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I have to show that given the matrix A equal to:

$$\pmatrix{a_{11} & \dots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \dots & a_{nn}}$$

Where:

$$ \sum_{j =1}^{n} a_{ij} = u, \forall i \in \{1, \dots, n\} $$

Then $ u $ is a eigenvalue, and then get his corresponding eigenvector....

I tried to show that checking if $ A \vec{v} = u \vec{v} $ or $ \det{(A - uI)} = 0 $, but I couldn't conclude anything....

How can I conclude that $ u $ is a eigenvalue?

OiciTrap
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1 Answers1

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Hint: The sum of the entries in the $i$-th row of a matrix $A$ is the same as the $i$-th entry of the vector

$$A \begin{pmatrix} 1 \\ 1 \\ \vdots \\ 1\end{pmatrix}$$

Daniel McLaury
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  • I understand now, but in first place, how did you come up with that? I mean, to do that you should know in the first place that (1, 1, 1, ..., 1) is an eigenvector, is there a hint or something I should notice to conclude that (1, 1, 1, ..., 1) is a eigenvector? Sorry if the question is too newbie, but I can't conclude that immediately just looking at the matrix – OiciTrap Nov 27 '16 at 13:55
  • Well, the thing I wrote as a hint was obvious to me, and once you think about that you realize that that's an eigenvector. – Daniel McLaury Nov 27 '16 at 16:20