Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 4 / 10B (Linear Algebra)

Apply the Gram-Schmidt process explicitly. Start with $f_1=e_1/\Vert e_1\Vert$, and for $i\ge 2$ setAssume inductively that the previous $f_j$ form an orthonormal basis for the span of $e_1,\dots ,e_{i-1}$. Inner products give $⟨v_i,f_j⟩=0$ for every $j<i$. Moreover $v_i\ne 0$: otherwise $e_i$ would lie in the span of its predecessors, contradicting that the $e_j$ form a basis. Thus $f_i$ is well defined and has unit norm. The definition expresses $f_i$ in the span of $e_1,\dots ,e_i$, and also expresses $e_i$ in the span of $f_1,\dots ,f_i$. The two spans therefore agree at each stage. Induction proves the required orthonormal, span-preserving basis.

The symmetric coefficient matrix of the quadratic form isThus $\boxed{q_a\text{ is nondegenerate if and only if }a\ne 1.}$ To obtain its signature, put $u=(y+z)/2$ and $v=(y-z)/2$. Completing the square yieldsThe transformation from $(x,y,z)$ to $(u+x,v,x)$ is invertible. The inertia of a bilinear form is consequentlyIf signature means the signed difference $n_{+}-n_{-}$, its value is $+1$ for $a>1$ and $-1$ for $a<1$. At $a=1$ the diagonal form has one zero direction, confirming degeneracy.