Show that a transformation is linear if and only if its restriction to subspaces of dimension 2 is linear. Let $V$ be a vector space over a field $\mathbb{K}$ with $\dim_\mathbb{K} \geq 3$. Show that a transformation $T : V \rightarrow V $ is linear if and only if the restriction of $T$ to each subspace of dimension $2$ of $V$ is linear.
(->) If $T$ is linear in $V$ then it's clear that it's also linear in any subspace of $V$.
(<-) Suppose $\dim_\mathbb{K} = n \geq 3$ and that $T$ is linear in any subspace of dimension $2$ of $V$. Let $\{b_1,b_2, \cdots, b_n\} \subset V$ be a basis for $V$. Now consider the following subspaces of $V$:
$$
W_i = \text{span}(\{b_i, b_{i+1}\})
$$
Now let $v = \big(\sum_{i=1}^n \alpha_i \cdot b_i\big) \in V$. Therefore:
$$
v = \sum_{i=1}^n \alpha_i \cdot b_i = \underbrace{(\alpha_1 b_1 + \alpha_2 b_2)}_{\in W_1} + \underbrace{(\alpha_3 b_3 + \alpha_4 b_4)}_{\in W_3} + \cdots + \underbrace{(\alpha_{n-1} b_{n-1} + \alpha_n b_n)}_{\in W_{n-1}}
$$
And from that it follows that if $n$ is even, then
$$
V = W_1 \oplus W_3 \oplus \cdots \oplus W_{n-1}
$$
and if $n$ is odd, then:
$$
V = W_1 \oplus W_3 \oplus \cdots \oplus W_{n-2} \oplus \text{span}(\{b_n\})
$$
It's clear to see that the sum is direct since $W_i \cap W_{i+2} = \{0\}$.
Now I need to prove the linearity of $T$ in $V$, so let $v = \sum_{i=1}^n \alpha_i \cdot b_i$, $u = \sum_{i=1}^n \beta_i \cdot b_i$ and $\lambda \in \mathbb{K}$.
So it remains to prove that $T(u+v) = T(u) + T(v)$ and $T(\lambda \cdot u) = \lambda \cdot T(u)$.
$$
\begin{align*}
T(u+v) = T\big( \sum_{i=1}^n \alpha_i \cdot b_i + \sum_{i=1}^n \beta_i \cdot b_i \big) = 
T\big( \sum_{i=1}^n (\alpha_i + \beta_i) \cdot b_i \big) = \cdots
\end{align*}
$$
And now I'm stuck because for me "the restriction of $T$ to each subspace of dimension $2$ of $V$ is linear" means is that $T$ is going to be linear in each of those $W_i$ that I've defined. That means that if $w = \alpha b_i + \beta b_{i+1} \in W_i$ then $T(w) = \alpha \cdot T(b_i) + \beta \cdot T(b_{i+1})$. But that does not implies that
$$
T(w_1 + w_3 + \cdots + w_{n-1}) = T(w_1) + T(w_3) + \cdots + T(w_{n-1})
$$
where $w_i \in W_i$.
Any help is highly appreciated.
Thanks!
 A: A suggestion without a complete proof
You're doing great so far. But you're right that maybe you've got the wrong 2D subspaces. If you look at a vector
$$
v = c_1 b_1 + \ldots + c_n b_n
$$
and $c_n \ne 0$ and not all of $c_1 ... c_{n-1}$ are zero, then
you might want to consider the subspace spanned by...
$$
p = (c_1 b_1 + \ldots c_{n-1}b_{n-1})
$$
and
$$
q = c_n b_n
$$
Linearity of $T$ on that subspace lets you inductively work on simplifying $T(p)$, and maybe this'll get you somewhere.
A: You are working too hard.  Suppose $T : V \to V$ is a function, and it is linear on each subspace of dimension $2$.  Then, by restriction, we know $T$ is also linear on each subspace of dimension less than $2$,
Part 1: Let $t$ be a scalar and $v$ a vector.  Then $T(tv) = tT(v)$ holds since $T$ in linear on the subspace spanned by $v$, which has dimension at most $1$.
Part 2: Let $u,v$ be vectors.  Then $T(u+v) = T(u)+T(v)$ holds since $T$ is linear on the subspace spanned by $\{u,v\}$, which has dimension at most $2$.
Perhaps (depending on your definition of vector space)  we also need a
Part 0:  $T(0)=0$ since $T$ is linear on the subspace $\{0\}$, whan has dimension $0$.
A: You assume finite dimension, which is not needed. In fact, it is much easier to not even work with a base.
You want to show that for any $v,w\in V$, $\alpha,\beta\in \Bbb K$, we have $$T(\alpha v+\beta w)=\alpha T(v)+\beta T(w).$$
It is enough to observe that $v,w$ are in a two-dimensional subspace of $V$ - namely the space spanned by $v$ and $w$ (which may even be just $1$- or $0$-dimensional, but that does not hurt)
