Proving decomposition $A=V^{\dagger}V$ for hermitian operators.

Question

How can I prove that for any hermitian operator $A$ I can decompose it in the form $A=V^{\dagger}V$?

Answer 1

You can use the fact that Hermitian operators have real eigenvalues and orthogonal eigenfunctions. Then, the Hermitian operator can be written as a function of its eigenvaules, $\lambda_j$ and projectors to the corresponding eigenspaces $P_j$, using the so-called spectral decomposition $$A=\sum_j \lambda_j P_{j}\;.$$ The operators you are interested in can be also written as a function of the same projectors as $$V=\sum_j \alpha_j P_{j}\;,$$ being $\alpha_j$ complex numbers, generally. Then $$V^\dagger V=\left(\sum_j\alpha_{j}P_j\right)^\dagger\left(\sum_k\alpha_{k}P_{k}\right)=\left(\sum_j\alpha_{j}^*P_j\right)\left(\sum_k\alpha_{k}P_{k}\right)$$ where $^*$ means complex conjugation and I have used the property of the projection operators $P_j^\dagger=P_j$. Then $$A=\sum_j\lambda_jP_j=\sum_j |\alpha_j|^2P_j$$ and we identify $\lambda_j=|\alpha_j|^2$. Thus, the decomposition is possible if the eigenvalues of the original operator are non-negative.

Answer 2

Since $A=A^{\dagger}$, using Spectral theorem, we can decompose it into $$ A=P^{\dagger}DP=P^{\dagger}\sqrt{D}DP $$ where $D$ is a diagonal matrix with real eigenvalues, which yields $$ A=P^{\dagger}(\sqrt{D})^{\dagger}DP=(\sqrt{D}P)^{\dagger}(\sqrt{D}P)\equiv V^{\dagger}V. $$