$T$ 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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$\begingroup$ I guess you might find a few similar questions if you have a look at questions tagged matrix-rank+inequality. $\endgroup$ – Martin Sleziak Sep 13 '16 at 5:16
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$\begingroup$ Some posts related to the left inequality: Sylvester rank inequality: $\operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n$ and Prove that $\text{rank}(AB)\ge\text{rank}(A)+\text{rank}(B)-n$. $\endgroup$ – Martin Sleziak Jul 9 '20 at 11:36
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Hints:
Note that for any linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^m$$ the rank of $T$ is less or equal than $\min(n,m)$.
A linear map $$T:\mathbb{R}^n\rightarrow \mathbb{R}^n$$ is bijective iff its rank is $n$.
$m\geq n$, due to $1$.
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.
