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Consider the real $n\times n$-matrix $A = (a_{ij})_{1\le i,j \le n}$ for which $\sum_{j=1}^n a_{ij} = 1$ for all $i \in \{1,\dots,n\}$. Show that $1$ is always an eigenvalue of $A$.

The sum of all elements per row equals $1$. Let's assume that $1$ is never an eigenvalue of $A$. This would mean that $\neg \exists x\in K^n: Ax=x \iff \forall x\in K^n: Ax\ne x.$ So we have: $Ax-x=x(A-I_n)\ne0, \forall x\in K^n$. This would imply that $x\ne0 \wedge A \ne I_n$ for all column matrices $x$. This is a contradiction, because the zero column matrix $O_{n\times 1} \in K^n$. Therefore $1$ is always an eigenvalue of $A$.

Or I could use this argument: $A \ne I_n$ means that $A$ cannot be the identity. However $I_n$ does satisfy the conditions stated problem. Therefore this is false.

Is this a valid proof?

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  • $\begingroup$ You cannot use $0$ as an eigenvector. Your contradiction is not a contradiction. If that were the case all matrices should have an eigenvalue of 1. $\endgroup$ – obareey Jan 12 at 17:47
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Just verify that (by hypothesis) vector $(1,1,\cdots,1)^{t}$ is an eigen vector with eigen value $1$.

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  • $\begingroup$ Since the row sums are 1. $\endgroup$ – Wuestenfux Jan 12 at 12:02
  • $\begingroup$ I misinterpreted the question and thought that all elements of $A$ were equal to $1$. I edited it and tried to prove the statement again. $\endgroup$ – Zachary Jan 12 at 12:23

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