Every $k$ dimensional subspace is $S$-invariant implies $S$ is a multiple of the identity Let $V$ be a $d$-dimensional real vector space. Let $S:V \to V$ be an invertible linear map, and let $2 \le k \le d-1$ be fixed. Suppose that every $k$-dimensional subspace is $S$-invariant.
Then, I claim that $S$ is a multiple of the identity. I present a proof below.
Question: Are there easier proofs? (or just other nice proofs, since "easier" or "elegant" is subjective). Also, does the claim hold for non-invertible maps? If not, can we characterize all the maps satisfying this property, for a given $k$?

My proof:
We reduce the general case to $k=1$, which is well-known and easy. We shall prove the following lemma:

Lemma: For every $w,v \neq 0 \in V$, $w \notin \text{span}\{v\} \Rightarrow Sv \notin \text{span}\{w\}$.

In particular, this implies $Sw \notin \text{span}\{w\} \Rightarrow Sw \notin \text{span}\{Sw\}$ which is a contradiction.
A proof of the lemma:
Suppose $w \notin \text{span}\{v\}$. Then, since $k \le d-1$ we can complete $w,v$ to an independent set $w,v,v_1,\dots,v_{k-1}$. Since $S$ is invertible, $Sw,Sv,Sv_1,\dots,Sv_{k-1}$ are independent. Thus,$$Sv \notin \text{span}\{Sw,Sv_1,\dots,Sv_{k-1}\}.$$
By assumption, we know $$\text{span}\{Sw,Sv_1,\dots,Sv_{k-1}\}\subseteq \text{span}\{w,v_1,\dots,v_{k-1}\},$$ and since both are $k$-dimensional there is an equality. Thus
$$Sv \notin \text{span}\{w,v_1,\dots,v_{k-1}\},$$ and in particular $Sv \notin \text{span}\{w\}$.
 A: Let $v\in V$ and let $X(v)$ be the collection of $k$-dimensional subspaces of $V$ that contain $v$. Then 
$$\langle v\rangle=\bigcap_{W\in X(v)}W,$$
and since each $W$ is $S$-invariant, also $\langle v\rangle$ is $S$-invariant, reducing the problem to the case $k=1$.

Note that this argument holds for any map, not necessarily invertible or even linear. To conclude that $S$ is a scalar multiple of the identity we do need that $S$ is linear, but not that it is invertible.
A: We want to prove that if every $k$-dimensional subspace is $T$-invariant for $2\leq k \leq d-1$, then every $(k-1)$-dimensional subspace is $T$-invariant as well.
Let $W$ be a $(k-1)$-dimensional subspace. Find vectors $u$ and $v$ such that $W\oplus\mathbb R u\oplus\mathbb R v$ is $(k+1)$-dimensional (we can do this because $k-1\leq d-2$). Since both $W\oplus \mathbb R u$ and $W\oplus\mathbb R v$ are $k$-dimensional, they are $T$-invariant, so for all $w\in W$, we have $Tw \in W\oplus \mathbb R u$ and $Tw\in W\oplus \mathbb R v$, which implies that $Tw\in (W\oplus \mathbb R u)\cap(W\oplus \mathbb R v) = W$. Thus, $W$ is $T$-invariant.
Therefore, we can reduce the problem to $k=1$, as you mentioned. Notice, however, that I've never assumed that $T$ is invertible. We can still conclude that $T = \lambda I$ and $\lambda$ might just as well be $0$.
A: An idea for you:
Take any basis $\;\{v_1,...,v_n\}\;$ of $\;V\;$ . Since $\;U_i:=\text{Span}\{v_i\}\;$ is $\;T\,-$ invariant, we get $\;Tv_i=k_iv_i\,,\,\,k_i\in\Bbb R\;$.
But also $\;W:=\text{Span}\{v_1+v_2\}\;$ is $\;T\,-$ invariant, which means that there exists $\;\alpha\in\Bbb R\;$ s.t.
$$T(v_1+v_2)=\begin{cases}\alpha(v_1+v_2)=\alpha v_1+\alpha v_2\;\;-\;\;\text{invariant}\\{}\\
Tv_1+Tv_2=k_1v_1+k_2v_2\;\;--\text{linearity of}\;T\end{cases}$$
and from here we get that (observe that $\;\{v_1,v_2\}\;$ is linearly independent!):
$$(\alpha-k_1)v_1+(\alpha-k_2)v_2=0\implies \;k_1=\alpha=k_2\implies\;k_1=k_2$$
and and easy inductive argument now finishes the proof that $\;T=k_1\cdot I\;$ 
