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# 急!線性代數問題 20點

u及e1,e2,en為向量

operator上面的"︿"也沒打

Ｌ+的"+"應該為上標

由於沒辦法用文字處理 有些符號都不是標準的數學表示法 請見諒

Let Ｌ: U → U be a linear operator acting on an n-dimensional

complex-valued inner product space U. It is known that Ｌ is a unitary

operator, i.e., Ｌ satisfies Ｌ* Ｌ＝ＬＬ* ＝Ｉ, where Ｉ is the identity

operator on U and Ｌ is the adjoint operator of Ｌ.

(a) Suppose ｕ is an eigenvector of Ｌ, i.e., Ｌ(ｕ)＝λｕ for

some number λ, and ｕ≠ 0. Let W be the orthogonal complement to ｕ.

Please show that W is an invariant subspace of Ｌ* , that is,

Ｌ (W) ⊂ W.

(b) Please use the fact of (a) (which says that W is an invariant

subspace of Ｌ) to prove that W is also an invariant subspace of Ｌ.

(Note: You can simply quote the result of (a) for this problem even if you

do not know how to prove (a).

(c) Please prove that we can find an orthonormal basis

{e1, e2, ..., en} of U such that Ｌ is diagonal, that is L(ej) ＝ λj*ej

for some number λj.

(d) Suppose ｕ is an eigenvector of Ｌ, please prove that ｕ is

necessarily an eigenvector of Ｌ+ , too

(e)Please prove that the absolute value of any eigenvalue of a

unitary operator is always unity.

sorry

Ｌ+=L*

我打錯了

### 1 Answer

- Scharze spaceLv 71 decade agoFavorite Answer
(a).For v€W, we will show that L*(v)€W

Note that 0=bar(λ)<v,u>

=<v,λu>

=<v,L(u)>

=<L*(v),u>

so L*(v)€W

(b)Similarly, given v€W,we will show that L(v)€W

=>0=bar(λ)<v,u>

=<v,λu>

=<v,L(u)>

=<v,L*(u)>

=<L(v),u>

=>L(v)€W

(c)Since L is self-adjoint, there is an orthonormal basic{e1,e2,....,en} such that L(ei)=λiei,i=1...n

(d)I don't know your mean for L^+

(e)Let λ be an eigenvalue of unitary operator L, there exists u≠0,such that L(u)=λu

=>|λ|^2<u,u>=<L(u),L(u)>=<u,L*L(u)>=<u,u>

=>(|λ|^2-1)<u,u>=0

=>|λ|^2=1 (∵<u,u>≠0)

=>|λ|=1

Q.E.D

2009-06-08 21:57:08 補充：

Claim: L is self-adjoint,all eigenvalne are real

proof of claim: Let λ be an eigenvalue of L, 存在x≠0 such that L(x)=λx

=>λ =<λx,x>= = = =

=bar(λ)

=>λ=bar(λ)(Since ≠0)

=>λ is real

Now let u be eigenvector of L,there exists a€C such that L(u)=au

=>(L-aI)(u)=0

2009-06-08 22:00:48 補充：

since L is normal, so is L-aI

=>0=<(L-aI)(u),(L-aI)(u)>

=

=

=<(L-aI)*(u),(L-aI)*(u)>

=>(L-aI)*(u)=L*(u)-bar(a)I(u)=0

=>L*(u)=bar(a)u

=>u is an eigenvector of L

2009-06-08 23:49:16 補充：

我真是受夠奇摩的爛系統

真的是受夠了

爛奇摩

被砍掉的字 自己想辦法吧