# Spectral radius and Dominant Eigenvalue

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

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.

• Put it simply, the spectral radius is the modulus of the dominant eigenvalue. – user1551 Jun 16 '14 at 15:46
• @user1551 False – Mark McClure Jun 16 '14 at 15:47
• @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. – Mark McClure Jun 16 '14 at 16:06
• @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? – yaska Jun 16 '14 at 16:59
• @Ranade Correct - such a matrix does not have a dominant eigenvalue. – Mark McClure Jun 16 '14 at 18:50