Existence of operator $B^2=A$

Question

Let $A$ be hermitian operator on finite-dimensional inner-space $V$ whose eigenvalues are positive real numbers. Show that there exists hermitian operator $B$ such that $B^2=A$. Is $B$ unique?

Answer 1

Since $A$ is hermitian it can be diagonalized: $A=PDP^{-1}$ where $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and $\lambda_i$ are the eigenvalues. By assumption, $\lambda_i>0$ so we can take their square root. Define $$\sqrt{D}=\operatorname{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})$$ and let $$B=P\sqrt{D}P^{-1}.$$ This satisfies $B^2=A$ since $$B^2=(P\sqrt{D}P^{-1})(P\sqrt{D}P^{-1})=P\sqrt{D}\sqrt{D}P^{-1}=PDP^{-1}=A.$$ It is not unique since, for example $-B$ has the same property.