Commutative property implies all 1-d subspaces are $\varphi$-invariant

Question

Given :

$(1)$ $\varphi$ commutes with all the endomorphisms on $\textsf V$.

$(2)$ Every $1$-dimensional subspace of $\textsf V$ is $\varphi$-invariant.

$(4)$ $\varphi = \lambda \textrm{id}_\textsf{V}$.

How do I show that $(1)$ implies $(2)$, which in turn implies $(4)$ ?

Answer 1

I presume that $\varphi$ is a linear map from $V$ to itself.

For a finite dimensional vector space over $k$, if you view $\varphi$ as a matrix $A$, then the fact that $\varphi$ commutes with all endomorphisms is exactly that $A$ is in the centre of the matrix ring $M_n(k)$, so must be a scalar times the identity (if not then you can just construct a matrix it doesn't commute with).

Then this implies both (2) and (4) directly.