Proving that Mx is an eigenvector of B with λ as the eigenvalue given that $B=MAM^{-1}$ and $Ax=λx$ > The square matrix A has λ as an eigenvalue with corresponding eigenvector x. The non-singular matrix M is of the sa...--prophetes.ai

Proving that Mx is an eigenvector of B with λ as the eigenvalue given that $B=MAM^{-1}$ and $Ax=λx$ > The square matrix A has λ as an eigenvalue with corresponding eigenvector x. The non-singular matrix M is of the same order as A. Show that Mx is an eigenvector of the matrix B, where $B = MAM^{−1}$, and that λ is the corresponding eigenvalue. Here is How I started: $$Ax=λx$$ $$BM=MAM^{-1}M$$ $$BM= MA$$ $$BM= Mλx$$ Something is not right. Can someone shed some light? Thanks.

We have

$$B(Mx)=(BM)x=(MA)x=M(Ax)=M(\lambda x)=\lambda(Mx)$$ hence the desired result.