 $V=WU$.
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 $V=WU$.
answered Nov 22 at 5:13
community wiki
user1551
Very neat. $ $
– Martin Argerami
Nov 22 at 5:27
add a comment |
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
add a comment |
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 $U=Z^T$.
answered Nov 22 at 4:20
Martin Argerami
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
add a comment |
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 $U=Z^T$.
answered Nov 22 at 4:20
Martin Argerami
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
add a comment |
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 $U=Z^T$.
answered Nov 22 at 4:20
Martin Argerami
123k1176174
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 $U=Z^T$.
answered Nov 22 at 4:20
Martin Argerami
123k1176174
edited Nov 22 at 13:34
answered Nov 22 at 4:20
Martin Argerami
123k1176174
answered Nov 22 at 4:20
Martin Argerami
123k1176174
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
add a comment |
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
add a comment |
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