The problem
Given the image space $$f(V)=\text{span}\{(1,0,2),(2,-1,0) \} $$
state and argue a vector in $\mathbb{R}^3$ that does not belong to $f(V)$
The instructors gave the following solution to the problem.
"For example, the vector $u=\begin{bmatrix}1 \\0 \\0 \end{bmatrix}$ does not belong to the image space, because you cannot arrive at $u$ through any linear combination of the spanning vectors, $v_1=\begin{bmatrix}1 \\0 \\2 \end{bmatrix}$ and $v_2 = \begin{bmatrix}2 \\-1 \\0 \end{bmatrix}$."
My question
Is another valid answer to this question that the zero vector, $v=\begin{bmatrix}0 \\0\\0 \end{bmatrix}$ doesn't belong to the image space $f(V)$ either, because it belongs to the kernel $\text{ker}(f)$?
I just think this solution is too easy and "straight forward".
Best Answer
Nope: clearly, $\begin{bmatrix}0\\0\\0\end{bmatrix} = 0v_1 + 0v_2 \in f(V)$.