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

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:

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

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 $$c_w=c_w$$.
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.