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$\begin{pmatrix} R_{11} & \cdots & R_{1A} \\ \vdots & \ddots & \vdots \\ R_{S1} & \cdots & R_{SA} \\ 1 & \cdots & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_A \end{pmatrix} = \begin{pmatrix} R_{11} & \cdots & R_{1A} \\ \vdots & \ddots & \vdots \\ R_{S1} & \cdots & R_{SA} \\ 1 & \cdots & 1 \end{pmatrix} \begin{pmatrix} w_1 \\ \vdots \\ w_A \end{pmatrix} \tag1$

Let $\mathcal{R}$ be the $(S+1)\times A$ coefficient matrix in $(1)$. System $(1)$ has solutions $\mathbf{x\neq w}$ if and only if there is a nonzero vector in the nullspace of $\mathcal{R}$.

I can understand the "if" part here, but can't get why "only if " part is true.(That is, why "if system $(1)$ has solutions $\mathbf{x\neq w}$ then there is a nonzero vector in the nullspace of $\mathcal{R}$")

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If there is a solution $x\neq w$, then $R(x-w)=0$ and $x-w\neq 0$.

If there exists $v\neq 0$ with $Rv=0$, then $Rv=R0$ and $(v,0)$ is a solution.

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