As Professor Crane mentioned in class, just because a function looks like a line, it does not necessarily mean that it's defined by a linear transformation. Recall that, roughly speaking, for $u, v \in \mathbb{R}^{n}, c_{1}, c_{2} \in \mathbb{R}$, a function $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ is defined as a linear transformation if $f(c_{1}u + c_{2}v) = c_{1}f(u) + c_{2}f(v) \forall u, v \in \mathbb{R}^{n}, c_{1}, c_{2} \in \mathbb{R}.$

Consider, however, that a simple line in $\mathbb{R}^{2}$ can be defined as $f(x) = mx + b$. Thus, for $x, w \in \mathbb{R}$, we see that $f(x + w) = m(x + w) + b$, while $f(x) + f(w) = mx + b + mw + b = m(x + w) + 2b \neq f(x + w)$ and the line is not defined by a linear transformation.

As Professor Crane mentioned in class, just because a function looks like a line, it does not necessarily mean that it's defined by a linear transformation. Recall that, roughly speaking, for $u, v \in \mathbb{R}^{n}, c_{1}, c_{2} \in \mathbb{R}$, a function $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ is defined as a linear transformation if $f(c_{1}u + c_{2}v) = c_{1}f(u) + c_{2}f(v) \forall u, v \in \mathbb{R}^{n}, c_{1}, c_{2} \in \mathbb{R}.$

Consider, however, that a simple line in $\mathbb{R}^{2}$ can be defined as $f(x) = mx + b$. Thus, for $x, w \in \mathbb{R}$, we see that $f(x + w) = m(x + w) + b$, while $f(x) + f(w) = mx + b + mw + b = m(x + w) + 2b \neq f(x + w)$ and the line is not defined by a linear transformation.