Eigenvalues and diagonalization of a matrix

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dyn
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Main Question or Discussion Point

When you diagonalize a matrix the diagonal elements are the eigenvalues but how do you know which order to put the eigenvalues in the diagonal elements as different orders give different matrices ?
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The same order as the matrix with your eigenvectors. If I recall correctly, so long as those two have their columns corresponding with each other, it's fine; you'll transform your matrix to a diagonal one, then later on you'll transform back. If it's self-consistent, the properties you're looking for should be conserved.
(But I'm not an expert on this, so hopefully there will be more input!)
 
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DrClaude
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ModestyKing needn't be so modest, he is correct :smile:
 
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HallsofIvy
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If an n by n matrix A has n independent eigenvectors, the there exist a matrix B such that D= B^{-1}AB is a diagonal matrix having the eigenvalues on the diagonal. B is the matrix having the corresponding eigenvectors as columns

What ModestyKing and DrClaude are saying is that the eigenvalues can be any order- as long as you have the eigenvectors, forming the columns of matrix B, in the same order.
 
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Fredrik
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An ordered basis for an n-dimensional vector space V is an n-tuple ##(e_1,\dots,e_n)## such that ##\{e_1,\dots,e_n\}## is a basis for V

The component n-tuple of a vector ##x## with respect to an ordered basis ##(e_1,\dots,e_n)## is the unique n-tuple of scalars ##(x_1,\dots,x_n)## such that ##x=\sum_{i=1}^n x_i e_i##.

The matrix of components of a linear operator ##A## with respect to an ordered basis ##(e_1,\dots,e_n)## is the n×n matrix [A] defined by ##[A]_{ij}=(Ae_j)_i##. (The right-hand side denotes the ##i##th component of ##Ae_j## with respect to ##(e_1,\dots,e_n)##). This matrix is diagonal if and only if the ##e_i## are eigenvectors of the linear operator ##A##.

Every matrix is the matrix of components of some linear operator, with respect to some ordered basis. To change the order of the non-zero numbers in a diagonal matrix, is to change the order of the vectors in the ordered basis. You end up with a representation of the same linear operator, with respect to a different ordered basis.
 
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