Sum of rank-one matrices equals identity

Question

Let $v_1,\dots,v_n\in \mathbb{C}^n$ be vectors satisfying $$ v_1v_1^* + \dots + v_n v_n^* = I $$where $I$ is the identity matrix and $v^*$ denotes the conjugate transpose. These vectors are clearly linearly independent. Is it necessarily the case that the $v_i$'s are orthonormal?

Answer 1

Yes, it is.

Considered as linear transformations one has $\,vv^*=v\,\langle v\mid\cdot\,\rangle$, assuming the scalar product to be conjugate-linear in its first slot.

Choose an ONB $\{\beta_k\}\subset\mathbb C^n$ and let $\,T\,$ be the linear transformation from that ONB to the given basis $\{v_1,\dots,v_n\}$, so that $T\beta_k = v_k\,$ for $\,k=1,2,\dots,n$. Then $$I\:=\:v_1v_1^*+\ldots + v_n v_n^* \:=\:\sum^n_{k=1} T\beta_k\langle T\beta_k\mid\cdot\,\rangle \:=\:\sum^n_{k=1} T\beta_k\langle\,\beta_k\mid T^*\cdot\,\rangle \:=\: TT^*,$$ hence $\,T\,$ is unitary and $\,\{v_1,\dots,v_n\}\,$ is an ONB as well.