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I have a matrix $\textbf{A}$ and form the diagonal matrix $\bar{\textbf{A}}$ from the diagonal entries of $\textbf{A}$. Is there a relationship between the eigenvalues of $\textbf{A}$ and $\bar{\textbf{A}}$, or at least between their spectral norms $||\textbf{A}||_2$ and $||\bar{\textbf{A}}||_2$?

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    $\begingroup$ The sum of eigenvalues is equal to the trace, which is the sum of diagonal elements. Without further information it's impossible to deduce more. $\endgroup$
    – user376343
    Jan 22, 2019 at 9:07
  • $\begingroup$ @user376343 What if $\textbf{A}$ is also positive definite? Does that have any implications on the eigenvalues of $\bar{\textbf{A}}$? $\endgroup$ Jan 22, 2019 at 9:10
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    $\begingroup$ I very much doubt that there’s anything beyond the trace that @user376343 mentioned. The only eigenvalue of $I_2$ is $1$, but the eigenvalues of a $2\times2$ matrix with $1$s on its main diagonal can be any two numbers that add up to $2$. The eigenvalues of $\small{\begin{bmatrix}1&1\\-1&1\end{bmatrix}}$ aren’t even real numbers. $\endgroup$
    – amd
    Jan 22, 2019 at 19:34
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    $\begingroup$ @amd it is exactly what I sayd - from the trace we know the sum of eigenvalues and nothing else. $\endgroup$
    – user376343
    Jan 22, 2019 at 19:58
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    $\begingroup$ @amd It might be interesting to you that the answers prove your suspicion wrong, both about the spectral norm (there is a strict inequality according to the OPs answer) and about all other eigenvalues (they are close to each other according to the Gershgorin circle theorem, depending on the size of the non-diagonal entries) $\endgroup$
    – Bananach
    Feb 2, 2019 at 21:54

2 Answers 2

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There actually is a relationship between their spectral norms. In general, the spectral norm of $\textbf{A}$ is greater than any of the entries of $\textbf{A}$, i.e. $$||\textbf{A}||_2\ge\textbf{A}_{ij}, \ \forall i,j$$ The proof for this is as follows: \begin{align*} ||\textbf{A}||_2&=\sqrt{\underset{\textbf{x}}{\text{sup}}\frac{\textbf{x}^T\textbf{A}\textbf{A}^T\textbf{x}}{||\textbf{x}||_2^2}}\\ &=\underset{||\textbf{y}||_2=1}{\text{sup}}\underset{||\textbf{x}||_2=1}{\text{sup}}\textbf{x}^T\textbf{A}\textbf{y}\\ &\ge\textbf{e}_i^T\textbf{A}\textbf{e}_j\\ &=\textbf{A}_{ij} \end{align*} where $\textbf{e}_i$ is a zero vector with a single $1$ in the $i$th entry. It follows that $$||\bar{\textbf{A}}||_2=\max\limits_{i}|A_{ii}|\le||\textbf{A}||_2$$

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The Gershgorin circle theorem bounds the distance of the eigenvalues to the diagonal elements

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  • $\begingroup$ Take that, delete-voter. The OP seems to disagree with you about the usefulness of this answer $\endgroup$
    – Bananach
    Mar 1, 2019 at 22:49

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