Linear operators and rk(AB)

  • Thread starter Zorba
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  • #1
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Main Question or Discussion Point

Studying old exam papers from my college I came across the following:

Given linear operators [tex]A,\,B: V\rightarrow V[/tex], show that:

[tex]\textrm{rk}AB\le \textrm{rk}A[/tex]

My solution:
Since all [tex]v \in \textrm{Ker}B[/tex] are also in [tex]\textrm{Ker}AB[/tex] (viz [tex]ABv=A(Bv)=A(0)=0[/tex]) and potentially there are [tex]w \in \textrm{Ker}A[/tex] such that [tex]w \in \textrm{Im}B[/tex] which implies [tex]w[/tex] must be also in [tex]\textrm{Ker}AB[/tex] (viz [tex]ABv=A(Bv)=A(w)=0[/tex] - either [tex]Bv[/tex] is inconsistant then we still have [tex]Bv=0[/tex], or we have [tex]Bv=w[/tex] and follows as earlier) thus we have [tex]\textrm{Ker}A \subset \textrm{Ker}AB[/tex] and thus from the rank-nullity theorem it follows that [tex]\textrm{rk}A \ge \textrm{rk}AB[/tex].

Is everything I've done there legal? I'm a bit iffy about the step where I conclude [tex]\textrm{Ker}A \subset \textrm{Ker}AB[/tex]...
 

Answers and Replies

  • #2
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Having w in Ker A and w in Im B does not imply that w is in Ker AB (although if w = Bv then v is in Ker AB, since ABv = Aw = 0).

Hint: You can solve this by considering only images, not kernels.
 

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