Is $\text{rk}L=\text{rk}L^*L  $ true for finite rank operators? Let $L$ be a  compact linear operator in an infinitedimensional space that has finite rank. Do the equations $$\text{rk}L=\text{rk}L^*L\ \text{and} \ \text{rk}L^*L=\text{rk}R,$$ where $R$ is the (unique) root of $L^*L$ have to hold ? If they do, why is this the case ?
The proof of this statement is easy in the finite dimensional case, but for finite rank operators I couldn't figure it out. 
 A: Let $H$ and $K$ be Hilbert spaces. For given $\xi\in K$ and $\eta\in H$ we denote by $\xi\bigcirc\eta$ the rank one operatror sending $\zeta\in H$ to $\langle\zeta,\eta\rangle\xi$. Since $L:H\to K$ is a finite rank operator then by collorary of Hilbert-Schmidt representation theorem we have
$$
L=\sum\limits_{k=1}^n s_k (e_k'\bigcirc e_k'')\tag{1}
$$
for some orthonormal systems $\{e_k':k=1,n\}\subset K$  and $\{e_k'':k=1,n\}\subset H$ and positive numbers $\{s_k:k=1,n\}$ that called singular values. One may check that
$$
L^*L=\sum\limits_{k=1}^n s^2_k(e_k''\bigcirc e_k'')\qquad
R=(L^*L)^{1/2}=\sum\limits_{k=1}^n s_k(e_k''\bigcirc e_k'')\tag{2}
$$
From $(1)$ and $(2)$ it follows that
$$
\mathrm{rk}(L)=\mathrm{dim}\;\mathrm{Im}(L)=\mathrm{dim}(\mathrm{span}\{e_k':k=1,n\})=n
$$
$$
\mathrm{rk}(L^*L)=\mathrm{dim}\;\mathrm{Im}(L^*L)=\mathrm{dim}(\mathrm{span}\{e_k'':k=1,n\})=n
$$
$$
\mathrm{rk}(R)=\mathrm{dim}\;\mathrm{Im}(R)=\mathrm{dim}(\mathrm{span}\{e_k'':k=1,n\})=n
$$
So
$$
\mathrm{rk}(L)=\mathrm{rk}(L^*L)=\mathrm{rk}(R)
$$
