Eigenvalues and diagonalization of a matrix

[dyn]

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 ?
Thanks

 

[ModestyKing]

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!)

 

[DrClaude]

ModestyKing needn't be so modest, he is correct

 

[HallsofIvy]

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.

 

[Fredrik]

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.