Show $K\cdot \sqrt{I_q-K^*K}=\sqrt{I_p-KK^*}\cdot K$

Question

Let $K\in\mathbb{C}^{p\times q}$ be a contractive matrix. I want to prove that $K\cdot \sqrt{I_q-K^*K}=\sqrt{I_p-KK^*}\cdot K$. I have already shown that since $K$ is contractive, $\sqrt{I_q-K^*K}$ is hermitian but I am not sure why that is helpful.

Answer 1

1. Contractivity of $K$ guarantees that $I_q - K^\ast K$ and $I_p - KK^\ast$ are positive semidefinite, so that $\sqrt{I_q - K^\ast K}$ and $\sqrt{I_p - KK^\ast}$ are well-defined.
2. Pick a singular value decomposition $K = U\Sigma V^\ast$ for $K$; let $\sigma_1,\dotsc,\sigma_r$ be the non-zero singular values of $K$ (with each $\sigma_i < 1$ by contractivity of $K$), and let $v_1,\dotsc,v_q$ be the columns of $V$, so that $K v_i = \sigma_i U v_i$ for $1 \leq i \leq r$ and $K v_i = 0$ for $r+1 \leq i \leq q$. Since $\sqrt{I_q - K^\ast K} = V\sqrt{I_q - \Sigma^\ast \Sigma}V^\ast$ and $\sqrt{I_p - KK^\ast} = U\sqrt{I_p-\Sigma\Sigma^\ast}U^\ast$, you can now easily check that $K\sqrt{I_q - K^\ast K}v_i = \sqrt{I_p - KK^\ast}Kv_i$ for $1 \leq i \leq q$, which does the job.