Why we can say that $T(\alpha_{i}+\alpha{j})=\lambda(\alpha_{i}+\alpha_{j})$ in a T-invariant subspace?

Question

My problem have the hypothesis that every subspace of V is T-invariant (with T a lineal operator over V). Then I've to prove that T is a scalar multiply of the identity operator. There are some questions about the proof here:

Let T be a linear operator on V.If every subspace of V is invariant under T,then T is a scalar multiple of the identity operator. If every subspace of V is T-invariant, prove that T is a multiple of the identity map.

My question is not about how to prove it, my question is about a step that I don't understand and I think it is something conceptual that I am not understanding well:

Let be a base ${\alpha_{i}}$ of V. Then, the subspace generated by one vector $\alpha_{i}$ (linear combinations of $\alpha_{i}$) satisfies $T\alpha_{i}=c_{i}\alpha_{i}$, this since every everysubspace of $V$ is $T$-invariant. (with $c_{i}$ a scalar)

My doubt is here:

If now we take the subspace generated by $\left \{ \alpha_{i},\alpha{j} \right \}$ , then $T(\alpha_{i}+\alpha{j})=T\alpha_{i}+T\alpha_{j}=c_{i}\alpha_{i}+c_{i}\alpha_{j}$

But I don't know how to use the fact that the subspace is T-invariant to conclude that: $T(\alpha_{i}+\alpha{j})=\lambda(\alpha_{i}+\alpha_{j})$ (with $\lambda$ a scalar)

I understand that $T(\alpha_{i}+\alpha{j})$ is in span$(\alpha_{i},\alpha{j})$, but that implies that $T(\alpha_{i}+\alpha{j})$ is equal to a linear combination of $\alpha_{i}$ and $\alpha{j}$, that is: $c_{i}\alpha_{i}+c_{i}\alpha_{j}$.

What I don't understand is why we can say $T(\alpha_{i}+\alpha{j})=\lambda(\alpha_{i}+\alpha_{j})$ (with $\lambda$ a scalar)

I really appreciate your help

Answer 1

You know that for each $v\in V$, there exists a constant $c_v$ such that $Tv=c_vv$. By linearity of $T$, $$ T(v+w)=Tv+Tw=c_vv+c_ww,$$ but of course also $$ T(v+w)=c_{v+w}(v+w).$$ Combined, $$(c_{v+w}-c_v)v+ (c_{v+w}-c_w)w=0.$$ Whenever $v,w$ are linearly independent (e.g., different members of a basis), this leads to $c_{v+w}-c_v=c_{v+w}-c_w=0$ and in particular $c_v=c_w$.

Answer 2

We have $$\left.\begin{array}{}T\alpha_i = c_i\alpha_i \\ T\alpha_j = c_j\alpha_j\end{array}\right\}\implies T(\alpha_i+\alpha_j) = c_i\alpha_i+c_j\alpha_j \\ T(\alpha_i+\alpha_j) = c_{ij}(\alpha_i+\alpha_j)$$

Compare the coefficients of $\alpha_i$ and $\alpha_j$ (using the fact they are linearly independent) to see $c_i = c_{ij} = c_j$ so all the coefficients are equal.