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$T$ is a linear transformation $T:\Bbb R^n\longrightarrow\Bbb R^m$ and $U:\Bbb R^m\longrightarrow\Bbb R^n$ where $m≠n$ such that $TU$ is bijective. Then where did this formula
$\mathrm{rank}(T)+\mathrm{rank}(U) -m \le \mathrm{rank}(TU) \le \min\{\mathrm{rank}(T), \mathrm{rank}(U)\}$ come from? Can someone please give its proof?

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Hints:

  1. Note that for any linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ the rank of $T$ is less or equal than $\min(n,m)$.

  2. A linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^n$$ is bijective iff its rank is $n$.

  3. $m\geq n$, due to $1$.

  4. The second inequality is standard.

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  • $\begingroup$ In point 2, you wrote T:R^n-> R^n, although T:R^n-> R^m. And i dont understand your point 3.Please explain $\endgroup$ – Parul Sep 12 '16 at 20:21
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This inequality is known as Sylvester's Rank inequality. The related properties are provided in the above link.

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