Why is the number of non zero eigenvalues equal to $x^T Sigma^{-1} x$
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I've been reading this code and I found that the number of non-zero eigenvalues of the estimated covariance is equal to $x_i^T Sigma^{-1} x_i$. I want to know how to arrive at this result.
Some background:
$x_i$ is a real column vector with dimension $d$ (one sample)
$X = [x_1, x_2, ..., x_n]$ with shape $d$x$n$ (all the samples)
I want to prove that:
$$sum_{i=0}^{i=n} x_i^T Sigma^{-1} x_i = n.len(s)$$
being $len(s)$ the number of non-zero singular values* of $Sigma$, that is defined as
$$Sigma = frac{1}{n} sum_{j=1}^{j=n} x_j x_j^T$$
If necessary, mean can be considered $0$
*Not necessarily mathematical $0$, this can also mean "not too small values" .
Actually non-zero is "non-negligible" depending on a threshold defined as the largest singular value times the square root of the machine epsilon.
In Python: s[0] * np.sqrt(np.finfo(np.float).eps) being s the singular values in descending order (see the code)
linear-algebra eigenvalues-eigenvectors covariance data-analysis maximum-likelihood
asked Nov 16 at 21:33
Franco Marchesoni
32
add a comment |
up vote
0
down vote
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I've been reading this code and I found that the number of non-zero eigenvalues of the estimated covariance is equal to $x_i^T Sigma^{-1} x_i$. I want to know how to arrive at this result.
Some background:
$x_i$ is a real column vector with dimension $d$ (one sample)
$X = [x_1, x_2, ..., x_n]$ with shape $d$x$n$ (all the samples)
I want to prove that:
$$sum_{i=0}^{i=n} x_i^T Sigma^{-1} x_i = n.len(s)$$
being $len(s)$ the number of non-zero singular values* of $Sigma$, that is defined as
$$Sigma = frac{1}{n} sum_{j=1}^{j=n} x_j x_j^T$$
If necessary, mean can be considered $0$
*Not necessarily mathematical $0$, this can also mean "not too small values" .
Actually non-zero is "non-negligible" depending on a threshold defined as the largest singular value times the square root of the machine epsilon.
In Python: s[0] * np.sqrt(np.finfo(np.float).eps) being s the singular values in descending order (see the code)
linear-algebra eigenvalues-eigenvectors covariance data-analysis maximum-likelihood
asked Nov 16 at 21:33
Franco Marchesoni
32
How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've been reading this code and I found that the number of non-zero eigenvalues of the estimated covariance is equal to $x_i^T Sigma^{-1} x_i$. I want to know how to arrive at this result.
Some background:
$x_i$ is a real column vector with dimension $d$ (one sample)
$X = [x_1, x_2, ..., x_n]$ with shape $d$x$n$ (all the samples)
I want to prove that:
$$sum_{i=0}^{i=n} x_i^T Sigma^{-1} x_i = n.len(s)$$
being $len(s)$ the number of non-zero singular values* of $Sigma$, that is defined as
$$Sigma = frac{1}{n} sum_{j=1}^{j=n} x_j x_j^T$$
If necessary, mean can be considered $0$
*Not necessarily mathematical $0$, this can also mean "not too small values" .
Actually non-zero is "non-negligible" depending on a threshold defined as the largest singular value times the square root of the machine epsilon.
In Python: s[0] * np.sqrt(np.finfo(np.float).eps) being s the singular values in descending order (see the code)
linear-algebra eigenvalues-eigenvectors covariance data-analysis maximum-likelihood
asked Nov 16 at 21:33
Franco Marchesoni
32
I've been reading this code and I found that the number of non-zero eigenvalues of the estimated covariance is equal to $x_i^T Sigma^{-1} x_i$. I want to know how to arrive at this result.
Some background:
$x_i$ is a real column vector with dimension $d$ (one sample)
$X = [x_1, x_2, ..., x_n]$ with shape $d$x$n$ (all the samples)
I want to prove that:
$$sum_{i=0}^{i=n} x_i^T Sigma^{-1} x_i = n.len(s)$$
being $len(s)$ the number of non-zero singular values* of $Sigma$, that is defined as
$$Sigma = frac{1}{n} sum_{j=1}^{j=n} x_j x_j^T$$
If necessary, mean can be considered $0$
*Not necessarily mathematical $0$, this can also mean "not too small values" .
Actually non-zero is "non-negligible" depending on a threshold defined as the largest singular value times the square root of the machine epsilon.
