Is the map sending everything to 0 conventionally considered a complex character?

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Fix a finite group $G$. Would the function sending each $g \in G$ to $0$ usually be considered a character over $\mathbb{C}$? (character in the sense of representation theory - I'm not sure whether there is any ambiguity but I feel like I've encountered things called characters that don't necessarily have the same structure)

My definition for representation is a group homomorphism $$ \rho : G \to \textrm{GL} (V) $$ where $V$ is a finite $\mathbb{C}$-dimensional vector space - which would allow $V = \{0\}$.

Then my definition of character is $\chi = \textrm{tr} \circ \rho: G \to \mathbb{C}$, which applied to $\{0\}$ yields $\chi(g) = 0$ for all $g \in G$.

(I'll write $\iota$ for the character sending $G$ to 1 - I'm not sure how to do a blackboard bold 1, if you do let me know!) But doing an exam past paper question, part c) effectively asked to show $\langle \iota,\chi\rangle$ is positive if and only if $\chi - 1$ was a character of $G$. It finishes asking to show that for any character $\chi$, then $-\chi$ is never a character of $G$.

But since $\langle \iota,\iota \rangle$ = 1, then according to the question $\iota - 1 = 0$ is a character of $G$, but the last part contradicts this (since $-0 = 0$).

This got me thinking where there's a missing assumption: are representations necessarily to positive dimensional vector spaces (and so part c has a missing assumption for $\chi-1$) or is the "zero character" usually considered alright (so part d would need to specify $\chi \not = 0$)?