Range space of linear mappings

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The discussion focuses on proving that the rank of the composition of two linear mappings, M and L, is less than or equal to the rank of L. It is noted that the range of M is a subspace of Rp, and the range of L is a subspace of Rm, which affects the rank of their composition. A participant suggests that rank(M o L) could equal rank(L), but clarifies that after applying L, the range may not cover all of Rm, impacting the subsequent mapping by M. For the second part of the homework, the discussion hints at using matrices from R^2 to R^2, specifically with rank 1 for both, resulting in a composition with rank 0. Overall, the key points emphasize the relationship between the ranks of linear mappings and their compositions.
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


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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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