The rank of a matrix of Gaussian random vectors
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If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?
linear-algebra probability random-matrices
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
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add a comment |
$begingroup$
If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?
linear-algebra probability random-matrices
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
$endgroup$
add a comment |
$begingroup$
If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?
linear-algebra probability random-matrices
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
$endgroup$
If I generate a matrix $Rinmathbb{R^{ktimes p}}$, $k<p$, with i.i.d. entries $R_{i,j}simmathcal{N}(0,sigma^2)$, is there a guarantee that this matrix will have rank $k$, i.e. its columns will be independent?
linear-algebra probability random-matrices
linear-algebra probability random-matrices
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
asked Jan 16 at 8:28
UndertherainbowUndertherainbow
307417
307417
add a comment |
add a comment |
1 Answer
1
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It is not 'surely' true, but we can say that
$$
P(;text{rank}(R)=k ;)=1,
$$ i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
$$ Thus, we have
$$
P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
$$ where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
$
{p(mathcal{R})=0}
$ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
$$
P(;text{rank}(R)<k ;)=0
$$ as desired.
answered Jan 16 at 8:42
SongSong
11.9k628
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
It is not 'surely' true, but we can say that
$$
P(;text{rank}(R)=k ;)=1,
$$ i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
$$ Thus, we have
$$
P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
$$ where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
$
{p(mathcal{R})=0}
$ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
$$
P(;text{rank}(R)<k ;)=0
$$ as desired.
answered Jan 16 at 8:42
SongSong
11.9k628
$endgroup$
add a comment |
$begingroup$
It is not 'surely' true, but we can say that
$$
P(;text{rank}(R)=k ;)=1,
$$ i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
$$ Thus, we have
$$
P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
$$ where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
$
{p(mathcal{R})=0}
$ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
$$
P(;text{rank}(R)<k ;)=0
$$ as desired.
answered Jan 16 at 8:42
SongSong
11.9k628
$endgroup$
add a comment |
$begingroup$
It is not 'surely' true, but we can say that
$$
P(;text{rank}(R)=k ;)=1,
$$ i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
$$ Thus, we have
$$
P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
$$ where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
$
{p(mathcal{R})=0}
$ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
$$
P(;text{rank}(R)<k ;)=0
$$ as desired.
answered Jan 16 at 8:42
SongSong
11.9k628
$endgroup$
It is not 'surely' true, but we can say that
$$
P(;text{rank}(R)=k ;)=1,
$$ i.e. $text{rank}(R)=k$ almost surely. Let $mathcal{R}=(R_{ij})inBbb R^{kp}$ denote the random vector consisting of $R_{ij}$'s. Note that $text{rank}(R)<k$ if and only if $RR^T$ is singular and $det(RR^T)=0$. Hence there is a polynomial $p:mathbb{R}^{kp}to Bbb R$ such that $$
text{rank}(R)<kLeftrightarrow p(mathcal{R})=0.
$$ Thus, we have
$$
P(;text{rank}(R)<k ;)=int_{p(mathcal{R})=0}f(mathcal{R})dmathcal{R},
$$ where $f(mathcal{R})$ is the pdf of $mathcal{R}$. Finally observe that
$
{p(mathcal{R})=0}
$ has a Lebesgue measure zero, which can be proved by induction on the dimesion and Fubini's theorem. This implies
$$
P(;text{rank}(R)<k ;)=0
$$ as desired.
answered Jan 16 at 8:42
SongSong
11.9k628
answered Jan 16 at 8:42
SongSong
11.9k628
answered Jan 16 at 8:42
SongSong
11.9k628
answered Jan 16 at 8:42
SongSong
11.9k628
11.9k628
add a comment |
add a comment |
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