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Let

$$A=\begin{pmatrix}3&1&1\\1&2&1\\0&1&2\end{pmatrix}.$$

I am asked to find its spectral radius, i.e., $$\rho(A) = \max \left\{ |\lambda| : \lambda \text{ eigenvalue of }A \right\}$$ without using the characteristic polynomial. I have tried Gershgorin circles, bounding by norms ($\rho(A)=\inf_{\|\cdot\|}{\| A \|}$), etc. I only got $\rho(A)\leq4$ with these methods. Any ideas?

(Practice exam of Numerical-Methods for Algebra, 2nd Grade in Mathematics).

moqui
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  • https://en.wikipedia.org/wiki/Power_iteration – Rodrigo de Azevedo Dec 29 '21 at 10:37
  • https://mathoverflow.net/questions/35445/estimating-the-spectral-radius-of-a-matrix-noniteratively – Golden_Ratio Dec 29 '21 at 10:44
  • To garantee the power iteration converges, how do you prove that $A$ has an eigenvalue that is strictly greater in magnitude than its other eigenvalues? – moqui Dec 29 '21 at 10:47
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    Perron-Frobenius? – markvs Dec 29 '21 at 10:55
  • Thanks, I didn't know that theorem. Now, by the power iteration, starting with the normalized vector $1/\sqrt{3}(1,1,1)$, I get 4 as the first approximation to $\rho(A)$, which is indeed the exact value. But how can I prove the power iteration converges to 4? – moqui Dec 29 '21 at 11:18

1 Answers1

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I found a solution, I post it in order to mark the question as answered.

As shown here, since $\sum_{i=1}^3a_{ij}=4$ for $j=1,2,3$, we have that 4 is eigenvalue of $A$. Therefore, $$\rho(A)\geq4$$ and we conclude $\rho(A)=4$.

moqui
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