Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Given a normal complex matrix $A$ and knowing that $A = U D U^\dagger$, how can we rewrite the spectral decomposition such that $$U D U^\dagger = \sum_i \lambda_i u_i u_i^\dagger$$ where $\lambda_i$ are the eigenvalues and $D$ is the matrix of eigenvalues?
Rewriting it likely is trivial, however I have not found a nice way to show this myself or online.
Let $x \in \mathbb{C}^n$. I am assuming that $u_1, u_2, \dots, u_n$ are orthonormal. This means $$x = (x, u_1)u_1 + \dots + (x, u_n)u_n.$$ Thus $$Ax = \lambda_1(x, u_1)u_1 + \dots + \lambda_n(x, u_n)u_n.$$ We can regard a vector $v \in \mathbb{C}^n$ as a linear map from $\mathbb{C}$ to $\mathbb{C}^n$ in the natural way: If $c \in \mathbb{C}$, then set $vc = cv$. Therefore there is an adjoint $v^* : \mathbb{C}^n \to \mathbb{C}$ satisfying $(x, v) = (v^*x, 1) = v^*x$. Thus $$Ax = \lambda_1(u_1^*x)u_1 + \dots + \lambda_n(u_n^*x)u_n = \lambda_1u_1u_1^*x + \dots + \lambda_nu_nu_n^*x.$$ Therefore $$A = \lambda_1u_1u_1^* + \dots + \lambda_nu_nu_n^*.$$
