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amjad-sh

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I'm reading "principles of quantum mechanics" by R.Shankar.

I reached a theorem talking about Hermitian operators.

The theorem says: " To every Hermetian operator Ω,there exist( at least) a basis consisting of its orthonormal eigenvectors.Its diagonal in this eigenbasis and has its eigenvalues as its diagonal entries".

I will talk here about the first part of the proof:" Let us start with the characteristic equation.It must have at least one root,call it w1.Corresponding to w1 there must exist at least one nonzero eigenvector |w1>.Consider the subspace v1(of n-1 dimention) of all vectors orthogonal to |w1 >.Let us choose as our basis the vector |w1> normalized to unity and any n-1 orthonormal vectors in v1.In this basis Ω has the following matrix{picture1}

The column 1 is just the image of |w1> after Ω has acted on it.Given the first column , the first row follows from the hermiticity of Ω.

The characteristic equation now takes the form (w1-w)*(determinant of the boxed submatrix)=0.

{picture3}

Now also the polynomial P (of degree n-1) must also generate one root w2 ,and a normalized eigenvector |w2>

Define the subspace v2(of n-2 dimentions ) of vectors in v1 orthogonal to |w2> (and automatically to |w1>).and repeat the procedure as before.Finally the matrix Ω becomes, in the basis |w1>,|w2>,,,,,,|wn>):

{picture2}"

My Question:As we can choose any n-1 orthonormal vectors in v1 and |w1> as a basis what guarantee that we will get the same eigenvalues if we changed the chosen n-1 orthonormal vectors in v1?