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Given $A \in\mathbb{S}^n$ positive semi-definite matrix where

$$ \lambda_1 \geq \cdots \geq \lambda_{n-1} > \lambda_n $$

I am trying to come up with an algorithm that uses the Power Iteration Method such that given $\epsilon>0$ will find a vector such that

$$ \lambda_n \leq x^TAx \leq \lambda_n + \epsilon $$ where $$\lambda_n >0$$

In other words I am trying to get the eigenvector relating to the smallest eigenvalue of the matrix $A$.

I saw the following question referring to positive definite, however, since eigenvalues can be equal to zero, the methods suggested such as shifting the eigenvalues won't work.

Any help would be greatly appreciated

Aa me
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  • There's nothing to stop you from applying the answer to the previous question. If there's an eigenvalue strictly greater than 0, then you can apply the approach suggested by the answers to the previous question. If all of the eigenvalues are 0, then any vector will do. – Brian Borchers Jan 01 '22 at 23:35
  • @BrianBorchers The question is if I have some (not strictly all) eigenvalues equal to 0, then when shifting them, wouldn't the the "zero eigenvalues" have the largest magnitude thus be perceived as the smallest? – Aa me Jan 02 '22 at 07:40

1 Answers1

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If $A$ is positive semi-definite, then $A+I$ is positive definite with the same eigenvectors, since $$ Av=\lambda v\iff (A+I)v=(\lambda+1)v. $$ Now, apply the power iteration method on $A+I$.

Edit

In case the power iteration method yields only the largest eigenvalue, perform the following transformation $$(A+I)^{-1}.$$

Mostafa Ayaz
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  • Suppose $\lambda_n = 0$, we first shift it by one to get $\lambda_n = 1$, when applying the power iteration method, the algorithm will return $\lambda_n = 1$,, and this isn't what I am looking for. Am I missing something? – Aa me Jan 02 '22 at 07:47
  • The power iteration method gives you the largest eigenvalue and the corresponding eigenvector. Since the eigenvectors of $A$ and $aA+bI$ are the same for $a\ne 0$, applying power iteration method gives you the eigenvectors of $A$ and the largest $a\lambda +b$ where $\lambda$ is an eigenvalue of $A$, Same is true about $(aA+bI)^{-1}$. – Mostafa Ayaz Jan 02 '22 at 08:31