# Spectral radius of a graph

I researched about the spectral radius and was confused. There are two definitions.

1. The spectral radius of a finite graph is defined to be the spectral radius of its adjacency matrix. the spectral radius of a square matrix is the largest absolute value of its eigenvalues.

2. The largest eigenvalue in the spectrum of a graph is the spectral radius of a graph.

When they are equivalent? If the graph is connected, these definitions are equivalent?

Thanks for the help.

By the Perron–Frobenius theorem:

• If $$A$$ is a $$n\times n$$ real matrix with strictly positive entries, it has a positive real eigenvalue $$r$$ such that any other eigenvalue $$\lambda$$ satisfies $$|\lambda| < r$$. There's also some other stuff about the eigenvector associated to $$r$$ that you don't need here.
• If $$A$$ is an $$n \times n$$ real matrix with nonnegative entries, then we can only say $$|\lambda| \le r$$ for other eigenvalues. The eigenvalue $$r$$ could appear multiple times, we could have complex eigenvalues with absolute value $$r$$, and so forth.

The second case applies in particular to adjacency matrices of graphs. (Also, these are symmetric, so all their eigenvalues are real). So the spectral radius is $$r$$, and it is both the largest eigenvalue, and the largest absolute value of an eigenvalue: your definitions are equivalent.

As a bonus, if the graph is connected, then the adjacency matrix is an irreducible matrix, and the Perron–Frobenius theorem additionally tells us that the eigenvalue $$r$$ is simple (it only appears once, both algebraically and geometrically). But we don't need the graph to be connected for the definitions to be equivalent.

• Dear Misha, thanks for your answer, but why they are equivalent? Sep 1, 2021 at 17:54
• One of your definitions says that the spectral radius is the largest eigenvalue. The other definition says that the spectral radius is the largest absolute value of any eigenvalue. I just explained why those are the same number. Sep 1, 2021 at 18:05

If the eigenvalue of largest norm was negative then $$\sum\limits_{i=1}^n \lambda_i^k$$ would be negative for large odd values of $$k$$.

However we know that $$\sum\limits_{i=1}^n \lambda_i^k = \operatorname{Tr}(A^k)$$, and this last quantity is non-negative as the diagonal of $$A^k$$ is clearly non-negative. We conclude there can't be a negative eigenvalue with norm strictly larger than all other eigenvalues.