Range space of linear mappings

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SUMMARY

The discussion focuses on the properties of linear mappings, specifically the relationship between the ranks of composed linear mappings L: Rn → Rm and M: Rm → Rp. It is established that rank(M o L) ≤ rank(L), with the possibility of rank(M o L) being less than both rank(M) and rank(L) under certain conditions. An example is provided where both mappings have rank 1, resulting in their composition having rank 0.

PREREQUISITES
  • Understanding of linear mappings and their properties
  • Familiarity with the concept of rank in linear algebra
  • Knowledge of vector spaces and subspaces
  • Experience with matrix operations and compositions
NEXT STEPS
  • Study the implications of the Rank-Nullity Theorem in linear mappings
  • Explore examples of linear transformations with varying ranks
  • Learn about the properties of subspaces in Rm and Rp
  • Investigate the construction of matrices that demonstrate specific rank conditions
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators looking for examples of linear mapping properties and rank relationships.

Jennifer1990
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Homework Statement


Let L : Rn --> Rm and M : Rm --> Rp be linear mappings.
a)Prove that rank( M o L) <= rank(L).
b)Give an example such that the rank(M o L) < rank(M) and rank(L)

Homework Equations


None


The Attempt at a Solution


a)I see that (M o L) takes all vectors in Rn and maps them to vectors in Rm then maps these vectors to vectors in Rp. (L) also takes all vectors in Rn and maps them to Rm. From this, i get the impression that rank(M o L) = rank (L) because the quantity of vectors should not change when (M o L) maps vectors in Rm to Rp.

b)Is there a method to get such a matrix or do I have to use trial and error?
 
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Hints:

For part a:
Note that the range of M is a subspace of Rp with dimension Rank(M).
Likewise, the range of L is a subspace of Rm with dimension Rank(L).
For the composition ML, notice that after L is applied, the range of L is not necessarily all of Rm. Moreover, when you next apply M, it is only acting on that subspace, range of L.

For part b, try matrices from R^2 to R^2. Make both of them with rank 1, yet the composition has rank 0.
 

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