Is the Rank-Nullity Theorem Always True for Linear Operators?

  • Thread starter geor
  • Start date
  • Tags
    Theorem
In summary, the conversation is about the generalized rank-nulity theorem and its application in linear algebra. The theorem states that if V and W are vector spaces and T: V -> W is a linear operator, then V is isomorphic with the direct sum of Im(T) and Ker(T). However, the professor gave a counter-example to show that this is not always true, even in the special case when V and W are the same vector space. The counter-example also highlights the difference between isomorphic and equal when it comes to direct sums.
  • #1
geor
35
0
Hello all,

In wikipedia, http://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem" a generalized rank-nulity theorem as below:

"If V, W are vector spaces and T : V -> W is a linear operator then
V is isomorphic with the direct sum of im(T) and ker(T)".

I had an exercise in Algebra which would be straightforward by using the theorem
above, but we had been given a somewhat complicated hint. When I mentioned this to the prof she said that this is not true and she also gave me the counter-example below:

T : R^2 -> R^2
T(e1)=0
T(e2)=e1

(R = the real numbers, e1, e2 the usual basis).

As she said, in this example, <e1>=ker(T)=Im(T) so the above theorem "is not true"..

I'm a bit confused, could you give some light please?!
I guess that this has to do with the fact that we say "isomorphic" and not "equal" (?!),
but still, that does not mean that the theorem is not correct.

In fact, the exercise we had to do was this:

If V is a v.s. and A: V -> V is a linear operator with Im(A^p) = Im( A^(p+1) ) for some p \in Z, prove <various stuff> and also prove that V = ker( A^p ) \directsum Im( A^p ).

Well, if you take in account that A^p is also a linear operator and by using the theorem above, this is straightforward.

Her proof is a almost a page...

Then I mentioned her this theorem and she said that it is not true and she gave me the above "counterexample"..

What do I miss here?!

Thanks in advance..
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Yup: "isomorphic" vs "equal to" is the problem. What wiki has is definitely true, as can be seen just by comparing dimensions. The direct sum in this case is the "exterior" direct sum of vector spaces, not the "interior" direct sum of subspaces. Your prof's counterexample shows that V is, in general, not an internal direct sum of kerT and imT when T is an operator V->V. On the other hand, notice that in that example kerT=~R and imT=~R, so that R^2=~kerT x imT.
 
  • #3
Just repeating what dvs said, what wiki said is true, but it's not helpful to the problem at hand because it is a statement about the external direct sum while you are asked to prove a statement about the internal direct sum.

Your professor gave a counter-example showing that even in the special case V=W, the formula [tex]V\cong Im(T)\oplus Ker(T)[/tex] (external direct sum) cannot be improved to [tex]V= Im(T)\oplus Ker(T)[/tex] (internal direct sum).
 
  • #4
Thanks a lot for the feedback! I see it now..
 

1. What is the rank-nullity theorem?

The rank-nullity theorem, also known as the dimension theorem, is a fundamental result in linear algebra that relates the dimensions of the kernel (null space) and image (column space) of a linear transformation.

2. How is the rank-nullity theorem used?

The rank-nullity theorem is used to calculate the dimension of the image and kernel of a linear transformation, and is also used to prove other important theorems in linear algebra, such as the invertible matrix theorem.

3. Can you provide an example of the rank-nullity theorem in action?

Yes, for example, if we have a linear transformation T: R^3 -> R^2, then the rank-nullity theorem tells us that the dimension of the kernel of T (nullity) plus the dimension of the image of T (rank) must equal 3, the dimension of the domain of T.

4. What is the significance of the rank-nullity theorem?

The rank-nullity theorem is significant because it provides a relationship between the dimensions of the kernel and image of a linear transformation, which helps us understand the structure of vector spaces and solve problems in linear algebra.

5. Are there any limitations to the rank-nullity theorem?

While the rank-nullity theorem is a powerful tool in linear algebra, it is limited to only working for finite-dimensional vector spaces. It also assumes that the linear transformation is defined over the same field.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
994
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
793
  • Linear and Abstract Algebra
Replies
1
Views
2K
  • Linear and Abstract Algebra
2
Replies
59
Views
8K
  • Linear and Abstract Algebra
Replies
8
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
2K
Replies
1
Views
1K
Back
Top