Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 3 / a

The rank-nullity theorem states that for a linear map $A:X\to Y$ with finite-dimensional domain,Equivalently, nullity plus rank equals the dimension of the domain. One proof extends a basis of $\ker A$ to a basis of $X$; the images of the added vectors form a basis of $\mathrm{im}A$.

The annihilator of a vector subspace $W\le V^{\ast }$ isIf $\dim W=k$, extend a basis of $W$ to one of $V^{\ast }$. In the resulting dual basis of $V^{\ast \ast }$, the last $n-k$ coordinate functionals span $W^{\circ }$. Hence $\boxed{\dim W+\dim W^{\circ }=n}$.

The bidual evaluation map $j:V\to V^{\ast \ast }$ is $j(x)(\varphi )=\varphi (x)$. It is a linear isomorphism in finite dimension. Its inverse image of the annihilator isThis identification uses evaluation only; no choice of inner product is involved.

Under the bidual evaluation map, a vector $y$ lies in $(\mathrm{im}\Phi {|}_U{)}^{\circ }$ exactly whenThat is precisely membership of the orthogonal complement for a bilinear form. ThereforeThe dimension formula in part (ii) givesso $\boxed{\dim U+\dim U^{\perp }\ge n}$. More precisely, the rank-nullity theorem givesThis is the orthogonal complement dimension for a degenerate bilinear form. The possible excess comes from vectors annihilated by the form against all of $V$.

The dimension formula for a sum of subspaces and the zero intersection implyBut $U+U^{\perp }$ is a vector subspace of the $n$-dimensional space $V$, so its dimension is at most $n$. Equality follows, and the sum is all of $V$. Since the intersection is zero, it is a direct sum:The argument remains valid for a degenerate ambient symmetric bilinear form; it uses the stated intersection hypothesis rather than assuming nondegeneracy everywhere.