Is $f: \mathbb{R^m} \rightarrow \mathbb{R^n}$ defined by $f(\vec x) \equiv 0 $ considered a linear map? If the definition for linear map is one that satisfies additivity and homogeneity, then $f$ obviously is a linear map. But I'd to confirm $f$ is not specifically excluded in the usual definitions of linear mappings.
2 Answers
Yes, the zero function is indeed a linear map.
Any function that is additive and homogeneous is a linear map by definition, and in your case:
- $f(x+y) = 0 = 0+0 = f(x) + f(y)$
- $f(\alpha x) = 0 = \alpha \cdot 0 = \alpha f(x)$
So $f$ is both homogeneous and additive.
In fact, as one commenter pointed out, you can define addition and multiplication with constant to the set of all linear functions.
- $(f+g)(x) = f(x) + g(x)$
- $(\alpha\cdot f)(x) = \alpha \cdot (f(x))$
These two definitions are well defined and map linear functions to linear functions.
Using these two operations, the set of all linear functions from $\mathbb R^m$ to $\mathbb R^n$ is a linear space (which, by the way, is isomorphic to $\mathbb R^{m\cdot n}$) and is more commonly viewed as the space of all $n\times m$ matrices) and the zero function is the zero element.
Yes, $f:\mathbb{R}^m \to \mathbb{R}^n, x \mapsto 0$ is a linear map.
Linearity: $$ f(\lambda x + \mu y) = 0 = \lambda 0 + \mu 0 = \lambda f(x) + \mu f(y) $$ for arbitrary $\lambda, \mu \in \mathbb{R}$ and $x, y \in \mathbb{R}^m$.
The linear maps from $\mathbb{R}^m$ to $\mathbb{R}^n$ can be represented as $n \times m$ matrices, with $0$ having all entries $0$.
