1212: Linear Algebra II

1212: Linear Algebra II

Dr. Vladimir Dotsenko (Vlad) Lecture 12

Today we shall start proving the theorems on quadratic forms that you used in the tutorial class last week.

Theorem 1. Let qbe a quadratic form on a vector spaceV. There exists a basis f1, . . . , fn ofV for which the quadratic formqbecomes a signed sum of squares:

q(x1f1+· · ·+xnfn) = Xn

i=1

εix²_i,

where all numbersε_i are either1 or−1or 0.

Proof. We shall prove the appropriate statement for symmetric bilinear forms; it is a bit more transparent that way. Suppose that bis a symmetric bilinear form. If q(v) = b(v, v) =0 for allv, then we have the representation with all εi = 0. Suppose that there exists a vector v with q(v) 6= 0. Consider some basis e1, e2, . . . , en=v ofV. We claim that we can change it into a basisf1, f2, . . . , fn with b(fn, fn) =±1, and b(fi, fn) =0 for alli=1, . . . , n−1. Indeed, we can replaceen byfn = √ ¹

|q(e_n)|en, and then replaceeiby f_i = e_i− _b(f^b(ei,fn)

n,f_n)f_n. Then we may consider the linear span of f₁, . . . , f_n−1, and proceed by induction on dimension.

Theorem 2 (Law of inertia). In the previous theorem, the triple (n+, n−, n0), where n± is the number of εi equal to ±1, andn0is the number of εi equal to 0, does not depend on the choice of the basisf1, . . . , fn. This triple is often referred to as the signatureof the quadratic formq.

Proof. Suppose that we have a basis

e₁, . . . , e_n+, f₁, . . . , f_n−, g₁, . . . , g_n0

which produces a system of coordinates whereqbecomes a signed sum of squares withn₊ofε_iare equal to1, n₋ofε_iare equal to−1, andn₀ofε_iare equal to0. Let us note that for the corresponding bilinear formb, the subspace spanned byg1, . . . , gn₀ is its kernel, that is the space of all vectorsufor whichb(u, v) =0for allv∈V. Indeed, those conditions readb(u, ei) =b(u, fj) =b(u, gk) =0 for alli, j, k, which by inspection shows thatuis a linear combination ofgk. Suppose now that there are two different bases

e1, . . . , en₊, f1, . . . , fn₋, g1, . . . , gn₀

and

e₁⁰, . . . , en₊⁰ , f₁⁰, . . . , fn₋⁰ , g₁⁰, . . . , g_n⁰₀

where q is a signed sum of squares, andn+ 6= n₊⁰ , so without loss of generalityn+ > n₊⁰. Note that this implies thatn−< n₋⁰. Consider the vectors

e₁, . . . , e_n+, f₁⁰, . . . , f_n⁰0

−, g₁, . . . , g_n0.

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The total number of those vectors exceeds the dimension ofV, so they must be linearly dependent, that is a₁e₁+· · ·+a_n+e_n++b₁f₁⁰ +· · ·+b_n⁰

−f_n⁰0

−+c₁g₁+· · ·+c_n0g_n0=0 for some scalarsa_i, b_j, c_k. Let us rewrite it as

a₁e₁+· · ·+a_n+e_n++c₁g₁+· · ·+c_n0g_n0 = −(b₁f₁⁰ +· · ·+b_n⁰

−f_n⁰0

−),

and denote the vector to which both the left hand side and the right hand side are equal to byv. Then a²₁+· · ·+a²_n+ =q(v) = −b²₁−· · ·−b²_n−,

which implies

a1=· · ·=an₊ =b1=· · ·=bn₋=0, and substituting it into

a1e1+· · ·+an+en++b1f₁⁰ +· · ·+bn₋⁰f_n⁰0

−+c1g1+· · ·+cn₀gn₀=0 we get

c₁g₁+· · ·+c_n0g_n0=0,

implying of course c₁ = · · · = c_n0 = 0, which altogether shows that these vectors cannot be linearly dependent, a contradiction.

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