Proving ker inequality of two transformations

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Let $ S, T: V rightarrow V $



prove: $ max dim ker T,dim ker S le dim ker (TS) le dim ker T + dim ker S $



I have no clue to prove it. Couldn't understand the solution given as well..




linear-algebra




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  • Please provide the solution and what specific point(s) you don't understand in it.
    – mathcounterexamples.net
    Jul 26 at 13:48











  • $ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
    – bm1125
    Jul 26 at 13:54














up vote
0
down vote

favorite












Let $ S, T: V rightarrow V $



prove: $ max dim ker T,dim ker S le dim ker (TS) le dim ker T + dim ker S $



I have no clue to prove it. Couldn't understand the solution given as well..




linear-algebra




share|cite|improve this question



















  • Please provide the solution and what specific point(s) you don't understand in it.
    – mathcounterexamples.net
    Jul 26 at 13:48











  • $ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
    – bm1125
    Jul 26 at 13:54












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $ S, T: V rightarrow V $



prove: $ max dim ker T,dim ker S le dim ker (TS) le dim ker T + dim ker S $



I have no clue to prove it. Couldn't understand the solution given as well..




linear-algebra




share|cite|improve this question











Let $ S, T: V rightarrow V $



prove: $ max dim ker T,dim ker S le dim ker (TS) le dim ker T + dim ker S $



I have no clue to prove it. Couldn't understand the solution given as well..





linear-algebra





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 26 at 13:38









bm1125

1216




1216











  • Please provide the solution and what specific point(s) you don't understand in it.
    – mathcounterexamples.net
    Jul 26 at 13:48











  • $ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
    – bm1125
    Jul 26 at 13:54
















  • Please provide the solution and what specific point(s) you don't understand in it.
    – mathcounterexamples.net
    Jul 26 at 13:48











  • $ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
    – bm1125
    Jul 26 at 13:54















Please provide the solution and what specific point(s) you don't understand in it.
– mathcounterexamples.net
Jul 26 at 13:48





Please provide the solution and what specific point(s) you don't understand in it.
– mathcounterexamples.net
Jul 26 at 13:48













$ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
– bm1125
Jul 26 at 13:54




$ dimImTS le min dimImT, dimImS $ - I couldn't prove it myself.
– bm1125
Jul 26 at 13:54










1 Answer
1






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up vote
2
down vote



accepted










I suppose V is finite-dimensional vector space and S,T are linear operators.



If $S(x)=0$ then $TS(x)=0$. Hence kerS $subseteq $ kerTS .



Now $ S(V) subseteq V $ so $ T(S(V))=TS(V) subseteq T(V) $ so dim Im $TS leq$ dim Im $T$ . So dim ker T $leq$ dim ker TS



This proves the first inequality.



Let $U: ker(TS) to kerT$ be defined by $U(x)=S(x)$. Then $U$ is linear transformation. And dim ker (TS) = dim ker (U) + rank (U) . Note that ker(U)=ker(S) and rank(U) $le$ dim ker (T). This proves the second claim.






share|cite|improve this answer























  • Thanks! much cleaner answer than I got!
    – bm1125
    Jul 26 at 15:50










  • Couldn't get on my own the last inequality. can you elaborate please?
    – bm1125
    Jul 26 at 17:52










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










I suppose V is finite-dimensional vector space and S,T are linear operators.



If $S(x)=0$ then $TS(x)=0$. Hence kerS $subseteq $ kerTS .



Now $ S(V) subseteq V $ so $ T(S(V))=TS(V) subseteq T(V) $ so dim Im $TS leq$ dim Im $T$ . So dim ker T $leq$ dim ker TS



This proves the first inequality.



Let $U: ker(TS) to kerT$ be defined by $U(x)=S(x)$. Then $U$ is linear transformation. And dim ker (TS) = dim ker (U) + rank (U) . Note that ker(U)=ker(S) and rank(U) $le$ dim ker (T). This proves the second claim.






share|cite|improve this answer























  • Thanks! much cleaner answer than I got!
    – bm1125
    Jul 26 at 15:50










  • Couldn't get on my own the last inequality. can you elaborate please?
    – bm1125
    Jul 26 at 17:52














up vote
2
down vote



accepted










I suppose V is finite-dimensional vector space and S,T are linear operators.



If $S(x)=0$ then $TS(x)=0$. Hence kerS $subseteq $ kerTS .



Now $ S(V) subseteq V $ so $ T(S(V))=TS(V) subseteq T(V) $ so dim Im $TS leq$ dim Im $T$ . So dim ker T $leq$ dim ker TS



This proves the first inequality.



Let $U: ker(TS) to kerT$ be defined by $U(x)=S(x)$. Then $U$ is linear transformation. And dim ker (TS) = dim ker (U) + rank (U) . Note that ker(U)=ker(S) and rank(U) $le$ dim ker (T). This proves the second claim.






share|cite|improve this answer























  • Thanks! much cleaner answer than I got!
    – bm1125
    Jul 26 at 15:50










  • Couldn't get on my own the last inequality. can you elaborate please?
    – bm1125
    Jul 26 at 17:52












up vote
2
down vote



accepted







up vote
2
down vote



accepted






I suppose V is finite-dimensional vector space and S,T are linear operators.



If $S(x)=0$ then $TS(x)=0$. Hence kerS $subseteq $ kerTS .



Now $ S(V) subseteq V $ so $ T(S(V))=TS(V) subseteq T(V) $ so dim Im $TS leq$ dim Im $T$ . So dim ker T $leq$ dim ker TS



This proves the first inequality.



Let $U: ker(TS) to kerT$ be defined by $U(x)=S(x)$. Then $U$ is linear transformation. And dim ker (TS) = dim ker (U) + rank (U) . Note that ker(U)=ker(S) and rank(U) $le$ dim ker (T). This proves the second claim.






share|cite|improve this answer















I suppose V is finite-dimensional vector space and S,T are linear operators.



If $S(x)=0$ then $TS(x)=0$. Hence kerS $subseteq $ kerTS .



Now $ S(V) subseteq V $ so $ T(S(V))=TS(V) subseteq T(V) $ so dim Im $TS leq$ dim Im $T$ . So dim ker T $leq$ dim ker TS



This proves the first inequality.



Let $U: ker(TS) to kerT$ be defined by $U(x)=S(x)$. Then $U$ is linear transformation. And dim ker (TS) = dim ker (U) + rank (U) . Note that ker(U)=ker(S) and rank(U) $le$ dim ker (T). This proves the second claim.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 27 at 6:51


























answered Jul 26 at 14:05









Aritra

1018




1018











  • Thanks! much cleaner answer than I got!
    – bm1125
    Jul 26 at 15:50










  • Couldn't get on my own the last inequality. can you elaborate please?
    – bm1125
    Jul 26 at 17:52
















  • Thanks! much cleaner answer than I got!
    – bm1125
    Jul 26 at 15:50










  • Couldn't get on my own the last inequality. can you elaborate please?
    – bm1125
    Jul 26 at 17:52















Thanks! much cleaner answer than I got!
– bm1125
Jul 26 at 15:50




Thanks! much cleaner answer than I got!
– bm1125
Jul 26 at 15:50












Couldn't get on my own the last inequality. can you elaborate please?
– bm1125
Jul 26 at 17:52




Couldn't get on my own the last inequality. can you elaborate please?
– bm1125
Jul 26 at 17:52












 

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