Let $X$ be a real vector space. Suppose that $T:X\to X$ is affine, that is, for $x_1,..,x_n\in X$ and $\alpha_1,...,\alpha_n\ge0,$ we have that $\sum_{i=1}^n\alpha_i=1$ is satisfied.
Show that the operator, $S:X\to X$ where $S(x)=T(x)-T(0)$ is linear.
I think this should be pretty easy, and I'm sure that I'm missing something minute. Here's what I have so far.
Consider,
$$S(\alpha_1x_1+\alpha_2x_2)=T(\alpha_1x_1+\alpha_2x_s)-T(0)$$
$$=\alpha_1Tx_1+\alpha_2Tx_2-T(0)$$
$$=(\alpha_1Tx_1-T(0))+\alpha_2Tx_2$$
$$=\alpha_1Sx_1+\alpha_2Sx_2+T(0)$$
To move from the first to the second line we have used that $T$ is affine, and from the third to the fourth we have introduced a $T(0)$ to balance the reintroduction of the $S$ formulation.
The problem is that this factor of $T(0)$ brought in to balance the reformulation breaks the linearity that we are trying to show. What am I missing here?
$$\alpha_1Tx_1 +\alpha_2Tx_2 - T(0) = \alpha_1(Tx_1 - T(0)) + \alpha_2(Tx_2 - T(0))$$ since $\sum\limits_{i=1}^2\alpha_i = 1$
but
$$\alpha_1(Tx_1 - T(0)) + \alpha_2(Tx_2 - T(0)) = \alpha_1Sx_1 + \alpha_2Sx_2$$