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Is it ok to assume that $\operatorname{dim}(\operatorname{Im}(T^*))=\operatorname{dim}[(\operatorname{Im}(T))^*]$, where $T$ is a linear map acting on a finite dimensional space. i.e. just taking the dual outside?

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Is your space finite dimensional? If so, the dimension of the dual equals the dimension of the original space, and the dimension of the image of $T^*$ is the rank of any matrix representation of $T^*$, which is the rank of the transpose conjugate of a matrix representation of $T$, which has the same rank as any matrix representation of $T$. – Arturo Magidin May 22 '12 at 18:56
@ArturoMagidin: Thank you! I never knew that interpretation in terms of matrices. – Blanc May 22 '12 at 18:57

To make it an "official" answer.

If $V,W$ are finite dimensional, $T\colon V\to W$, then let $\beta$ be a basis for $V$, $\gamma$ a basis for $W$, and let $\beta^*$, $\gamma^*$ be the dual bases. Then $\dim(\mathrm{Im}(T^*)) = \mathrm{rank}([T^*]_{\gamma^*}^{\beta^*})$, where $[T^*]_{\gamma^*}^{\beta^*}$ is the matrix of $T^*$ with respect to $\gamma^*$ and $\beta^*$. But $$[T^*]_{\gamma^*}^{\beta^*} = \left([T]_{\beta}^{\gamma}\right)^*,$$ the conjugate transpose of the matrix of $T$ with respect to $\beta$ and $\gamma$. Since $\mathrm{rank}(A) = \mathrm{rank}(A^*)$ for any matrix, it follows that $$\begin{align*} \dim(\mathrm{Im}(T^*)) &= \mathrm{rank}([T^*]_{\gamma^*}^{\beta^*})\\ &= \mathrm{rank}(([T]_{\beta}^{\gamma})^*)\\ &=\mathrm{rank}([T]_{\beta}^{\gamma})\\ &= \dim(\mathrm{Im}(T))\\ &= \dim((\mathrm{Im}(T))^*), \end{align*}$$ the last equality because in the finite dimensional case, the dual has the same dimension as the original space.

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