Bilinear forms: linear in both dimension.
Symmetric form: \(\langle u,v \rangle = \langle v,u\rangle\); skew symmetric: \(\langle u,v \rangle = -\langle v,u\rangle\).
Hermitian form: linear in second dimension, conjugate linear in first dimension. Conjugate symmetric. (E.g. \(\langle u,v \rangle = u^* v\). ) (Hermitian matrix: \(P^*= P\). )
Orthogonal matrix in \(\R^n\): \(PP^t = I\). Unitary matrix in \(\C^n\): \(P^*P = I\).
Thm: Eigenvalues of Hermitian Mat \(\in \R\). Proof: \(Pv = \lambda v\), then \(v^*P = \bar \lambda v^*\). Compute \(v^* P v\) in two ways, know \(\lambda =\bar \lambda\).
Def of ortho: in herm/sym form: \(u\perp v\) iff \(\langle u,v \rangle = 0\).
\(W^\perp\): \(\perp\) to every \(w\in W\). Nullspace: \(V^\perp\)
(Note: When the form is \(x^* A y\), then \(V^\perp = \ker A\). )
Def: non degenerate on \(W\) iff \(W\cap W^\perp = \empty\).
Important Prop: non-deg on \(W\), then \(W\oplus W^\perp = V\). Proof: LHS is direct sum. To show dim: pick a basis \(w_1, \cdots, w_k\) for \(W\), then \(x\to (\langle x, w_1 \rangle, \cdots, \langle x, w_k \rangle)\) linear trans, consider dim ker and dim image: done.
Thm. \(V\) has orthogonal basis. Pf: Induction on \(\dim V\). If any \(v\) we have \(\langle v, v \rangle=0\): any basis is orthogonal. Ow: pick \(\langle v, v \rangle \ne 0\), then \(V = \langle v \rangle \oplus v^\perp\).
(Cor: Every symmetric form has an orthog basis with \(\langle v_i, v_i \rangle \in \{0, 1, -1\}\). (The numbers of each: Signature, determined by \(V\) and form(Sylvester's Law) ))
Matrix form: \(P^*AP\) with some \(1, -1, 0\) on diag. For positive definite: \(=I\).
(Note: Orthogonal projection: By computing the form. Given basis \((v_1, \cdots v_n)\), let \(V_i = span(v_1, \cdots, v_n)\), finding orthogonal basis for all \(V_i\): set \(t_2 = V_1^\perp \) component of \(v_2\) (\(v_2-v_1(v_2, v_1)/(v_1, v_1)\)) and so on.$)
For any linear trans \(T: V\to W\), existsm unique \(T^*\) such that \(\langle Tv, w \rangle =\langle v, T^*w \rangle\).
From now on, let \(T:V\to V\). Call \(T\) hermitian if \(T^*=T\), call \(T\) unitary if \(TT^*=I\) or \(\langle u,v\rangle =\langle Tu,Tv \rangle\). Call \(T\) normal if \(T^*T = TT^*\) or \(\langle Tu, Tv\rangle=\langle T^*u, T^*v \rangle\).
Two propositions and the spectral thm.
Prop1. For normal \(T\), If \(TW\subset W\), then \(T^*W^\perp \subset W^\perp\). (Proof: pick \(u \in W^\perp\), \(v\in W\), \(\langle T^*u, v \rangle=0\)).
Prop2. For normal \(T\), if \(Tv = \lambda v\), then \(T^* v = \bar \lambda v\). (Proof: When \(\lambda = 0\), since \((Tv, Tv) = (T^*v, T^*v)\), is true. Now by \(T-\lambda I\): Done. )
Spectral thm. \(V\) is a Hermitian space, $T:V\to V $ normal transformation. Then exists orthognomal basis of \(V\) consists of eigen vectors of \(V\). (Matrix version: Diagonalizable by unitary matrix. )
Proof: Pick \(Tw = \lambda w\), then \(T^*w =\bar \lambda w\), so \(Tw^\perp\subset w^\perp\), by induction ( \(T\) is also a normal form on \(w^\perp\)): done.
For real case: need eigenvalues \(\in \R\). (Or the entries might in \(\C\), Prop 2 won't hold since not definite. ) So hold for symmetric matrices (Can be diagonalized by orthogonal matrix. )