 Solution

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.