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Rank of a matrix

  1. May 9, 2009 #1
    My name is Jennifer and I'm new to Physics Forum. I was googling algebraic terms when I came across this site. It looks very helpful and I will greatly appreciate it if someone can help me answer this question:-

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

    Thank you~~
  2. jcsd
  3. May 9, 2009 #2


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    The rank of a matrix is the dimension of its range space:

    rank(M) = dim(range(M))
    rank(M o L) = dim(range(M o L))

    range(M) = the set of all vectors of the form Mx with x in Rm
    range(M o L) = the set of all vectors of the form MLx with x in Rn

    Compare the two and it should be obvious why (a) is true. Proving (b) is done similarly.
  4. May 9, 2009 #3
    Hi jbunni,

    I'm confused by your reply. We don't have specific values for m and n so I don't understand how we can prove a) is true :S
  5. May 9, 2009 #4


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    Try showing that range(M o L) is a subset of range(M), i.e., if y is in range(M o L) then y is also in range(M). To show this, write down what a typical element of range(M o L) looks like. It should be immediately apparent that it is also an element of range(M).

    Then answer this: given that you know that range(M o L) is contained in range(M), what can you say about the dimensions of range(M o L) and range(M)?
  6. May 9, 2009 #5
    Ohhhh I think I get it now...

    M o L takes vectors in Rn, maps them to vectors in Rm and then takes those vectors and map them to vectors in Rp. It does not necessarily map a vector in Rn to each vector in Rm right? Therefore, M o L can be a subset of Rp. On the otherhand, M maps all vectors in Rm to Rp. Is this right?
  7. May 9, 2009 #6


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    That's exactly right. M o L means you are applying first L, then M. Notice that to form M o L, the dimensions of the spaces have to agree: if L maps Rn to Rs, and M maps Rt to Rp, then M o L can only be defined if s = t. In matrix terms, this means that you must have

    # of columns of [L] = # of rows of [M]

    where [L] and [M] are matrix representations of L and M, with respect to some bases of the appropriate spaces.

    P.S. For future reference, the preferred place to ask homework questions is in the "Homework & Coursework Questions" section. The linear algebra stuff is typically asked in the "Calculus and Beyond" subsection.
    Last edited: May 9, 2009
  8. May 9, 2009 #7
    Thanks! you were a big help. I'll ask future questions in the Homework & Coursework Questions. However, um, cud i ask you one more question?

    Don't you mean # rows of [L] = # columns of [M]? :S...and how is this relevant to the answer other than telling u the dimensions of the matrix M o L which is pxn?
  9. May 9, 2009 #8


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    Yes, sorry, I got it reversed.

    It's relevant only because sometimes it's easier to visualize what is happening with linear maps between (finite-dimensional) vector spaces if you consider what it means in terms of matrices, because many students already have some experience with matrices by the time they start learning about abstract vector spaces.

    For example, the range of a matrix is the set of all linear combinations of its columns, so the dimension of the range (which is equal to the rank) is simply the number of linearly independent columns. This fact can help give you some concrete insight into why inequalities like the ones you are trying to prove are true, even though if possible the proofs themselves shouldn't depend on matrices.
  10. May 9, 2009 #9

    I was looking up a similar question also :)

    Umm... I was wondering how you knew that M maps all vectors in Rm to Rp, and not just some of the vectors?

    Also, for one of the rank properties I read about, it says that rank (ML) <= min(rank M, rank L). (I'm not sure if this is exactly relevant), but does this property imply that in this specific question, M maps ALL vectors in Rm to Rp?
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