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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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    $\begingroup$ What does $L^{\star}L$ mean ? The composition is not well defined, unless we have some canonical identification between the space and its dual. $\endgroup$ – Ahriman Aug 26 '12 at 17:11
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    $\begingroup$ @Ahriman Well $L:H_a \rightarrow H_b$, so $L^*:H_b \rightarrow H_a$, where $L^*$ is the adjoint of $L$. Thus $L*L:H_a \rightarrow H_a$, right ? $\endgroup$ – user36772 Aug 26 '12 at 17:26
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    $\begingroup$ 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. $\endgroup$ – Ahriman Aug 26 '12 at 18:20
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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 $$ \mathrm{rk}(L)=\mathrm{rk}(L^*L)=\mathrm{rk}(R) $$

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