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Uniqueness (up to isometries) of a $L^TL$ factorization of a PSD matrix.
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I'm trying to see that measuring in $mathbbR^d$ with a metric given by a PSD matrix s equivalent to measuring with the usual euclidean metric after making a linear transformation. This is consequence from a result about PSD matrices that says that every PSD matrix $M$ of dimension $d$ can be decomposed as $M = L^TL$, where $L$ is a square matrix of order $d$. I can prove this by taking a spectral decomposition of $M$, $M = U^TDU$, with $U$ orthogonal, and taking $L = U^TD^1/2U$, where $D^1/2$ denotes the matrix with the square root of the (non negative) elements of the diagonal matrix $D$ (in fact, this $L$ is symmetric).
I was wondering about the uniqueness of the decompositions $M = L^TL$. As $M$ defines a metric, it seems logic that all the linear maps that define that decomposition must be equal up to an isometry, that is, if $M = L^TL = K^TK$, exists an orthogonal matrix $O$ so that $K = OL$. I'm trying to prove this, but I don't know how should I start. Any hints?
matrix-decomposition isometry positive-semidefinite
asked Jul 24 at 11:39
gtf
205
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I'm trying to see that measuring in $mathbbR^d$ with a metric given by a PSD matrix s equivalent to measuring with the usual euclidean metric after making a linear transformation. This is consequence from a result about PSD matrices that says that every PSD matrix $M$ of dimension $d$ can be decomposed as $M = L^TL$, where $L$ is a square matrix of order $d$. I can prove this by taking a spectral decomposition of $M$, $M = U^TDU$, with $U$ orthogonal, and taking $L = U^TD^1/2U$, where $D^1/2$ denotes the matrix with the square root of the (non negative) elements of the diagonal matrix $D$ (in fact, this $L$ is symmetric).
I was wondering about the uniqueness of the decompositions $M = L^TL$. As $M$ defines a metric, it seems logic that all the linear maps that define that decomposition must be equal up to an isometry, that is, if $M = L^TL = K^TK$, exists an orthogonal matrix $O$ so that $K = OL$. I'm trying to prove this, but I don't know how should I start. Any hints?
matrix-decomposition isometry positive-semidefinite
asked Jul 24 at 11:39
gtf
205
mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45
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I'm trying to see that measuring in $mathbbR^d$ with a metric given by a PSD matrix s equivalent to measuring with the usual euclidean metric after making a linear transformation. This is consequence from a result about PSD matrices that says that every PSD matrix $M$ of dimension $d$ can be decomposed as $M = L^TL$, where $L$ is a square matrix of order $d$. I can prove this by taking a spectral decomposition of $M$, $M = U^TDU$, with $U$ orthogonal, and taking $L = U^TD^1/2U$, where $D^1/2$ denotes the matrix with the square root of the (non negative) elements of the diagonal matrix $D$ (in fact, this $L$ is symmetric).
I was wondering about the uniqueness of the decompositions $M = L^TL$. As $M$ defines a metric, it seems logic that all the linear maps that define that decomposition must be equal up to an isometry, that is, if $M = L^TL = K^TK$, exists an orthogonal matrix $O$ so that $K = OL$. I'm trying to prove this, but I don't know how should I start. Any hints?
matrix-decomposition isometry positive-semidefinite
asked Jul 24 at 11:39
gtf
205
I'm trying to see that measuring in $mathbbR^d$ with a metric given by a PSD matrix s equivalent to measuring with the usual euclidean metric after making a linear transformation. This is consequence from a result about PSD matrices that says that every PSD matrix $M$ of dimension $d$ can be decomposed as $M = L^TL$, where $L$ is a square matrix of order $d$. I can prove this by taking a spectral decomposition of $M$, $M = U^TDU$, with $U$ orthogonal, and taking $L = U^TD^1/2U$, where $D^1/2$ denotes the matrix with the square root of the (non negative) elements of the diagonal matrix $D$ (in fact, this $L$ is symmetric).
I was wondering about the uniqueness of the decompositions $M = L^TL$. As $M$ defines a metric, it seems logic that all the linear maps that define that decomposition must be equal up to an isometry, that is, if $M = L^TL = K^TK$, exists an orthogonal matrix $O$ so that $K = OL$. I'm trying to prove this, but I don't know how should I start. Any hints?
matrix-decomposition isometry positive-semidefinite
asked Jul 24 at 11:39
gtf
205
asked Jul 24 at 11:39
gtf
205
asked Jul 24 at 11:39
gtf
205
asked Jul 24 at 11:39
gtf
205
205
mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45
add a comment |Â
mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45
mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45
mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45
add a comment |Â
1 Answer
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I think I have found an answer, making use of the polar decomposition and the uniqueness of the square root of positive semidefinite matrices. Every matrix $L in mathcalM_d(mathbbR)$ can be decomposed as $L = U|L|$, where $U$ is orthogonal, and $|L| = (L^TL)^1/2$ is a PSD matrix. So, if we have $M = L^TL = K^TK$, and polar decompositions $L = U|L|, K = V|K|$, with $U$ and $V$ orthogonal matrices, then
beginalign*
L^TL = K^TK &implies |L|^TU^TU|L| = |K|^TV^TV|K| \
&implies |L|^T|L| = |K|^T|K| \
&implies |L|^2 = |K|^2,
endalign*
so we have two PSD matrices whose squares are equal. By the uniqueness of the square root of positive matrices, we have $|L| = |K|$. We call that matrix $N$. Going back, we have
$$N = U^TL = V^TK implies K = VU^TL. $$
So taking $O = VU^T$, we have the orthogonal matrix we looked for. However, I find the polar decomposition a too powerful tool for a result like this. So if anyone finds a simple way to prove this I'm still interested.
