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Proof for $\dim(R(T))=\dim(R(T^{*}))$ for a linear operator in a Hilbert space. $T$ is the operator and $T^{*}$ is its adjoint.

I would like to know about the authenticity of the following line of proof of the above fact and get directed to a reference which uses provides this line of proof (if at all it is correct).

Here goes:

Proof: From rank nullity theorem $\dim(T)+\dim(\ker(T))=\dim(V)$, where $T:V \to W$. From the fact that $\ker(T)$ and $\dim(T^{*})$ are orthogonal complements, and the fact that $\ker(T)$ is a subspace of $V$, $\dim(T^{*})+\dim(\ker(T))=\dim(V)$. Thus, $\dim(T^{*})+\dim(\ker(T))=\dim(T)+\dim(\ker(T))$. $\dim(\ker(T))$ cancels out and we are left with, $\dim(T^{*})=\dim(T)$...Q.E.D.

Patrick Da Silva
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Hamza
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  • That works only in the finite-dimensional setting. If the dimension of $\ker T$ is infinite, you cannot cancel (and the orthogonal complement of $\ker T$ is the closure if $\operatorname{im} T^\ast$ if $V$ is infinite-dimensional). – Daniel Fischer Aug 15 '14 at 20:21
  • Thank you for your answer Sir. It means that as long as I am dealing with matrices in $R^{m\times n}$ I am safe($m$ and $n$ are finite). It can used to prove $rank(A) = rank(A^{T})$. – Hamza Aug 15 '14 at 20:31

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