SVD - reconstruction from U,S,V
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I am learning some linear algebra for image compression and I am stuck at this point:
Suppose I have a matrix $R$,
$$ begin{bmatrix}
5 & 7\
2 & 1\end{bmatrix} $$
Then I compute the covariance matrix s.t. $$Sigma =frac12R^TR$$
And I performed SVD with a Matlab function s.t. $ [U, S, V] = svd(Sigma) $
I can see that $USV = Sigma$ but how can I solve this equation below for $R$:
$Sigma=frac12R^TR$
linear-algebra matrices svd
edited Nov 17 at 9:29
Parcly Taxel
41k137199
asked Nov 17 at 9:24
michcs
1
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michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
I am learning some linear algebra for image compression and I am stuck at this point:
Suppose I have a matrix $R$,
$$ begin{bmatrix}
5 & 7\
2 & 1\end{bmatrix} $$
Then I compute the covariance matrix s.t. $$Sigma =frac12R^TR$$
And I performed SVD with a Matlab function s.t. $ [U, S, V] = svd(Sigma) $
I can see that $USV = Sigma$ but how can I solve this equation below for $R$:
$Sigma=frac12R^TR$
linear-algebra matrices svd
edited Nov 17 at 9:29
Parcly Taxel
41k137199
asked Nov 17 at 9:24
michcs
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
1
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
1
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
1
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am learning some linear algebra for image compression and I am stuck at this point:
Suppose I have a matrix $R$,
$$ begin{bmatrix}
5 & 7\
2 & 1\end{bmatrix} $$
Then I compute the covariance matrix s.t. $$Sigma =frac12R^TR$$
And I performed SVD with a Matlab function s.t. $ [U, S, V] = svd(Sigma) $
I can see that $USV = Sigma$ but how can I solve this equation below for $R$:
$Sigma=frac12R^TR$
linear-algebra matrices svd
edited Nov 17 at 9:29
Parcly Taxel
41k137199
asked Nov 17 at 9:24
michcs
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am learning some linear algebra for image compression and I am stuck at this point:
Suppose I have a matrix $R$,
$$ begin{bmatrix}
5 & 7\
2 & 1\end{bmatrix} $$
Then I compute the covariance matrix s.t. $$Sigma =frac12R^TR$$
And I performed SVD with a Matlab function s.t. $ [U, S, V] = svd(Sigma) $
I can see that $USV = Sigma$ but how can I solve this equation below for $R$:
$Sigma=frac12R^TR$
linear-algebra matrices svd
linear-algebra matrices svd
edited Nov 17 at 9:29
Parcly Taxel
41k137199
asked Nov 17 at 9:24
michcs
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 17 at 9:29
Parcly Taxel
41k137199
asked Nov 17 at 9:24
michcs
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 17 at 9:29
Parcly Taxel
41k137199
edited Nov 17 at 9:29
Parcly Taxel
41k137199
edited Nov 17 at 9:29
Parcly Taxel
41k137199
41k137199
asked Nov 17 at 9:24
michcs
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 17 at 9:24
michcs
1
asked Nov 17 at 9:24
michcs
1
1
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
michcs is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
1
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
1
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
1
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52
|
show 1 more comment
You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
1
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
1
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
1
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52
You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
1
1
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
1
1
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
1
1
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
Given $Sigma = frac{1}{2}R^TR$, you first let $Sigma = LL^T$. Find out $L$ by Cholesky factorization.
Now $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$. Can you continue from here?
answered Nov 17 at 9:42
tonychow0929
15612
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Given $Sigma = frac{1}{2}R^TR$, you first let $Sigma = LL^T$. Find out $L$ by Cholesky factorization.
Now $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$. Can you continue from here?
answered Nov 17 at 9:42
tonychow0929
15612
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
add a comment |
up vote
0
down vote
Given $Sigma = frac{1}{2}R^TR$, you first let $Sigma = LL^T$. Find out $L$ by Cholesky factorization.
Now $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$. Can you continue from here?
answered Nov 17 at 9:42
tonychow0929
15612
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
add a comment |
up vote
0
down vote
up vote
0
down vote
Given $Sigma = frac{1}{2}R^TR$, you first let $Sigma = LL^T$. Find out $L$ by Cholesky factorization.
Now $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$. Can you continue from here?
answered Nov 17 at 9:42
tonychow0929
15612
Given $Sigma = frac{1}{2}R^TR$, you first let $Sigma = LL^T$. Find out $L$ by Cholesky factorization.
Now $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$. Can you continue from here?
answered Nov 17 at 9:42
tonychow0929
15612
answered Nov 17 at 9:42
tonychow0929
15612
answered Nov 17 at 9:42
tonychow0929
15612
answered Nov 17 at 9:42
tonychow0929
15612
15612
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
add a comment |
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
2
2
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
The solution is not unique though, and quite probably not the original $R$.
– Qidi
Nov 17 at 10:03
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Thank you, with the $R$ above, I computed $Sigma$ to be; $$ begin{bmatrix} 14.5 & 18.5\ 18.5 & 25\end{bmatrix} $$ and the cholesky factorization gives L as; $$ begin{bmatrix} 3.8079 & 0\ 4.8583 & 1.1813\end{bmatrix} $$ From $(frac{1}{sqrt{2}})^2(R^T)(R^T)^T=LL^T$, it seems like I am back at the same problem of $R^T*R$, can you give me a little more guidance please? Thank you -
– michcs
Nov 19 at 7:49
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
Additionally, I have noticed that my toy example $R$ is non-Hermitian positive definite matrix. Suppose that I find the closest approximation to make it positive definite, how can I proceed?
– michcs
Nov 19 at 8:08
add a comment |
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You can't for $R =begin{pmatrix} 1&0\0&1end{pmatrix}$ and for $R=begin{pmatrix}0&1\1&0end{pmatrix}$, you have $Sigma=frac{1}{2}begin{pmatrix}1&0\0&1end{pmatrix}$
– Charles Madeline
Nov 17 at 9:29
@CharlesMadeline That is true, I just did the maths on paper and I see your point. What I was wondering about is, given $Sigma = 1/2 R^T R$, how can I solve for $R$?
– michcs
Nov 17 at 9:47
1
@michcs This comments just explained with a couterexample that you can't in general.
– Jean-Claude Arbaut
Nov 17 at 9:49
1
@Jean-ClaudeArbaut Thank you, if I understand this correctly, if I perform an SVD on a covariance matrix to get $U, S, V$, I can only reconstruct the covariance matrix, but not the $R$ I used in $Sigma = 1/2 R^T R$.
– michcs
Nov 17 at 9:52
1
Yes, that's right. Actually it has nothing to do with the SVD: once you compute $Sigma=frac12R^TR$, $R$ is lost, whatever you do on $Sigma$, since the mapping $Rto R^TR$ is not injective.
– Jean-Claude Arbaut
Nov 17 at 9:52