answered Jul 24 at 15:32
gtf
205
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think I have found an answer, making use of the polar decomposition and the uniqueness of the square root of positive semidefinite matrices. Every matrix $L in mathcalM_d(mathbbR)$ can be decomposed as $L = U|L|$, where $U$ is orthogonal, and $|L| = (L^TL)^1/2$ is a PSD matrix. So, if we have $M = L^TL = K^TK$, and polar decompositions $L = U|L|, K = V|K|$, with $U$ and $V$ orthogonal matrices, then
beginalign*
L^TL = K^TK &implies |L|^TU^TU|L| = |K|^TV^TV|K| \
&implies |L|^T|L| = |K|^T|K| \
&implies |L|^2 = |K|^2,
endalign*
so we have two PSD matrices whose squares are equal. By the uniqueness of the square root of positive matrices, we have $|L| = |K|$. We call that matrix $N$. Going back, we have
$$N = U^TL = V^TK implies K = VU^TL. $$
So taking $O = VU^T$, we have the orthogonal matrix we looked for. However, I find the polar decomposition a too powerful tool for a result like this. So if anyone finds a simple way to prove this I'm still interested.
answered Jul 24 at 15:32
gtf
205
add a comment |Â
up vote
0
down vote
I think I have found an answer, making use of the polar decomposition and the uniqueness of the square root of positive semidefinite matrices. Every matrix $L in mathcalM_d(mathbbR)$ can be decomposed as $L = U|L|$, where $U$ is orthogonal, and $|L| = (L^TL)^1/2$ is a PSD matrix. So, if we have $M = L^TL = K^TK$, and polar decompositions $L = U|L|, K = V|K|$, with $U$ and $V$ orthogonal matrices, then
beginalign*
L^TL = K^TK &implies |L|^TU^TU|L| = |K|^TV^TV|K| \
&implies |L|^T|L| = |K|^T|K| \
&implies |L|^2 = |K|^2,
endalign*
so we have two PSD matrices whose squares are equal. By the uniqueness of the square root of positive matrices, we have $|L| = |K|$. We call that matrix $N$. Going back, we have
$$N = U^TL = V^TK implies K = VU^TL. $$
So taking $O = VU^T$, we have the orthogonal matrix we looked for. However, I find the polar decomposition a too powerful tool for a result like this. So if anyone finds a simple way to prove this I'm still interested.
answered Jul 24 at 15:32
gtf
205
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think I have found an answer, making use of the polar decomposition and the uniqueness of the square root of positive semidefinite matrices. Every matrix $L in mathcalM_d(mathbbR)$ can be decomposed as $L = U|L|$, where $U$ is orthogonal, and $|L| = (L^TL)^1/2$ is a PSD matrix. So, if we have $M = L^TL = K^TK$, and polar decompositions $L = U|L|, K = V|K|$, with $U$ and $V$ orthogonal matrices, then
beginalign*
L^TL = K^TK &implies |L|^TU^TU|L| = |K|^TV^TV|K| \
&implies |L|^T|L| = |K|^T|K| \
&implies |L|^2 = |K|^2,
endalign*
so we have two PSD matrices whose squares are equal. By the uniqueness of the square root of positive matrices, we have $|L| = |K|$. We call that matrix $N$. Going back, we have
$$N = U^TL = V^TK implies K = VU^TL. $$
So taking $O = VU^T$, we have the orthogonal matrix we looked for. However, I find the polar decomposition a too powerful tool for a result like this. So if anyone finds a simple way to prove this I'm still interested.
answered Jul 24 at 15:32
gtf
205
I think I have found an answer, making use of the polar decomposition and the uniqueness of the square root of positive semidefinite matrices. Every matrix $L in mathcalM_d(mathbbR)$ can be decomposed as $L = U|L|$, where $U$ is orthogonal, and $|L| = (L^TL)^1/2$ is a PSD matrix. So, if we have $M = L^TL = K^TK$, and polar decompositions $L = U|L|, K = V|K|$, with $U$ and $V$ orthogonal matrices, then
beginalign*
L^TL = K^TK &implies |L|^TU^TU|L| = |K|^TV^TV|K| \
&implies |L|^T|L| = |K|^T|K| \
&implies |L|^2 = |K|^2,
endalign*
so we have two PSD matrices whose squares are equal. By the uniqueness of the square root of positive matrices, we have $|L| = |K|$. We call that matrix $N$. Going back, we have
$$N = U^TL = V^TK implies K = VU^TL. $$
So taking $O = VU^T$, we have the orthogonal matrix we looked for. However, I find the polar decomposition a too powerful tool for a result like this. So if anyone finds a simple way to prove this I'm still interested.
answered Jul 24 at 15:32
gtf
205
answered Jul 24 at 15:32
gtf
205
answered Jul 24 at 15:32
gtf
205
answered Jul 24 at 15:32
gtf
205
205
add a comment |Â
add a comment |Â
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mathoverflow.net/questions/155147/…
– RHowe
Jul 24 at 15:45