corniness9a

2022-09-06

The vector ${\mathbf{u}}_{\lambda}$ is assumed to be normalized.

Then I want to understand How these two conditions are satisfied?

$\begin{array}{rl}& \sum _{k}{u}_{k\kappa}^{\ast}{u}_{k\lambda}={\delta}_{\kappa \lambda}\\ & \sum _{\kappa}{u}_{k\kappa}^{\ast}{u}_{l\kappa}={\delta}_{kl}\end{array}$

Then I want to understand How these two conditions are satisfied?

$\begin{array}{rl}& \sum _{k}{u}_{k\kappa}^{\ast}{u}_{k\lambda}={\delta}_{\kappa \lambda}\\ & \sum _{\kappa}{u}_{k\kappa}^{\ast}{u}_{l\kappa}={\delta}_{kl}\end{array}$

dheasca8d

Beginner2022-09-07Added 6 answers

M is positive definite, therefore it has orthogonal set of eigenvectors. Since we assume them to be normalized, the matrix $U=[{\mathbf{u}}_{1},{\mathbf{u}}_{2},...,{\mathbf{u}}_{n}]$ (we treat columns as vectors) is orthonormal. The first formula (U^*U = I) comes from the definition:

Vectors are normalized -> ones on diagonal

Orthogonal -> zeros off diagonal

This makes ${U}^{\ast}={U}^{-1}$, therefore $U{U}^{\ast}=I$. Take element-wise complex conjugate and you get the second formula.

Vectors are normalized -> ones on diagonal

Orthogonal -> zeros off diagonal

This makes ${U}^{\ast}={U}^{-1}$, therefore $U{U}^{\ast}=I$. Take element-wise complex conjugate and you get the second formula.

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