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I've a question regarding to eigenvalues, sinve I am not very familiar with the concept. Suppose I've a matrix $X'X$ in the case of an OLS regession. And lets assume that the regarding eigenvalues are very small. This imlpies strongly multicollinearity and a high variance of the OLS estimators. Is there any relationship between the eigenvalues of $X'X$ and the ones of $(X'X)^{-1}$ ?

The following point should be correct: The smaller the eigenvalues of $X'X$ or the larger the eigenvalues of $(X'X)^{-1}$ the nearer we are to multicollinearity and $Var(\beta)$ will be large.

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Let's say $\lambda$ is one of the eigen values of $X'X$.

For eiegen vector $v$, this equation is hold : $X'X v= \lambda v$

And this can be rewritten like this : $(X'X)^{-1} v= \lambda^{-1} v$

which implies that the eiegen values of the inverse matrix is reciprocal of those of the original matrix.

I hope this is helpful for you.

I am not a native English speaker. Any improvement is welcome.Thank you in advance.

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  • $\begingroup$ Thank you. This would underline my last sentence. And the larger the eigenvalues of the inverse are the large the variance of the coefficients, or am I wrong ? $\endgroup$
    – Leo96
    Commented Aug 29, 2018 at 9:19
  • $\begingroup$ How about using spectral decomposition? That allows us to write like this : $(X'X)^{-1}=A' \Lambda A=\sum_{i=1}^n \lambda_i v_i v_i'$ $\endgroup$
    – KDG
    Commented Aug 29, 2018 at 9:29
  • $\begingroup$ and the variance covariance matrix of beta vector is $\sigma^2 (X'X)^{-1}$ $\endgroup$
    – KDG
    Commented Aug 29, 2018 at 9:30
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    $\begingroup$ I think this would show that your statement is true clearly. $\endgroup$
    – KDG
    Commented Aug 29, 2018 at 9:31
  • $\begingroup$ I am not familiar with specral decomposition, still a beginner in statistics^^. But it seems as larger the the eigenvalues of the inverse as larger the variance, via spectral decompositoin $\endgroup$
    – Leo96
    Commented Aug 29, 2018 at 9:31

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