Studying old exam papers from my college I came across the following:(adsbygoogle = window.adsbygoogle || []).push({});

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]...

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Linear operators and rk(AB)

Loading...

Similar Threads - Linear operators | Date |
---|---|

A Nonlinear operators vs linear | Jan 16, 2018 |

A Extending a linear representation by an anti-linear operator | Dec 21, 2017 |

A Commutation and Non-Linear Operators | Nov 15, 2017 |

I Can a shear operation introduce a new linear dependency? | Oct 11, 2017 |

B Proof of elementary row matrix operation. | Jun 6, 2017 |

**Physics Forums - The Fusion of Science and Community**