Proving existence of common eigenvalue using invariant subspace properties

Question

Let $T: \Bbb R^n \rightarrow \Bbb R^n$ be a linear transformation such that every subspace $U \subseteq V$ of dimension $n-1$ is $T$-invariant.

I want to prove that there exists a $\lambda$ such that for every $v \in \Bbb R^n$ then $Tv = \lambda v$.

I managed to prove that every vector is an eigenvector, but got kind of stuck there, perhaps this is not the road i should take...

Any help much appreciated!

Answer 1

Take $n=3$ to ease up on notation a little bit. Let $\{e_1,e_2,e_3\}$ be a basis for $\mathbb R^3$.

Let $V = \mathrm{span} \{e_1,e_2\}$ and $W = \mathrm{span}\{e_1,e_3\}$. Since $V$ and $W$ are $T$-invariant there are constants $\alpha,\beta,\gamma,\delta$ with $$Te_1 = \alpha e_1 + \beta e_2$$ and $$Te_1 = \gamma e_1 + \delta e_3.$$ It follows that $\beta = \delta = 0$ and $ \alpha = \gamma$ so that $Te_1 = \alpha_1e_1$ for $\alpha_1 = \alpha$. Of course there similarly exist $\alpha_2$ and $\alpha_3$ so that $Te_2 = \alpha_2 e_2$ and $Te_3 = \alpha_3 e_3$.

Now let $U = \mathrm{span} \{e_1 + e_2, e_1+e_3\}$. Again invoking invariance there are constants $\epsilon, \zeta$ such that $$T(e_1+e_2) = \epsilon(e_1+e_2) + \zeta(e_1 + e_3)$$ but also by the previous argument $T(e_1 + e_2) = \alpha_1 e_1 + \alpha_2 e_2$. Thus $\zeta = 0$ and consequently $\alpha_1 = \epsilon$ and $\alpha_2 = \epsilon$. In particular $\alpha_1 = \alpha_2$.

A very similar argument shows $\alpha_1 = \alpha_3$ so that $\alpha_1=\alpha_2=\alpha_3 = \alpha$ and thus $Te_i = \alpha e_i$ for each $i=1,2,3$. It follows that $Tv = \alpha v$ for all $v$.

Generalizing to $n \not= 3$ isn't difficult but the notation will get tedious. I'm sure you don't mind if I leave it to you.

Answer 2

Consider the standard inner product on $\mathbb{R}^n$.

Claim:

If $W \subseteq \mathbb{R}^n$ is a $T$-invariant subspace, then $W^\perp$ is a $T^t$-invariant subspace.

Proof: Let $v \in W^\perp$. For any $w \in W$ we have $$\langle T^tv,w\rangle = \langle v,Tw\rangle = 0$$ because $Tw \in W$. Hence $T^tv \in W^\perp$.

Now let $v \in \mathbb{R}^n, v \ne 0$ be arbitrary. The subspace $U = \{v\}^\perp$ is $(n-1)$-dimensional so it is $T$-invariant. Therefore, $\operatorname{span}\{v\} = U^\perp$ is $T^t$-invariant which means that $T^tv = \lambda v$ for some $\lambda \in \mathbb{R}$.

Consider the standard basis vector $e_1, \ldots, e_n$ and let $T^te_i = \lambda_i e_i$. Also $\exists \lambda \in \mathbb{R}$ such that $$\lambda(e_1+\cdots + e_n) = T^t(e_1+\cdots + e_n) = T^te_1 + \cdots + T^te_n = \lambda_1e_1 + \cdots + \lambda_ne_n$$

Since $e_1, \ldots, e_n$ are linearly independent, we conclude that $\lambda = \lambda_1 = \cdots = \lambda_n$ so $T^t =\lambda I$.

But then of course we also have $T = \lambda I$.