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A function $T:V \rightarrow W$ is additive if $T(x+y) = T(x) + T(y)$ for every $x, y \in V$. If $V$ and $W$ are vector spaces over the field of $\mathbb{Q}$, then show that every additive function is linear.

My attempt: To be explicit, denote $+_{V}: V \times V \rightarrow V$ and $\cdot_{V}: \mathbb{Q} \times V \rightarrow V$ as the vector addition and scalar multiplication in the vector space $V$ over $\mathbb{Q}$ and similarly, denote $+_{W}: W \times W \rightarrow W$ and $\cdot_{W}: \mathbb{Q} \times W \rightarrow W$ as the vector addition and scalar multiplication in the vector space $W$ over $\mathbb{Q}$. Let $T: V \rightarrow W$ be any additive function. To show that $T$ is linear, I need to show that $T(\alpha_1 \cdot_V x_1 +_{V} \alpha_2 \cdot_V x_2) = \alpha_1 \cdot_W T(x_1) +_W \alpha_2 \cdot_W T(x_2)$ for all $x_1, x_2$ in $V$ and $\alpha_1, \alpha_2$ in $\mathbb{Q}$.

Now since $T$ is additive, we have $T(\alpha_1 \cdot_V x_1 +_{V} \alpha_2 \cdot_V x_2) = T(\alpha_1 \cdot_V x_1 ) +_{W} T(\alpha_2 \cdot_V x_2)$. Now I am stuck. Any help would be appreciated.

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From additivity: $T(0) + T(0) = T(0+0) = T(0)$, so $T(0) = 0$.

For any positive integer $n$:

  • $T( n \cdot v ) = T( \underbrace{v+\ldots+v}_n ) = \underbrace{T(v)+\ldots+T(v)}_n = n \cdot T(v)$

  • $T( -n \cdot v ) + T( n \cdot v ) = T( -nv + nv ) = T(0) = 0$, so $T( -n \cdot v ) = -T(n \cdot v) = -n \cdot T(v)$.

Thus $T(k \cdot v) = k \cdot T(v)$ for every integer $k$.

  • By the above $T(v) = T\left( n \cdot \frac{1}{n} v \right) = n \cdot T \left( \frac{1}{n} v \right)$, therefore $T\left( \frac{1}{n} v \right) = \frac{1}{n} T(v)$

  • Finally $T \left( \frac{m}{n} v \right) = T \left( m \cdot \frac{1}{n} v \right) = m \cdot T \left( \frac{1}{n} v \right) = \frac{m}{n} \cdot T(v)$,

    i.e. $T(\alpha v) = \alpha T(v)$ for each $\alpha \in \mathbb{Q}$.

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  • $\begingroup$ Thanks, this makes sense, but what about if $n$ is negative? For example, say we have $T(-3 \cdot v)$, then accordingly, we would have $T(-3 \cdot v) = T(-v + -v + -v) = 3T(-v)$, where as we should want to have $-3T(v)$? $\endgroup$ – elbarto Jul 31 '17 at 11:17
  • $\begingroup$ @elbarto Hey, you're right. I shall adjust my answer accordingly. $\endgroup$ – Adayah Jul 31 '17 at 11:32

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