Theorem (Spectral). Let A be an n×n real symmetric matrix, then A is orthogonally diagonalizable.In other words, there is an orthonormal basis {v1,…,vn} such that each of vi's is an eigenvector. Througout the proof the following will be assumed:
Theorem (Lagrange's Multiplier). Let n≥m, suppose that:A detailed proof of this multiplier method has been recorded in my lengthened version of Math3033 tutorial note 3.
Then there are λ1,λ2,…,λm∈R such that ∇f(x0)=λ1∇g1(x0)+λ2∇g2(x0)+⋯+λm∇gm(x0).
- G=(g1,…,gm):Rn→Rm is C1 on Rn and G′(x0) has full rank.
- f is a real-valued function defined near x0 and is differentiable at x0.
- f has a local extreme at x0 under the constraint G(x0)=0.
Proof of Spectral Theorem. The maximum of ψ(x)=xTAx w.r.t. the constraint xTx=1 can be attained since Sn−1 is compact. Now it is easy to show that ψ′(x)=2xTAand∇(xTx)(x)=2xT, therefore the existence of the constrained extreme implies that there is λ∈R and a vector v∈Rn such that
- vTv=1.
- ∇ψ(v)=λ∇(xTx−1)(v)⟺2vTA=2λvT⟺Av=λv.
Therefore we have obtained our first eigenvalue and eigenvector.
Now in view of the statement of Spectral Theorem let's suppose that there are already λ1,…,λk∈R and v1,…,vk orthonormal such that Avi=λivi for each i=1,…,k. We try to find μ∈R and u∈Sn−1 such that Au=μu and vi⋅u=0 for all 1≤i≤k. After that our proof is completed by induction.
Following the same idea in the first paragraph of the proof we know that the constrained extreme problem with the following data can be attained: xTAx=max,xTx=1,vTix=0,∀i=1,…,k. Therefore there are μ1,…,μk,μ∈R and a u∈Rn satisfying the constraints such that ∇(xTAx)(u)=μ∇(xTx−1)(u)+k∑i=1μi∇(vTix)(u), on simplification we get 2uTA=2μuT+k∑i=1μivTi. From our hypothesis in the last paragraph, by applying vi's on both sides above we have for i=1,2,…,k, 0=0+μi‖vi‖2, therefore μ1=μ2=⋯=μk=0, and we have 2uTA=2μuT⟺Au=μu. Recall that u satisfies the constraints vi⋅u=0 for i=1,…,k and u⋅u=1.◼
No comments:
Post a Comment