Show when $x \notin R(A)$, then $x \in N(A^T)$? Let $A \in \mathbb{R}^{m \times n}$. Can we say when $x \in \mathbb{R}^n$ is not in range of $A$, then it is in the null space of $A$?
My try:
We know that $R(A)$ is a subspace in $\mathbb{R}^{m}$ whose dimension is $r \leq m$ and its complement is $N(A^T)$, i.e., $R(A) \bigoplus N(A^T) = \mathbb{R}^{m}$. Therefore, when $x$ is not in $\mathbb{R}^{m}$, it would be in the other one.
Am I right? If so, can you please provide me with other proofs and explanations?

I think your question boils down to the following statement.
Let $V$ a vector space and $W_1$ and $W_2$ be two subspaces of $V$. Given:
\begin{equation}
V = W_1 \oplus W_2
\end{equation} For some $v \in V$, if $ v \notin W_1$ does that say anything about $v$ being in $W_2$. The answer is no.
Let us look at one consequence of the direct sum decomposition above. It tells us there exist bases of $W_1$ and $W_2$, let them be $\mathcal{B}_1$ and $\mathcal{B}_2$, such that:
\begin{equation}
\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2
\end{equation} where $\mathcal{B}$ is a basis for $V$.
Now if an element cannot be written as a linear combination of elements of $\mathcal{B}_1$, does that mean it can be written a linear combination of elements of $\mathcal{B}_2$?
For a simple counter example, take
\begin{equation}
\mathcal{B}_1 = \{b_1,b_2\} \text{ and } \mathcal{B}_2 = \{b_3\}
\end{equation}
Let $v=b_1+b_2+b_3$. As evident, $v \notin W_1$ and $v \notin W_2$.
