Diagonal Matrix & Perturbation Theory in Quantum Mechanics

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What does it mean for a matrix to be diagonal, especially in Quantum Mechanics, where we get to Perturbation theory (Degeneracy).
I don't get it. Please if you can explain in 'simple' language.
 
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For finite dimensional vector spaces, a "diagonal matrix" is something like
[tex]\begin{bmatrix}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{bmatrix}[/tex]
having non-zero entries only on the main diagona. But I suspect you already knew that!

More generally, a square matrix represents a linear operator on some vector space. If that that vector space has finite dimension, n, then we can represent the operator as a n by n matrix. If the vector space is infinite dimensional is, as is typically the case in Quantum theory, we can't really write it as a "matrix" but the same ideas work.

The matrix is "diagonal" in a particular basis, [itex]\{v_1, v_2, ..., v_n\}[/itex] then the basis vectors are eigenvectors: [itex]Av_i= a_iv_i[/itex] where [itex]a_i[/itex] is the number, on the diagonal, at the ith row and column. To say that an operator in Quantum Mechanics is "diagonal" also means that the basis vectors are all eigenvectors (eigenstates). Physically, "eigenstates" are those that give specific values to whatever the property is associated to the state while vectors that are not eigenstates can be written as linear combinations of the eigenstates and then give "mixtures" of those values.
 
Your question is pretty vague, so it's going to be hard to offer any concrete help. Here is what a diagonal matrix is in general: http://en.wikipedia.org/wiki/Diagonal_matrix

In QM, we say an operator ##\hat{A}## is diagonal in some basis of states ##| \psi_i\rangle## (where i is some index labeling the states) if ##\langle \psi_i | \hat{A} | \psi_j \rangle## is only nonzero when ##i = j##.
 
Thank you, but I suppose it was really vague. I am having a hard time understanding Perturbation theory in its degenerate case.
Anything on that matter could help especially the usage of Matrix in that section..