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

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In summary, the conversation discusses ways to prove rank(A+B)<=rank(A)+rank(B). One method is to show that the span of A+B's columns is a subset of the span of the vectors (a_i+b_i), which can be written as a linear combination of the base vectors a_i and b_j. However, it is noted that the number of columns of A does not necessarily equal its rank. The conversation also addresses the misconception that a choice of r columns will always form a basis, and clarifies that this is not always the case. Instead, the replacement lemma is suggested as a method for proving the original statement.
  • #1
boombaby
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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
 
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  • #2
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.)
 
  • #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...
 
  • #4
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)
 
  • #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
 
  • #6
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.
 
  • #7
matt grime said:
there is nothing that says r of them will form a basis.

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
 
  • #8
I was being dense - this is just a replacement lemma.
 

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

How do you define rank?

Rank refers to the number of linearly independent rows or columns in a matrix. It is also known as the dimension of the vector space spanned by the rows or columns of the matrix.

What is the formula for calculating rank(A+B)?

The rank of the sum of two matrices, A and B, can be calculated using the formula rank(A+B) = rank(A) + rank(B) - rank(A∩B), where A∩B is the intersection of the two matrices.

Is the rank of a matrix always equal to the number of its non-zero rows or columns?

No, the rank of a matrix is not always equal to the number of its non-zero rows or columns. It depends on the linear independence of the rows or columns.

How do you prove that rank(A+B) is less than or equal to rank(A) + rank(B)?

This can be proved using the rank-nullity theorem, which states that the rank of a matrix plus the nullity (dimension of the null space) is equal to the number of columns in the matrix. By substituting this in the formula for calculating rank(A+B), we can show that rank(A+B) is less than or equal to rank(A) + rank(B).

Can you provide an example to illustrate the concept of rank(A+B)<=rankA+rankB?

Consider two matrices, A = [1 2 3; 4 5 6; 7 8 9] and B = [10 11 12; 13 14 15; 16 17 18]. The rank of A is 2, as the first two rows are linearly independent. The rank of B is also 2, as the first two columns are linearly independent. When we add A and B, the result is a matrix with rank 3, as all three rows are linearly independent. Therefore, the rank of A+B is less than or equal to rank(A) + rank(B) (3 is less than or equal to 2+2).

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