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I have come across a definition of rank in some lecture notes I am using to prepare for a linear algebra qualifying exam and the notes define the rank of a linear map in a somewhat different manner than I am used to.

let $V$ be a finite dimensional vector space and $f \in End(V)$. Since $End(V) \cong V^* \otimes V$ we can provide a new definition of the rank of $f$.

How do we show $dim(Im(f)) = min \{ t : f = \sum_{i=1}^{t} v_{i}^{*} \otimes v_i \}$ where $v_{i}^{*} \in V^*$ and $ v_i \in V$?

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up vote 2 down vote accepted

It should be immediate that $\dim(\mathrm{Im}\;f))$ (let's call that the "geometrical rank") cannot be more than the tensor rank, because $v_i^*\otimes v_i$ represents an endomorphism that always produces a multiple of $v_i$.

To show that the tensor rank cannot be more than the geometrical rank, choose a basis $(v_1,\ldots,v_t)$ for the image of $f$. Then define $v_i^*$ as the linear map that sends $w\in V$ to the coefficient of $v_i$ of $f(w)$. Then $f$ is represented by $\sum_{i=1}^t v_i^* \otimes v_i$, so the tensor rank is at most $t$.

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