Linear Algebra - Dimension of Kernel

  • #1
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Homework Statement


Suppose that U and V are finite-dimensional vector spaces and that S is in L(V, W), T is in L(U, V). Prove that

dim[Ker(ST)] <= dim[Ker(S)] + dim[Ker(T)]

Homework Equations


(*) dim[Ker(S)] = dim(U) - dim[Im(T)]
(**) dim[Ker(T)] = dim(V) - dim[Im(S)]

The Attempt at a Solution


I know that ST is in L(U, W), so dim[Ker(ST)] = dim(U) - dim[Im(ST)]. So now I need to show:

dim(U) - dim[Im(ST)] <= dim(U) + dim(V) - dim[Im(T)] - dim[Im(S)]

But that just boils down to showing that dim[Im(ST)] is greater than dim[Im(T)] + dim[Im(S)]. It seems like I didn't really get anywhere. What am I missing? Thanks for your help.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement


Suppose that U and V are finite-dimensional vector spaces and that S is in L(V, W), T is in L(U, V). Prove that

dim[Ker(ST)] <= dim[Ker(S)] + dim[Ker(T)]

Homework Equations


(*) dim[Ker(S)] = dim(U) - dim[Im(T)]
(**) dim[Ker(T)] = dim(V) - dim[Im(S)]

The Attempt at a Solution


I know that ST is in L(U, W), so dim[Ker(ST)] = dim(U) - dim[Im(ST)]. So now I need to show:

dim(U) - dim[Im(ST)] <= dim(U) + dim(V) - dim[Im(T)] - dim[Im(S)]

But that just boils down to showing that dim[Im(ST)] is greater than dim[Im(T)] + dim[Im(S)]. It seems like I didn't really get anywhere. What am I missing? Thanks for your help.

Are you and rjw5002 taking the same course? My response to his identical question is here:
https://www.physicsforums.com/showthread.php?t=215288
 
  • #3
159
0
We must be taking the same course. It's the second course in Linear Algebra at Penn State. I read your post where you said the following:

Suppose v is in the null space of T. Then Tv= 0 so STv= 0 and v is in the null space of ST. That shows that the null space of T is a subspace of the null space of ST. Now suppose v is in U but NOT in the null space of T. Then Tv is a non-zero vector in V and, in order that STv=0, must be in the null space of T. That is, v is in T-1(null space of V). That set cannot have dimension larger than dim(null space of S). Putting those together, the dimension of the null space of TS cannot be larger than the dim(null space of S)+ dim(null space of T).

So I understand how you show that T is a subspace of Ker(ST), but what do you mean by "in order that STv=0, must be in the null space of T. That is, v is in T-1(null space of V)."? I'm not really sure what this is saying.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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Did I say that? Of course, that's a typo. If Tv is such that S(Tv)= 0, then Tv is in the null space of S. So that v itself is in T-1(null space of S).
 

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