= Solution
Let $\dim V=n$ and $\dim W=k$. Extend a <basis> $w_1,\ldots,w_k$ of the <vector subspace> $W$ to $w_1,\ldots,w_n$ of $V$, and let $f_1,\ldots,f_n$ be its <dual basis>. Every <linear functional> is uniquely $\sum_i c_i f_i$. It vanishes on $W$ exactly when $c_1=\cdots=c_k=0$. Thus the <annihilator of a vector subspace> has <basis> $f_{k+1},\ldots,f_n$, and
$$
\boxed{\dim\alpha(W)=n-k}.
$$
To apply this to a <matrix>, regard its rows as elements of $(\mathbb R^n)^*$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ <matrix> to a block identity $\operatorname{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification $\mathbb R^n\cong((\mathbb R^n)^*)^*$ sends $x$ to evaluation $f\mapsto f(x)$: it is injective because coordinate functionals separate points, and is bijective by equal dimensions. Consequently the solution <vector space> is the annihilator of $R$ under this identification. The dimension formula just proved gives
$$
\boxed{\dim\ker A=n-r}.
$$
A <basis> of this <kernel> consists of exactly $n-r$ <linearly independent> solutions, and every solution is their linear combination.
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