The rank of the Sum of two matrices

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SUMMARY

The discussion centers on determining the rank of the sum of two square matrices A and B, both of rank 1. It establishes that the rank of A+B can only be 0, 1, or 2, with a definitive conclusion that it cannot exceed 2. The participants explore the relationship between the dimensions of the column spaces of A and B, using the formula for the dimension of the sum of vector spaces. They emphasize the importance of providing general examples to illustrate these rank possibilities.

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  • Knowledge of linear algebra concepts, particularly the properties of square matrices
  • Ability to apply the dimension formula for vector spaces
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Dank2
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Homework Statement


Let A,B be square matrices of order n. n>=2
lets A and B be matrices of Rank 1. What are the options of the Rank of A+B ?

Homework Equations

The Attempt at a Solution


I know that there are 3 possibilities, 2, 1 , 0. Just having trouble with coming up with a formula. i tried:
Rank(A+B) = dim(SP{Ac+Bc}) , (where Ac is the columns of A.)
I know also this equation
Rank of (A+B) = n - dim(P(A+B)) (where P denotes the solution space of (A+B)x = 0.
 
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There is no formula. For all three cases there can easily be found an example. How did you show that it cannot be more than two?
 
let Ac denote the span of the columns of Matrix A.
A+B is contained in Ac+Bc
therefore, dim(A+B) is smaller or equal to dim(Ac+Bc)

but dim(Ac+Bc) = dim(Ac) + dim(Bc) - dim(Ac intersection Bc)
 
fresh_42 said:
There is no formula. For all three cases there can easily be found an example. How did you show that it cannot be more than two?
Fixed typing errors.
 
Dank2 said:
let Ac denote the span of the columns of Matrix A.
A+B is contained in Ac+Bc
therefore, dim(A+B) is smaller or equal to dim(Ac+Bc)

but dim(Ac+Bc) = dim(Ac) + dim(Bc) - dim(Ac intersection Bc)
Yes. But your A+B should both be (A+B)c and maybe a ≤ 1+1=2 at the end.
If you give examples for the three cases that is all you can do.
 
fresh_42 said:
(A+B)c
i wanted to use Ac + Bc, because then, i can use the fact that Ac is a vector space, and usehe sum of the dimensions formula of vector spaces, which is dim(A+B)= dim(A)+dim(B) - dim(AintersectB)
 
I was thinking if i could show it with nxn general matrices, but i can't think about a way,
 
Dank2 said:
I was thinking if i could show it with nxn general matrices, but i can't think about a way,
Two by two are enough. But you can attach zeros as many as you want or other linear dependencies. But why? A minimal example will be fine.
 
fresh_42 said:
Two by two are enough. But you can attach zeros as many as you want or other linear dependencies. But why? A minimal example will be fine.
I have been told a numerated example is not enough.I need to show it the possibilities in general. I've shown how the Rank of A + B is less or equal to 2. not sure if that's the correct way.
 
  • #10
Dank2 said:
I have been told a numerated example is not enough.I need to show it the possibilities in general. I've shown how the Rank of A + B is less or equal to 2. not sure if that's the correct way.
Beside what I've said in post #5 about A+B (which should be (A+B)c) and the unusual notation for a linear span by c it is ok.
What does it mean for a (n,n)-matrix to have rank 0,1 or 2?
 
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  • #11
fresh_42 said:
What does it mean for a (n,n)-matrix to have rank 0,1 or 2?
not sure if i got the question right.
Rank = number of linearly independent vectors in columns or A or the the rows of A.

or that the columns of A span the 0 vector in Rn, a line in Rn or a plain in Rn respectively.
 
  • #12
Let's say rows. (Columns would work as well.)
So rank 0 means no linear independent row. That leaves only one possible vector.
Rank 1 thus means exactly one linear independent row. Let us take any fixed row vector as first. What does it mean for all others?
And at last two linear independent rows, but only two. All others must be in their span.
If you've found 3 matrices C with these properties, then you will have found C = A+B. All it needs then is to find summands A, B which add up to your C.
 
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