
#1
Feb507, 07:13 AM

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P: 3,225

Let A and B me matrices such that the product AB is defined. One has to proove that r(AB) <= r(A) and r(AB) <= r(B).
My first thoughts are: let A be 'mxn' and B be 'nxp', so AB is 'mxp'. Further on, we know that r(A) <= min{m, n}, r(B) <= min{n, p} and r(AB) <= min{m, p}. I'm stuck here, although I tried to work something out with the inequalities, but without success. Any hints would be appreciated. 



#2
Feb507, 08:18 AM

P: 3,177

you should look on the vector space of the homogenous solutions of Bx=0.
and use the fact that dim of this space equals nrankB, also you should look on ABx=0. from this you can easily prove that rank(AB)<=rank(B). now i forgot how to prove the second inequality. 



#3
Feb507, 09:04 AM

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P: 3,225

The only way that r(AB) <= r(A) would follow from this is that d2 >= d1, where d2 is the dimension of the vector space of solutions to ABx=0, and d1 the dimension of the vector space of solutions to Ax=0. But why would I conclude that d2>=d1? 



#4
Feb507, 10:05 AM

P: 3,177

Another matrix rank proof
ok, you need to show that U={xBx=0} is a subset of V={xABx=0}, then
dim U<=dim V and dim U=nrankB and dimV=nrankAB, n is the number of columns of the matrix, the number of columns of AB is the same as of B. the same way you can prove that rankAB<=rankA, with the use of rankAB=rank(AB)^t=rank(B^tA^t). and rankA=rankA^t. 



#5
Feb507, 10:07 AM

P: 3,177

if you ask why dimU<=dimV, then it follow from a thoerem that if U is subspace of V then (and V is finite) the inequality holds.




#7
Feb507, 12:56 PM

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P: 9,398

Woah, boys, you're making this way way to complicated. What is the rank? Just the dimension of the image. This is cleary a submultiplicative function (take some linearly independent vectors, applying a linear map can only make it more dependent).




#8
Feb507, 01:11 PM

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#9
Feb507, 02:43 PM

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The image of AB is a subet of the image of A. So r(AB) <= r(A).
The second inequality follows from the first, as r(AB) = r((AB)^t) = r(B^t A^t) <= r(B^t) = r(B) 



#11
Feb507, 03:11 PM

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The set { Av in R^m : v is in R^n }, if A is mxn, which is the same as the image of the linear transformation T : R^n > R^m given by Tv = Av. So a matrix can simply be thought of as a linear transformation.




#12
Feb507, 03:12 PM

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#13
Feb607, 04:26 AM

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P: 3,225





#14
Nov1507, 07:51 AM

P: 1

let C=AB [ A=m*k , B= k*n, C= m*n]
now let the rows be C and B be: {c[1],c[2],....c[m]} , { b[1],b[2]....b[k] thus we can write : c[i]= A[i][1]*b[1] + A[i][2]*b[2] +...+ A[i][m]*b[m] therefore the row space of C is a subspace of row space of B therefre Rowrank(C)=rank(C) <= Rank (B) similarly proceed wid column rank argument for Column rank (C)=Rank(C) <= Colum rank (A)=Rank(A) hence Rank(C)<= {Rank(A),Rank(B)} 



#15
Nov2907, 03:30 PM

P: 341

You can think this as ''how can the dimension of the nullspace of the product increase ?" For a little intuition, think about how can (AB)x = 0 be satisfied with a nonzero x vector, either it is because x is in the nullspace of B or (Bx) is in the nullspace in A. It is very useful to experiment these kind of stuff with full rank rectangular matrices...



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