Generating a rectangular matrix $A_{mxn}$ where $AA^{T}$ is positive semidefinite
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Let, $m>n$. I want to generate rectangular matrices $A_{mtext{x}n}$ where $AA^{T}$ is always positive semidefinite (PSD).
When I say 'generate', I mean something like this: Take any lower triangle matrix $L$, the product of $LL^{T}$ will be positive semidefinite. So, no matter how I populate the matrix $L$ I can be sure that $LL^{T}$ is always PSD.
I am looking for something similar in this case.
Thanks in advance.
Here is what I was thinking:
Say, $A$ can be decomposed using SVD such that $A=USigma V^{T}$ where $U$ and $V$ are orthonormal matrices. So, may be we need to generate $U$ and $V$?
linear-algebra matrices
asked Dec 3 '18 at 17:27
kasakasa
328114
$endgroup$
add a comment |
$begingroup$
Let, $m>n$. I want to generate rectangular matrices $A_{mtext{x}n}$ where $AA^{T}$ is always positive semidefinite (PSD).
When I say 'generate', I mean something like this: Take any lower triangle matrix $L$, the product of $LL^{T}$ will be positive semidefinite. So, no matter how I populate the matrix $L$ I can be sure that $LL^{T}$ is always PSD.
I am looking for something similar in this case.
Thanks in advance.
Here is what I was thinking:
Say, $A$ can be decomposed using SVD such that $A=USigma V^{T}$ where $U$ and $V$ are orthonormal matrices. So, may be we need to generate $U$ and $V$?
linear-algebra matrices
asked Dec 3 '18 at 17:27
kasakasa
328114
$endgroup$
2
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
2
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37
add a comment |
$begingroup$
Let, $m>n$. I want to generate rectangular matrices $A_{mtext{x}n}$ where $AA^{T}$ is always positive semidefinite (PSD).
When I say 'generate', I mean something like this: Take any lower triangle matrix $L$, the product of $LL^{T}$ will be positive semidefinite. So, no matter how I populate the matrix $L$ I can be sure that $LL^{T}$ is always PSD.
I am looking for something similar in this case.
Thanks in advance.
Here is what I was thinking:
Say, $A$ can be decomposed using SVD such that $A=USigma V^{T}$ where $U$ and $V$ are orthonormal matrices. So, may be we need to generate $U$ and $V$?
linear-algebra matrices
asked Dec 3 '18 at 17:27
kasakasa
328114
$endgroup$
Let, $m>n$. I want to generate rectangular matrices $A_{mtext{x}n}$ where $AA^{T}$ is always positive semidefinite (PSD).
When I say 'generate', I mean something like this: Take any lower triangle matrix $L$, the product of $LL^{T}$ will be positive semidefinite. So, no matter how I populate the matrix $L$ I can be sure that $LL^{T}$ is always PSD.
I am looking for something similar in this case.
Thanks in advance.
Here is what I was thinking:
Say, $A$ can be decomposed using SVD such that $A=USigma V^{T}$ where $U$ and $V$ are orthonormal matrices. So, may be we need to generate $U$ and $V$?
linear-algebra matrices
linear-algebra matrices
asked Dec 3 '18 at 17:27
kasakasa
328114
asked Dec 3 '18 at 17:27
kasakasa
328114
asked Dec 3 '18 at 17:27
kasakasa
328114
asked Dec 3 '18 at 17:27
kasakasa
328114
asked Dec 3 '18 at 17:27
kasakasa
328114
328114
2
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
2
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37
add a comment |
2
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
2
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37
2
2
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
2
2
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37
add a comment |
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2
$begingroup$
This problem is way asier than you think...
$endgroup$
– kimchi lover
Dec 3 '18 at 19:49
2
$begingroup$
$AA^T$ is always PSD when $A$ is a real matrix.
$endgroup$
– darij grinberg
Dec 3 '18 at 19:54
$begingroup$
@darijgrinberg I understand that $A^{T}A$ is $ntext{x}n$ square matrix, and will be of min(rank $n$,rank $m$) = $n$, hence PSD. But, I am not sure, how $AA^{T}$ is PSD. Can you please throw some light?
$endgroup$
– kasa
Dec 5 '18 at 19:37