# Proof explanation about $T=\lambda\,\mathit{Id}_V$

Let $V$ a vector space of finite dimension an let $T:V\rightarrow V$ be a linear operator such that every hyperplane of $V$ is $T$-stable. Prove $T=\lambda\,\mathit{Id}_{V}$ for some $\lambda$.

Note: $\mathit{Id}_{V}$ is the identity operator.

Proof:

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 $(n-1)$-dimensional subspaces $W_i = \operatorname{span}( v, v_2, \dots, v_{i-1}, v_{i+1}, v_n )$, for $i=2,\dots,n$. It is clear that the subspace $\operatorname{span}(v)= W_2 \cap \dots \cap W_n$ 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 $\lambda_v = \lambda_{v+v'} = \lambda_{v'}$.

My questions:

1) Why does $\operatorname{span}(v)= W_2 \cap \dots \cap W_n$?

2) Why $\operatorname{span}(v)$ is $T$-invariant?

3) I can find $v_2, \dots, v_n \in V$ for create a basis for the theorem of completion of basis, don't I?

Question 1. It is clear that $v\in W_2\cap\dots\cap W_n$. Conversely, suppose $w\in W_2\cap\dots\cap W_n$. Then $$w=\alpha v+\alpha_2v_2+\dots+\alpha_{n-1}v_{n-1}$$ due to $w\in W_n$. Due to $w\in W_2$, we have $w=\beta v+\beta_3v_3+\beta_4v_4+\dots+\beta_nv_n$. Since the set $\{v,v_2,\dots,v_n\}$ is linearly independent, we conclude, $\alpha_2=0$. Similarly for $\alpha_3,\dots,\alpha_{n-1}$.
Question 2. The intersection of $T$-invariant subspaces is also $T$-invariant. It suffices to show it for two subspaces, say $U_1$ and $U_2$. If $v\in U_1\cap U_2$, then $T(v)\in U_1$, because $U_1$ is $T$-invariant and similarly $T(v)\in U_2$; therefore $T(v)\in U_1\cap U_2$.
Since it is assumed that every hyperplane is $T$-invariant, then also every intersection of hyperplanes is $T$-invariant.
• egreg, if $\{v,v_2,\dots,v_n\}$ is linearly independent then not happen $\alpha=0$?, i say in the question 1. – Bvss12 Nov 27 '17 at 3:33
• And other question, i don't see how you obtain this: $w=\beta v+\beta_3v_3+\beta_4v_4+\dots+\beta_nv_n$ and for what do you use that to the proof? – Bvss12 Nov 27 '17 at 3:49
• @Bvss12 Why should $\alpha=0$? Look at the case $w=v$. The second expression for $w$ is because it belongs to $W_2$ and what the spanning set is. – egreg Nov 27 '17 at 7:10