Question

rankA+rankB

Answer 1

@imranmeah91 i don't remember my lin alg, maybe u can help

Answer 2

@TuringTest : bahrom7893 and GOODMAN suggested you might be able to help. If you can, much appreciated! :)

Answer 3

what are you trying to prove?
that \[\text{rank}(A+B)\ge\text{rank}(A)+\text{rank}(B)\]for all mxn matrices ?

Answer 4

determine the relationship, the >= I think,... then prove that it is true.

Answer 5

for an nxm matrix A we have that
rank(A)+nullity(A)=m
so rank(A)=<m
that should help us get started maybe

Answer 6

Hmm. I might be able to use that and figure out something.

Answer 7

let matrix C=A+B

Answer 8

I think we will get
rank(A)+rank(B) =< rank(A+B)
but I want to prove it rigorously....

Answer 9

this theorem:\[rank(A)\le\min(n,m)\]\[rank(B)\le\min(n,m)\]\[rank(A+B)\le\min(n,m)\]seems to definitely imply that\[rank(A)+rank(B)\le rank(A+B)\]but that is not a 'proof' by any means, lol

Answer 10

by the theorem above, let us assume that\[rank(A)=m\]and that\[rank(B)=n\]then\[rank(A)+rank(B)=n+m>rank(A+B)=\min(n,m)\]which suggests you were right the first time

Answer 11

That makes sense. I think. :)

Answer 12

now let rank (B) take on the minimum of 0 (when nullity(B)=m)
we then have\[rank(A)=m\]\[rank(B)=0\]and let\[rank(A+B)=\min(n,m)=m\]we then have\[rank(A)+rank(B)=m+0=rank(n,m)=m\]which means that\[rank(A)+rank(B)\ge rank(A+B)\]that is definately not a formal proof
(this is not easy for me either frankly)

Answer 13

*that line second-to last should have rank(A+B)=m at the end

Answer 14

welcome, hope it helps!
if you do get the answer from your prof please let me know, I'm curious as well

-just for good measure:
@Zarkon @JamesJ help with a linear algebra proof?

Answer 15

@phi linear algebra help?
@mysesshou you're welcome :D

Answer 16

I hope this is not too much gobblygook:

Let A and B be m x n matrices
Let M= A+B


The rank of matrix A = the dimension of the column space of A, denoted as C(A). In other words, the rank is the number of independent column vectors in A.
Similarly for matrix B.

Define the sum of two vector spaces as the span of the union of their vectors.

The dimension of the vector space spanned by the sum of the C(A) and C(B) is the number of independent vectors in the sum. This must be less than or equal to rank(A)+rank(B). (If all rank(A) columns of A are independent of the rank(B) independent columns of B, the rank of C(A)+C(B) is equal to the sum of the ranks. If any columns of A are linear combinations of the columns of B, the rank of C(A)+C(B) is less than the sum of the ranks.) So,
rank ( C(A)+C(B)) ≤ rank(A)+rank(B)


The columns of M are a specific linear combination of the columns of A and B. i.e.
column i has Mi= Ai+Bi That is, M represents a subspace of C(A)+C(B)
therefore
rank(M) ≤ rank ( C(A)+C(B)) ≤ rank(A)+rank(B)