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

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.

Use the $m$-dimensional space $V_m$ for this change of basis. With the columns of $P$ recording the new basis in the old one, bilinearity givesThus the matrices of the bilinear form satisfyThis is matrix congruence, rather than the similarity rule for an endomorphism: both arguments of the bilinear form change basis.

The restriction of a positive definite form to each $V_m$ is positive definite. By the spectral theorem for real symmetric matrices, choose an orthogonal matrix $P$ such that $P^T{A}_{m}P=\mathrm{diag}(d_1,\dots ,d_m)$.

Each $d_i=\phi (f_i,f_i)>0$, since a basis vector is nonzero. Taking determinants in part (i) givesbecause an orthogonal matrix has determinant $\pm 1$. ConsequentlyThese determinants are the leading principal minors in Sylvester's criterion.

Work inside $V_{m+1}$, and take the orthogonal complement for a bilinear form of $U=V_m$ relative to this ambient space. If $z\in U\cap U^{\perp }$, then $\phi (z,z)=0$, so positive definiteness on $U$ forces $z=0$.

Part (a)(iv) therefore gives $V_{m+1}=U\oplus U^{\perp }$, with $\dim U^{\perp }=1$. Choose $w\ne 0$ spanning that complement and use the basis $e_1,\dots ,e_m,w$. Symmetry and orthogonality make its Gram matrixBy part (b)(i), $\det B=(\det P{)}^2\det A_{m+1}>0$. Part (ii) applied to the already positive definite restriction gives $\det A_m>0$. ThereforeFor a nonzero vector $u+tw$ with $u\in U$,This proves positive definiteness on $V_{m+1}$.

Finally $\det A_1>0$ says $\phi (e_1,e_1)>0$, so the restriction to $V_1$ is positive definite. Induction using all successive positive leading principal minors reaches $V_n=V$. Combined with part (ii), this proves Sylvester's criterion in both directions.