If $f$ is an affine map and $f(0) = 0$, then $f$ is a linear map.

Question

If $f$ is affine and $f(0) = 0$, then $f$ is a linear map. Define $f(x) = T(x) - T(0)$, where $T$ is an affine map from $R^n$ to $R^k$.

I see that I would have to show that $f(ax) = af(x)$, which means $T(ax) - T(0) = a[T(x)-T(0)]$, but how would I show this equality?

I see that $f(x + y) = f(x) + f(y)$ iff $T(0) = 0$, but I don't see how I would show $f(ax) = af(x)$ since $aT(x) \neq T(ax).$

The definition of an affine map:

Let $\{p_0, .., p_n\} \subseteq R^n$ be affine independent and let $A$ denote the affine set is spans. An affine map $T: A \rightarrow R^k$ (for some $k \gt 1)$ is a function satisfying $T(\sum t_ip_i) = \sum t_i T(p_i)$ whenever $\sum t_i = 1$. The restriction to $[p_0, ..., p_n]$, which is the convex set spanned by $\{p_0, ..., p_n\}$, is also an affine map.

Answer 1

Your thoughts are conspicuously missing the key step of writing down what it means for $T$ to be an affine map.

Answer 2

let $(p_n)_n$ be the standard basis in $\mathbb{R^n}$, ie $\sum p_n = 1$, and $p_n$ spans $\mathbb{R^n}$. Thus $T$ must be in the form of $T(x) = kx + m.$ Can you finish from here?