Proving a Theorem based on Gerschgorin Theorem

1k Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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