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Prove that if $f$ is a linear map between two finite-dimensional vector spaces, then rank of $f$ is equal to the rank of the dual map $f^*$.

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Anyone who has forgotten the definition of the dual map may find the discussion at Wikipedia will jog your memory.

It also says there, "If the linear map $f$ is represented by the matrix $A$ with respect to two bases of $V$ and $W$, then $f^*$ is represented by the transpose matrix $A^t$ with respect to the dual bases of $W^*$ and $V^*$." If you are allowed to use that, then the rank of $f$ equals the rank of $f^*$ because the rank of any matrix equals the rank of its transpose.

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Here's another proof. Let us consider the kernel of $f^*$, where $f:V\to W$ is a linear map between finite dimensional vector spaces: for any $\alpha\in W^*$, you have $$\begin{array}{rcl} \alpha\in\mathrm{Ker}(f^*)&\text{iff}& f^*(\alpha)=\alpha\circ f=0_{V^*}\\ &\text{iff}&\forall v\in V,~\alpha(f(v))=0\\ &\text{iff}&\mathrm{Im}(f)\subset\mathrm{Ker}(\alpha)\\ &\text{iff}&\alpha\in\mathrm{Ann}(\mathrm{Im}(f)) \end{array} $$ $\mathrm{Ann}(U)$ (of a subspace $U\subset V$) stands for the subspace of $V^*$ of all linear functionals that vanish on $U$. Thus $\mathrm{Ker}(f^*)=\mathrm{Ann}(\mathrm{Im}(f))$. It is a fact that for any subspace $B$ of a finite dimensional vector space $A$ we have $$\dim(B)+\dim(\mathrm{Ann}(B))=\dim(A),$$ so by the rank theorem it follows that $$\begin{array}{rcl}\mathrm{rk}(f^*)&=&\dim(W^*)-\dim(\mathrm{Ker}(f^*))\\ &=&\dim(W)-\dim(\mathrm{Ann}(\mathrm{Im}(f)))\\ &=&\dim(W)-\big(\dim(W)-\dim(\mathrm{Im}(f))\big)\\ &=&\dim(\mathrm{Im}(f))\\ &=&\mathrm{rk}(f) \end{array}$$ Therefore $f$ and its dual have the same rank.

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