In Python: s[0] * np.sqrt(np.finfo(np.float).eps) being s the singular values in descending order (see the code)
linear-algebra eigenvalues-eigenvectors covariance data-analysis maximum-likelihood
linear-algebra eigenvalues-eigenvectors covariance data-analysis maximum-likelihood
asked Nov 16 at 21:33
Franco Marchesoni
32
asked Nov 16 at 21:33
Franco Marchesoni
32
asked Nov 16 at 21:33
Franco Marchesoni
32
asked Nov 16 at 21:33
Franco Marchesoni
32
asked Nov 16 at 21:33
Franco Marchesoni
32
32
How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07
add a comment |
How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07
How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07
add a comment |
1 Answer
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0
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Assuming that the sample ${x_i}$ spans the whole space $mathbb{R}^d$, the value is always $nd$. Here is the proof:
Using your notation for $X$, we know $Sigma = frac{1}{n} XX^T$ and therefore $Sigma^{-1} = n left(XX^T right)^{-1}$. Then
$$
x_i^T Sigma^{-1} x_i = tr(x_i^T Sigma^{-1} x_i) =
tr(Sigma^{-1} x_i x_i^T)
$$
and therefore
$$
sum_{i = 1}^n x_i^T Sigma^{-1} x_i = tr(Sigma^{-1} sum_{i=1}^n x_i x_i^T)
= tr(Sigma^{-1} (XX^T)) = n tr(I_d) = nd
$$
where $I_d$ is the $d$-dim identity matrix.
answered Nov 19 at 23:41
Hans Engler
9,98411836
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming that the sample ${x_i}$ spans the whole space $mathbb{R}^d$, the value is always $nd$. Here is the proof:
Using your notation for $X$, we know $Sigma = frac{1}{n} XX^T$ and therefore $Sigma^{-1} = n left(XX^T right)^{-1}$. Then
$$
x_i^T Sigma^{-1} x_i = tr(x_i^T Sigma^{-1} x_i) =
tr(Sigma^{-1} x_i x_i^T)
$$
and therefore
$$
sum_{i = 1}^n x_i^T Sigma^{-1} x_i = tr(Sigma^{-1} sum_{i=1}^n x_i x_i^T)
= tr(Sigma^{-1} (XX^T)) = n tr(I_d) = nd
$$
where $I_d$ is the $d$-dim identity matrix.
answered Nov 19 at 23:41
Hans Engler
9,98411836
add a comment |
up vote
0
down vote
Assuming that the sample ${x_i}$ spans the whole space $mathbb{R}^d$, the value is always $nd$. Here is the proof:
Using your notation for $X$, we know $Sigma = frac{1}{n} XX^T$ and therefore $Sigma^{-1} = n left(XX^T right)^{-1}$. Then
$$
x_i^T Sigma^{-1} x_i = tr(x_i^T Sigma^{-1} x_i) =
tr(Sigma^{-1} x_i x_i^T)
$$
and therefore
$$
sum_{i = 1}^n x_i^T Sigma^{-1} x_i = tr(Sigma^{-1} sum_{i=1}^n x_i x_i^T)
= tr(Sigma^{-1} (XX^T)) = n tr(I_d) = nd
$$
where $I_d$ is the $d$-dim identity matrix.
answered Nov 19 at 23:41
Hans Engler
9,98411836
add a comment |
up vote
0
down vote
up vote
0
down vote
Assuming that the sample ${x_i}$ spans the whole space $mathbb{R}^d$, the value is always $nd$. Here is the proof:
Using your notation for $X$, we know $Sigma = frac{1}{n} XX^T$ and therefore $Sigma^{-1} = n left(XX^T right)^{-1}$. Then
$$
x_i^T Sigma^{-1} x_i = tr(x_i^T Sigma^{-1} x_i) =
tr(Sigma^{-1} x_i x_i^T)
$$
and therefore
$$
sum_{i = 1}^n x_i^T Sigma^{-1} x_i = tr(Sigma^{-1} sum_{i=1}^n x_i x_i^T)
= tr(Sigma^{-1} (XX^T)) = n tr(I_d) = nd
$$
where $I_d$ is the $d$-dim identity matrix.
answered Nov 19 at 23:41
Hans Engler
9,98411836
Assuming that the sample ${x_i}$ spans the whole space $mathbb{R}^d$, the value is always $nd$. Here is the proof:
Using your notation for $X$, we know $Sigma = frac{1}{n} XX^T$ and therefore $Sigma^{-1} = n left(XX^T right)^{-1}$. Then
$$
x_i^T Sigma^{-1} x_i = tr(x_i^T Sigma^{-1} x_i) =
tr(Sigma^{-1} x_i x_i^T)
$$
and therefore
$$
sum_{i = 1}^n x_i^T Sigma^{-1} x_i = tr(Sigma^{-1} sum_{i=1}^n x_i x_i^T)
= tr(Sigma^{-1} (XX^T)) = n tr(I_d) = nd
$$
where $I_d$ is the $d$-dim identity matrix.
answered Nov 19 at 23:41
Hans Engler
9,98411836
answered Nov 19 at 23:41
Hans Engler
9,98411836
answered Nov 19 at 23:41
Hans Engler
9,98411836
answered Nov 19 at 23:41
Hans Engler
9,98411836
9,98411836
add a comment |
add a comment |
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How do you propose to compute $Sigma^{-1}$ if $Sigma$ has zero singular values? Otherwise, this is just a trace identity. The answer should always be $n$.
– Hans Engler
Nov 16 at 21:54
In fact, the question is about the proposal itself. The code I'm looking into, and this other R package, assume the proposition that involves $n. len(s)$. I want to know when it's valid, and what approximations and assumptions are being made. Sorry if it wasn't clear
– Franco Marchesoni
Nov 19 at 21:07