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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asked Jul 26 at 13:38

bm1125

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
  

  
  
  
  

add a comment | 

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

asked Jul 26 at 13:38

bm1125

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
  

  
  
  
  

add a comment | 

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

asked Jul 26 at 13:38

bm1125

1216

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

asked Jul 26 at 13:38

bm1125

1216

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 13:38

bm1125

1216

asked Jul 26 at 13:38

bm1125

1216

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

1 Answer
1

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

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

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
  

  
  
  
  

add a comment | 

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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

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

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
  

  
  
  
  

add a comment | 

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

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

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
  

  
  
  
  

add a comment | 

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

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

1018

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

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

1018

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 6:51

edited Jul 27 at 6:51

edited Jul 27 at 6:51

answered Jul 26 at 14:05

Aritra

1018

answered Jul 26 at 14:05

Aritra

1018

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

 

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