# Diagonalize operator A by matrix S

In summary, diagonalizing an operator A by matrix S means finding a new basis for the vector space in which A acts, such that A can be represented as a diagonal matrix. This process is useful because it simplifies calculations and reveals important properties of the operator. To diagonalize A, we need to find its eigenvalues and eigenvectors, which form the columns and diagonal elements of matrix S, respectively. However, this is only possible if A is a square matrix with distinct eigenvalues. Diagonalization is not unique, as there can be multiple matrices and orders of eigenvalues that can be used to diagonalize A.
Suppose, i want to diagonalize operator A by matrix S (A'= S^\\dagger A S). Do i need to form S from "NORMALIZED" eigenvectors? I checked and found that even S formed from not-normalized eigenvectors works.

If S isn't normalized, you will get a diagonal matrix. However, the diagonal elements of the matrix will not be the eigenvalues of A.

## What does it mean to diagonalize an operator A by matrix S?

Diagonalizing an operator A by matrix S means finding a new basis for the vector space in which A acts, such that A can be represented as a diagonal matrix with respect to this new basis.

## Why is diagonalizing an operator A by matrix S useful?

Diagonalization simplifies the calculations involving the operator A, making it easier to analyze and manipulate. It also reveals important properties of the operator, such as its eigenvalues and eigenvectors.

## How do you diagonalize an operator A by matrix S?

To diagonalize an operator A by matrix S, we need to find the eigenvalues and corresponding eigenvectors of A. The eigenvectors then form the columns of the matrix S, and the eigenvalues form the diagonal elements of the diagonal matrix. The resulting diagonal matrix is the diagonalized form of A.

## Under what conditions can an operator A be diagonalized by matrix S?

An operator A can be diagonalized by matrix S if it has a full set of linearly independent eigenvectors. This is only possible if A is a square matrix with distinct eigenvalues.

## Is diagonalization unique?

No, diagonalization is not unique. There can be multiple matrices S that can be used to diagonalize an operator A. Furthermore, the diagonal matrix obtained may also not be unique, as the order of the eigenvalues can be changed without affecting the diagonalization.

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