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How do I prove that $$ \dim (\ker T^* T)=\dim \ker(T T^* )$$

where $T^*$ is the hermitian conjugate of $T$. These operators are acting on a Hilbert space $H.$

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This is false: $T^{*}Tx=0$ iff $Tx=0$. [one way is obvious. If $T^{*}Tx=0$ the $ \langle T^{*}Tx, x \rangle=0$ which means $\|Tx||^{2}=0$ or $Tx=0$]. Hence $\dim (\ker (T^{*}T))=\dim(\ker (T))$. Similarly, $\dim (\ker (TT^{*}))=\dim(\ker (T^{*}))$. Hence this question reduces to $\dim (\ker (T)=\dim (\ker (T^{*})$ which is not true.

See Counter example for infinite dimensional vector space

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