Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 1 / 9C (Linear Algebra)

Let $\dim V=n$ and $\dim W=k$. Extend a basis $w_1,\dots ,w_k$ of the vector subspace $W$ to $w_1,\dots ,w_n$ of $V$, and let $f_1,\dots ,f_n$ 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 $f_{k+1},\dots ,f_n$, and

To apply this to a matrix, regard its rows as elements of $({\mathbb{R}}^n{)}^{\ast }$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ matrix to a block identity $\mathrm{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification ${\mathbb{R}}^n\cong (({\mathbb{R}}^n{)}^{\ast }{)}^{\ast }$ 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 givesA basis of this kernel consists of exactly $n-r$ linearly independent solutions, and every solution is their linear combination.