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What is the difference between the spectral radius and dominant eigenvalue? If they are one and the same then why do both get mentioned, for instance here http://reference.wolfram.com/mathematica/tutorial/NDSolveStiffnessTest.html

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Let $A$ be a matrix, and $\sigma(A)$ signifies the set of all eigenvalues$(\lambda_i)$ of $A$. Then

An eigenvalue of $A$ that is larger in absolute value than any other eigenvalue is called the dominant eigenvalue.

But

Spectral radius of $A$, which is denoted by $\rho(A)$ is defined as: $\rho(A) = \max\{|\lambda|:\lambda\ \epsilon\hspace{1mm}\sigma(A)$

Thus, spectral radius is more widely applicable; every matrix has a well defined spectral radius. Not every matrix has a dominant eigenvalue but there are theorems guaranteeing the existence of a dominant eigenvalue under appropriate conditions; first among these is the Perron-Frobenius theorem.

Matrices with dominant eigenvalues often arise in numerical approximation schemes for differential equations and the "stiffness" of a system can be quantified in terms of the size of the dominant eigenvalue. Rather than compute the exact value of the dominant eigenvalue, a numerical scheme might use a cheaper estimate of the spectral radius to determine stiffness. This is why both terms are mentioned in your link.

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    $\begingroup$ Put it simply, the spectral radius is the modulus of the dominant eigenvalue. $\endgroup$ – user1551 Jun 16 '14 at 15:46
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    $\begingroup$ @user1551 False $\endgroup$ – Mark McClure Jun 16 '14 at 15:47
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    $\begingroup$ @user1551 Maybe - Every time I've seen the term 'dominant eigenvalue', it's uniqueness is essential. The power method is a classic example. More to the point, the uniqueness is an essential part of the answer as stated. $\endgroup$ – Mark McClure Jun 16 '14 at 16:06
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    $\begingroup$ @lavkush if a matrix has 2 eigenvalues with the same modulus, say complex conjugates, which is the largest, then do we say that the matrix does not have a dominant eigenvalue? $\endgroup$ – yaska Jun 16 '14 at 16:59
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    $\begingroup$ @Ranade Correct - such a matrix does not have a dominant eigenvalue. $\endgroup$ – Mark McClure Jun 16 '14 at 18:50

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