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Given some $n \times n$ matrix with $m\le n$ distinct eigenvalues, I usually see the spectrum denoted as distinct eigenvalues $ \lambda=\{ \lambda_1, \ldots,\lambda_m \} $ with multiplicities $ \mu=\{ \mu_1, \ldots,\mu_m \} $. For example, given $$ A=\begin{bmatrix} 5 & 0 & 0 & 0\\ 0 & 4 & 0 & 0\\ 0 & 0 & 4 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} $$ would denote $\sigma(A)$ as $ \lambda=\{5,4,1\} $ with multiplicities $ \mu=\{1,2,1\}$.

How is the eigenvalue corresponding to the $i$th eigenvector usually denoted? i.e., how do I denote both of:

  1. the third (distinct) eigenvalue of $A$ is $1$?
  2. the third (including non-distinct) eigenvalue of $A$ is $4$?

I am currently thinking:
$\lambda^{(i)}$ refers to the $i$th (distinct) eigenvalue such that $\lambda^{(1)}>\dots>\lambda^{(m)}$
$\lambda_i$ refers to the $i$th (including non-distinct) eigenvalue such that $\lambda_1\ge\dots\ge\lambda_n$
According to this schema $\lambda^{(3)}=1$ and $\lambda_3=4$.
Is this reasonable?

Context of question:
I am writing an analysis of an algorithm which utilises eigenprojections. I frequently need to denote both the $i$th distinct eigenvalue and the eigenvalue corresponding to the $i$th eigenvector.

Thanks in advance for your patience.

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  • $\begingroup$ Eigenvalue or eigenspace? $\endgroup$ Jun 4 at 15:17
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    $\begingroup$ Welcome to MSE. I think you have at least one typo in your question or it requires some clarification. You state yourself that the eigenvalues are usually denoted with lambdas but then ask how the eigenvalues corresponding to eigenvectors are denoted. And in what manner do you enumerate the eigenvectors when you're referring to the $i$-th one? How do you handle the multiplicities here? Also: it's just "spectrum" without the "eigen" prefix and conventionally we'd call that set $\sigma(A)$ (assuming the matrix is called $A$) rather than $\lambda$. $\endgroup$
    – SV-97
    Jun 4 at 15:20
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    $\begingroup$ People typically say $\lambda_1\leq \lambda_2 \leq \dots \lambda_n$ listed with multiplicity. No special notation is introduced. $\endgroup$ Jun 5 at 8:58

1 Answer 1

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I would say $\lambda_{i}$ the $i$th eigenvalue, which has $E_{i}$ as eigenspace.

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  • $\begingroup$ Under this schema, I feel that in my example $\lambda_3$ could refer to the value 1 rather than the value 4, as desired. $\endgroup$
    – user160623
    Jun 4 at 18:43
  • $\begingroup$ $\lambda_{i}$ is the $a_{i \times i}$ element in A. $\endgroup$ Jun 4 at 18:49

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