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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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    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 ;)2011-09-22

1 Answers 1

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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 $|\lambda| \leq \sum_{j} |a_{ij}| \leq \max_{i} \sum_{j} |a_{ij}|.$