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How to prove rank(A+B)<=rankA+rankB ?thanks

  1. Nov 16, 2007 #1
    can anyone give some hints to prove rank(A+B)<=rankA+rankB ?
    i just don't know how to find the rank(A+B) from the column vectors of (A+B) ... or there's another way to prove?
    thanks a lot
     
  2. jcsd
  3. Nov 16, 2007 #2

    matt grime

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    You don't have to find the rank of A+B. You just need to show it is less than something else. Let let a_i and b_j, 1<=i,j<=n be the columns of A and B. What is the span of A+B's columns? It is the span of the vectors (a_i+b_i). What is the dimension of this space in relation to the span of hte a_i and the b_j? (You don't need to work anything out.)
     
  4. Nov 16, 2007 #3
    and so <a_i+b_i> = <a_1,a_2,..a_rankA,b_1,b_2...b_rankB> as every a_i+b_i can be written as a linear combination of those vectors.....this is really easy..thanks a lot!
    without your help i might still be stuck with finding out which a_i+b_i is the base vector of A+B...
     
  5. Nov 16, 2007 #4

    matt grime

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    No, that is not the case. The span <a_i+b_i> is definitely not the span <a_i,b_j>. If it were then the question would be: show the rank of A+B equals the rank of A plus the rank of B. That is false - just consider B=-A to see this.

    But, anything in the span of <a_i+b_i> is in the span of <a_i,b_j>, which gives the answer.


    Also, the number of columns of A is not its rank - you don't just take the columns a_1,..,a_rank(A)
     
  6. Nov 16, 2007 #5
    maybe I should write a_x(1) , a_x(2) , ...a_x(r) where r=rank(A) and x(i) shows the actual column number of the chosen base vectors...(sorry for my poor english...)

    but i think... the rank of <a_i,b_j> dose not equal to rank(A)+ rank(B). if A=B ,then <a_i,b_j>=<a_i> .am i right?

    so can i say rank(A+B)<= rank<A,B> <=rank A+rank B ? thanks for help
     
  7. Nov 16, 2007 #6

    matt grime

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    You shouldn't write some choice of r of the columns - there is nothing that says r of them will form a basis. But that is completely immaterial.

    You are on the right lines - think about how to write things clearly.
     
  8. Nov 16, 2007 #7
    i'm not quite clear about this.
    rank(A) is defined (in my book) to be dim<a_1 ,a_2,a_n>. and so definitely r of them will form a basis of the span <a_1,a_2,...a_n> ,so why can't i make some choice of a_i to be the basis? sorry for asking if i misunderstand something.....but i just want to make sure of it.
    thanks again
     
  9. Nov 17, 2007 #8

    matt grime

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    I was being dense - this is just a replacement lemma.
     
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