Proving Disjoint Range & Null Space of Linear Operator T

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SUMMARY

The discussion centers on proving that the range and null space of a linear operator T are disjoint, specifically under the condition that rank(T^2) equals rank(T). Participants clarify that both the range and null space are subspaces containing the zero vector, thus they are not entirely disjoint. The key conclusion is that the only common vector in both spaces is the zero vector, which can be demonstrated through a proof by contradiction involving the existence of a non-zero vector in both spaces.

PREREQUISITES
  • Understanding of linear operators and their properties
  • Knowledge of vector spaces, specifically range and null space
  • Familiarity with the concept of rank in linear algebra
  • Experience with proof techniques, particularly proof by contradiction
NEXT STEPS
  • Study the properties of linear operators in detail
  • Learn about the relationship between rank and nullity in linear algebra
  • Explore proof techniques in mathematics, focusing on proof by contradiction
  • Investigate the implications of the rank-nullity theorem
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Students of linear algebra, mathematicians, and educators seeking to deepen their understanding of linear operators, vector spaces, and proof techniques in mathematics.

guroten
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Homework Statement



Given a linear operator T, show that if rank(T^2)=rank(T), then the range and null space are disjoint.

So I know that I can form a the same basis for range(T^2) and range(T), but I'm not sure where to go from there.
 
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guroten said:

Homework Statement



Given a linear operator T, show that if rank(T^2)=rank(T), then the range and null space are disjoint.

So I know that I can form a the same basis for range(T^2) and range(T), but I'm not sure where to go from there.

I recommend you go back and reread the problem. The range and nullspace of any linear operator are subspaces- they both include the 0 vector and so are never disjoint. Perhaps the problem is to show that the only vector in both the range and null space is the 0 vector?

If so try a proof by contradiction. Let v be a non-zero vector in both range and null space of T. Since v is in the range of T there exist u such that v= Tu. Now, what is T2u?
 
You're right, I meant disjoint other than 0. I get that T2u=Tv=0, but what does that say about the rank?
 

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