# 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!

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)

• Of course the assumption can be relaxed to $\dim V\ge2$. Aug 7 '20 at 21:46
• Ooh... I feel stupid now. Makes total sense... Aug 7 '20 at 21:46

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.

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 $$0$$.

• Yeah... I went to the wrong direction! Thank you GEdgar :) Aug 7 '20 at 21:47