MHB An Overview of the History of Tea

[enogjini]

 

[Evgeny.Makarov]

Welcome to the forum. Have you tried simplifying $(A-cI)x=(\lambda-c)x$?

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[Kansas Boy]

First. $\lambda$ is an eigenvalue of A if and only if there exist a vector, v, such that $Av= \lambda v$. Of course, for any vector v, Iv= v so for any number c, cIv= cv. Then $(A- cI)v= Av- cIv= \lambda v- cv= (\lambda- c)v$