Analysis Theorems

Lemma 7.1. If \(f:\Sym ^2V\to F\) is a map of \(F\)-vector spaces which is nontrivial, and \(\ch (F)\neq 2\), there is an element so that \(f(v\otimes v)\neq 0\).

Proof. There is some \(v_1,v_2\) such that \(f(v_1\otimes v_2)\neq 0\). Now WLOG, \(v_1\neq v_2\) \(f(v_1\otimes v_1)=f(v_2\otimes v_2)=0\), so we have \(\sum _{1\leq i,j\leq 2}f(v_i\otimes v_j) = f((v_1+v_2)\otimes (v_1+v_2))\neq 0\) □

Proposition 7.2 (Decomposition Theorem). If \(f:\Sym ^2V\to F\) is a map of \(F\)-vector spaces which is nontrivial, for any element \(v \in V\) so that \(f(v\otimes v)\neq 0\), \(V \cong V'\oplus vF\), where \(V'\) consists of vectors \(f\)-orthogonal to \(v\).

Proof. \(\pi (x) = x-v\frac {f(v\otimes x)}{f(v\otimes v)}\) is a projection onto \(V'\). □

Corollary 7.3 (Graham-Schmidt Theorem). If \(f:\Sym ^2V\to F\) is a map of \(F\)-vector spaces, \(V\) is finite dimensional, and \(\ch (F)\neq 2\), then \(V\) has an \(f\)-orthogonal basis. Hence \(f\) is represented by a diagonal matrix.

Proof. If \(f\) is \(0\), any basis will do. If not, by induction, Lemma 7.1 and Proposition 7.2 we are done. □

Corollary 7.4 (Sylvester’s Law of Inertia). If \(f:\Sym ^2V\to \RR \) is a map of \(\RR \)-vector spaces, \(V\) is finite dimensional, then \(f\) is represented by a unique matrix of the form \(I_n\oplus -I_m\oplus 0_r\). (Congruent real symmetric matrices have the same rank and signature).

Proof. Once we have diagonalized a matrix representing \(f\) from Corollary 7.3, we can scale each diagonal by a square, giving \(1,-1,0\). This form is unique as \(V\) decomposes into \(V_+\),\(V_-\),\(V_0\) where \(V_+\) is the span of vectors \(v\) with \(f(v\otimes v)=1\) and similarly for the other two. By orthogonality, \(f\) is positive definite in \(V_+\) and negative definite in \(V_-\), so these subspaces, and hence the rank and signature, are invariant. □

Proposition 7.5 (Sylvester’s Criterion). A symmetric real matrix \(A\) is positive definite iff the principle minors are positive, and negative iff the odd principle minors are negative and the even ones positive.

Proof. If \(A\) is positive definite, it is of the form \(P^tP\) for invertible \(P\), so \(\det (A)=\det (P)^2>0\). Every principle submatrix is positive definite on the corresponding subspace so the principle minors are positive. Conversely if \(A\) has positive principle minors, we induct. Write \(A = BA'B^t\),

\[A = \begin {pmatrix} A_0&a\\ a^t&\alpha \end {pmatrix}, B = \begin {pmatrix} I_{n-1}&0\\ a^tA_0^{-1}&1 \end {pmatrix}, A' = \begin {pmatrix} A_0&0\\ 0&\alpha -a^tA_0a \end {pmatrix}\]

So \(\alpha -a^tA_0a\) is positive by the fact that the determinant is positive and \(A_0\) is positive definite by induction, so we are done. For the negative definite case note \(A\) is negative definite iff \(-A\) is positive definite. □

Corollary 7.6 (Hessian Test). A critical point \(c\) of a \(\cC ^2\) function \(f:\RR ^n\to \RR \) is a local maximum iff the Hessian matrix \((\partial _i\partial _jf(c))\) is negative definite, and a local minimum iff the Hessian matrix is positive definite.

Proof. If this matrix is positive definite, then by the Taylor expansion, \(f(x-c)-f(c)\) is closely approximated by positive definite quadratic function, so is larger than \(0\) in a deleted neighborhood of \(0\). The corresponding argument yields the other result. □