Show that a row vector (not 0) is not in the nullspace of A $A \in \Bbb R^{m\times n}$
How can I show that a row vector, that is not null, is not in the null space?
Is it possible to show that the vector is not orthogonal?
 A: Building upon what was mentioned in the comments, there are two straightforward ways to verify that $\mathbf {Ax} \neq \mathbf 0$ whenever $x$ is a non-zero row of $\mathbf A$. The first deals with the fundamental theorem of linear algebra, and the second with how matrix multiplication works, as eluded to in the comments.


*

*One part of the Fundamental Theorem of Linear Algebra states that for any matrix $\mathbf A$ in $\mathbb{R}^{m \times n}$, the nullspace is the orthogonal compliment of the row space (in $\mathbb{R}^n$), meaning that a vector is in the row space of $\mathbf {A}$ if and only if it is not in the nullspace. In symbols,
\begin{equation*}N(\mathbf{A}) = C(\mathbf{A}^\mathrm T)^\perp \\
    \mathbf{x} \in N(\mathbf{A}) \iff \mathbf{x} \notin C(\mathbf{A}^\mathrm T)
\end{equation*}
From this result, we can conclude with confidence that if some vector $\mathbf x$ is a (non-zero) row of $\mathbf A$, the product $\mathbf{Ax}$ cannot be $\mathbf {0}$, since that would imply that $\mathbf x$ is also in the nullspace of $\mathbf A$.

*One way to represent multiplication of two matrices $\mathbf A$ and $\mathbf B$ is by using the dot products of their rows and columns. The entry in row $i$ and column $j$ of $\mathbf{AB}$ is simply $(\mathrm{row}\: i \:\mathrm{of}\: \mathbf{A}) \cdot (\mathrm{column} \: j \:\mathrm{of} \: \mathbf{B})$:
$$
    \begin{bmatrix}
        \text{---} & \mathbf{a_1} & \text{---} \\
        & \vdots & \\
        \text{---} & \mathbf{a_m} & \text{---}
    \end{bmatrix} 
    \begin{bmatrix}
        \vert & & \vert \\
        \mathbf{b_1} & \cdots & \mathbf{b_p} \\
        \vert & & \vert
    \end{bmatrix}
    = \begin{bmatrix}
        \mathbf{a_1 \cdot b_1} & \cdots & \mathbf{a_1 \cdot b_p} \\
        \vdots & & \vdots \\
        \mathbf{a_m \cdot b_1} & \cdots & \mathbf{a_m \cdot b_p}
    \end{bmatrix}
$$
Now, let us fix $\mathbf r_p$ to be the vector in the $p$th row of $\mathbf A \in \mathbb{R}^{m \times n}$. Then the product $\mathbf{Ar}_p$ looks like
\begin{equation}
\mathbf{Ar}_p =
    \begin{bmatrix}
        \text{---} & \mathbf{r_1} & \text{---} \\
        & \vdots & \\ \text{---}& \mathbf{r}_p &\text{---} \\ & \vdots & \\
        \text{---} & \mathbf{r}_m & \text{---}
    \end{bmatrix} \mathbf{r}_p =
    \begin{bmatrix}
        \mathbf{r_1} \cdot \mathbf{r_p} \\
        \vdots \\
        \mathbf{r_p} \cdot \mathbf{r_p} \\
        \vdots \\ 
        \mathbf{r_m} \cdot \mathbf{r_p} \\
\end{bmatrix}
\end{equation}
From this, it is fairly easy to see that the $p$th entry of $\mathbf{Ar}_p$ cannot be 0 (given that $\mathbf{r}_p \neq \mathbf{0}$), and so the product $\mathbf{Ar}_p$ cannot be $\mathbf{0}$.
