# Proof that $1$ is always an eigenvalue of following matrix:

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?

• 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. – obareey Jan 12 at 17:47

Just verify that (by hypothesis) vector $$(1,1,\cdots,1)^{t}$$ is an eigen vector with eigen value $$1$$.
• 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. – Zachary Jan 12 at 12:23