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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.

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    So you're assuming the space to be a Hilbert space ? This is not explicitely said in your question, and that's why I talked about an identification between the space and its dual, as it is the case for Hilbert space.2012-08-26

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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) $