首页 > 其他分享> > Spectral thm review

Spectral thm review

2021-10-25 03:00:06 作者：互联网

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. )

标签：Spectral,langle,form,basis,review,thm,perp,rangle,lambda
来源： https://www.cnblogs.com/cauchysheep/p/15456702.html