show that $\operatorname{rank}(g\circ f) \leq \operatorname{rank}(f)+\operatorname{rank}(g)-n$

by Wadiaro Wapoo   Last Updated August 10, 2018 11:20 AM

Let $E$ a vector space and $\dim(E)=n$

and let $f,g \in L(E)$

show that $\operatorname{rank}(f\circ g) \leq \operatorname{rank}(f)+\operatorname{rank}(g)-n$

I can see that $\operatorname{Ker}(g) \subset \operatorname{Ker}(f\circ g)$

so $\dim \operatorname{Ker}(g) \leq \dim \operatorname{Ker}(f\circ g)$

by the rank-nullity theorem $\operatorname{rank}(g) \leq \operatorname{rank}(f\circ g)$

I am stuck here.



Answers 1


The reverse inequality is true. To see it, apply the rank-nullity theorem twice. $\DeclareMathOperator{\Im}{Im}$

Observe in the first place that $$\DeclareMathOperator{\rk}{rank}\rk(g\circ f)=\rk\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)\quad\text{and}\quad \ker\Bigl(g_{\,\bigm\vert_{\,\scriptstyle\Im f}}\Bigr)=\ker g\cap\Im f, $$ so, by the rank-nullity theorem $$\rk(g\circ f)= \dim(\Im f)-\dim(\ker g\cap\Im f)=\rk f-\dim(\ker g\cap\Im f). $$ Now $\;\ker g\cap\Im f\subset \ker g$, whence $$\dim(\ker g\cap\Im f)\le \dim(\ker g)=n-\rk g, $$ and therefore $$\rk(g\circ f)\ge\rk f-(n-\rk g).$$

Bernard
Bernard
August 10, 2018 11:14 AM

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