Why is the number of non zero eigenvalues equal to $x^T Sigma^{-1} x$












0














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)










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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 '18 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 '18 at 21:07
















0














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)










share|cite|improve this question






















  • 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 '18 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 '18 at 21:07














0












0








0







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)










share|cite|improve this question













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






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asked Nov 16 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 21:07










1 Answer
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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.






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    0














    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 '18 at 23:41









        Hans Engler

        10.1k11836




        10.1k11836






























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