$T:V \to V$-every sub-space of $n-1$ dimension is T-invariant, then $T=\lambda I$ for $\lambda \in F$ Let $T:V \to V$ be linear transformation, $\dim V=n$. If I know that every subspace of dimension $n-1$ is $T$-invariant, then how can I prove that $T=\lambda I$ for $\lambda \in F$?
$V$ is the vector space and $F$ is the field we work in. 
Thank you.
 A: Every $k$-dimensional subspace of $V$ can be seen as the intersection of two $(k+1)$-dimensional subspaces (for $k < n$).
Therefore, by induction, $T$ stabilizes every subspace of $V$, in particular the $1$-dimensional ones.
Thus there exists a function $\lambda : V\setminus\{0\} \to F$ such that  for all $x\in V\setminus\{0\}$ , $Tx=\lambda(x) x$.
Taking $x,y$ independent, we obtain $\lambda(x)=\lambda(x+y)=\lambda(y)$, which finally shows that $\lambda$ is constant. This concludes.
A: Assume that there is $v\in V$ such that $Tv$ is linearly independent of $v$. Then we could find a subspace of dimension $n-1$ containing $v$ but not $Tv$, which contradicts the fact that $T$ leaves such a subspace invariant. Thus there is no such $v$, and $T$ acts as a scalar on all vectors. This must be the same scalar for all vectors, since otherwise applying $T$ to the sum of two non-zero vectors multiplied by different scalars would yield a linearly independent vector.
Note that $n-1$ could be replaced by any other integer $m$ with $0<m<n$.
A: Pick a vector $v \in V \setminus \{ 0 \}$. You can find $v_2, \dots, v_n \in V$ such that $\{v,v_2, \dots,v_n \}$ is basis of $V$. Now consider the $(n-1)$-subspaces $W_i = \langle v, v_2, \dots, v_{i-1}, v_{i+1}, \dots, v_n \rangle$, for $i=2,\dots,n$. It is clear that the subspace $\langle v \rangle = W_2 \cap \cdots \cap W_n$ is $T$-invariant. So there exists $\lambda_v \in F$ such that $Tv = \lambda_v v$. (Pay attention: $\lambda_v$ depends on $v$.)
Now we must prove the scalars $\{ \lambda_v \}_{v \in V \setminus \{ 0 \}}$ are all the same. Choose two vectors $v, v' \neq 0$; if $v$ and $v'$ are linearly dependent then it is clear that $\lambda_v = \lambda_{v'}$. If $v$ and $v'$ are linearly independent, then $\lambda_{v+v'} (v + v') = Tv + Tv' = \lambda_v v + \lambda_{v'} v'$, so $\lambda_v = \lambda_{v+v'} = \lambda_{v'}$.
