Chapter 7 - Eigenvalues and Eigenvectors - 7.7 Chapter Review - Additional Problems - Page 490: 20

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Let $A$ and $B$ be $n \times n$ matrices. We have $Av=\lambda_1 .v\\ Bv=\lambda.v$ Obtain: $(AB-BA)v=ABv-BAv\\=A(\lambda_2v)-B(\lambda_1 v)\\ =\lambda_2(Av)-\lambda_1(Bv)\\ =\lambda_2 \lambda_1v-\lambda_1 \lambda_2 v\\ =0$ We know that $v$ is an eigenvector of $A$ o $v \ne 0$ and since $(AB-BA)v=0$, $AB-BA$ is not invertible.