All subspaces $W$ of $V$ are $T$-invariant, is $T = cI:V\to V$?

Question

$V$ is a vector space over $\mathbb{F}$. Any subspace $W$ of $V$ is $T$-invariant, is $T = cI:V\to V$ for some $c \in \mathbb{F}$?

If $T=cI:V\to V$ for some $c \in \mathbb{F}$, then any subspace $W$ of $V$ is $T$-invariant (obviously). Is the converse true, i.e. if any subspace $W$ of $V$ is $T$-invariant, is $T = cI:V\to V$ for some $c \in \mathbb{F}$?

I'm unable to prove this or find a counterexample. Any hints?

Answer 1

Yes, that is true.

Take a vector $v\ne 0$. Since all 1-dimensional subspaces are invariant, then $$Tv = c_vv$$ If now $0\ne w\ne v$, for the same reason $$Tw = c_ww$$
If $w,v$ are dependent, then $w=\lambda v$ and $$Tw = \lambda Tv = \lambda c_vv = c_vw$$ so $c_w=c_v$ necessarily. If $w,v$ are independent, $$T(w+v) = c_{w+v}(w+v) = c_ww + c_vv \implies (c_w - c_{w+v})w + (c_v - c_{w+v})v=0$$ but the coefficient must be zero, so $c_w = c_{w+v} = c_v$.

This proves that $c_w=c_v$ for any couple of nonzero vectors $v,w$, so there's a unique constant $c$ such that $T = cI$