# Proving a Theorem based on Gerschgorin Theorem

Here is the theorem as it appears in my textbook. I am so lost with it.

For $A=(a_{ij}) \in \mathbb C^{n\times n}$ we have

$$\rho(A) \leq \max_i\sum_j^n | a_{ij}|$$

where $\rho(A)$ is the spectral radius.

I need to prove this. I have no idea how to do it. :(

The textbook I am using is Matrices and Linear Transformations by Cullen.

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You need to say what p is. – Ted Sep 21 '11 at 7:21
What is $p(A)$, did you mean the spectral radius $\rho(A)$? In that case perhaps it is helpful to you that your right hand side is the norm of $A$ induced by the maximum norm on $\mathbb R^n$. – martini Sep 21 '11 at 7:23
You might want to mention what textbook you're using, so people can steer you to a proof that builds from what you've already learned from the book. – J. M. Sep 21 '11 at 7:39
Im sorry, i just added the additional info now. For some reason, my page wasnt refreshing, so I only saw the comments and the answer now. Thanks ;) – sarah jamal Sep 22 '11 at 13:47

Using just the Gershgorin circle theorem and some basic facts about $|\cdot|$ and $\rho(A)$ is the spectral radius of $A$ (i.e., $\max\{|\lambda|\}$ of matrix $A$). Here is a sketch of the proof.
If $\lambda$ is an eigenvalue of A then the Gershgorin circle theorem states that there exists $i$, $|\lambda - a_{ii}| \leq \sum_{j\neq i} |a_{ij}|$.
As $|\lambda - a_{ii}| \geq |\lambda| - |a_{ii}|$ we have $$|\lambda| - |a_{ii}| \leq \sum_{j\neq i} |a_{ij}|.$$ Now $$|\lambda| \leq \sum_{j} |a_{ij}| \leq \max_{i} \sum_{j} |a_{ij}|.$$