Suppose I have finite dimensional Hilbert spaces $V$, $W$, and an operator $A$ acting on vectors in $V$ such that it has eigenvectors/values $Ax_a=\lambda_ax_a$. In the tensor product space I want to find the most general eigenvectors of operator $A\otimes I$. I think that they are all of the form $x_a\otimes w$ where $w\in W$ is any vector. It's easy to show that $x_a\otimes w$ is an eigenvector: $$(A\otimes I)(x_a \otimes w)=(Ax_a)\otimes w=\lambda_a (x_a\otimes w)$$
Moreover, if $v\otimes w$ is a general pure tensor, and we suppose it is an eigenvector, then we have $$(A\otimes I)(v\otimes w)=(Av)\otimes w=\lambda v\otimes w=(\lambda v)\otimes w$$ which is iff $v$ is an eigenvector of $A$. But how can I show that the only eigenvectors of $A\otimes I$ are of the form $x_a\otimes w$?
Best Answer
Your conjecture is not true. Take e.g. $A=I$. Then every tensor $\sum_i v_i\otimes w_i\in V\otimes W$ is an eigenvector of $A\otimes I=I\otimes I=I$.