Why operators on complex vector spaces have eigenvalues There is one thing I can't grasp about the proof given in the Linear Algebra Done Right book by Sheldon Axler (attached below).
In the last part it says that $(T - \lambda_1I)...(T - \lambda_mI)v = 0$, hence $T - \lambda_jI$ is not injective for some $j$.
What I don't understand is why the following reasoning is not correct:

*

*the factors in the equation can be reordered.

*suppose $\lambda_j$ is the only eigenvalue $T$ has. Let's put it at the end: $(T - \lambda_1I)...(T - \lambda_mI)(T - \lambda_jI)v = 0$

*the only way for the expression above to be equal to $0$ is if $(T - \lambda_jI)v = 0$ (because $\lambda_j$ is the only eigenvalue, so the other $T - \lambda_iI$ are injective).

*hence $v$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda_j$. But $v$ was chosen arbitrarily, so it can't be true.

I know that my logic is flawed but I can't see where. Would appreciate it if someone pointed out to me where I'm wrong.


If $\lambda_j$ is the only eigenvalue then $(T−\lambda_j)^n$ is the full expression and reordering doesn't change anything.
