Imagine discrete orthonormal basis made of infinite set of kets $|\phi_1> , ..., |\phi_n>,...$ Completeness or closure of the basis is given by:

$\sum_{n=1}^{\infty} | \phi_n > < \phi_n | = \hat {I}$ (From Zetilli's Quantum mechanics book)

So what I did manage is that the eigenvalues of the operator $|\phi_m><\phi_m|$ are all zeros except one which is the square of norm of $|\phi_m>$. That means the matrix can be diagonalized but I don't really understand how you diagonalize matrices like this. Since the norm is 1, all entries should be zero except the $(n,n)$ one? Also doesn't diagonalizing a matrix change it? I'm a bit confused, please guide me.

Imagine discrete orthonormal basis made of infinite set of kets $|\phi_1\rangle , ..., |\phi_n\rangle,...$ Completeness or closure of the basis is given by:

$\sum_{n=1}^{\infty} | \phi_n \rangle \langle \phi_n | = \hat {I}$ (From Zetilli's Quantum mechanics book)

So what I did manage is that the eigenvalues of the operator $|\phi_m\rangle\langle\phi_m|$ are all zeros except one which is the square of norm of $|\phi_m\rangle$. That means the matrix can be diagonalized but I don't really understand how you diagonalize matrices like this. Since the norm is 1, all entries should be zero except the $(n,n)$ one? Also doesn't diagonalizing a matrix change it? I'm a bit confused, please guide me.