The rank of the Sum of two matrices

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Homework Help Overview

The discussion revolves around the rank of the sum of two square matrices, specifically when both matrices are of rank 1. Participants are exploring the possible ranks of the resulting matrix A+B and the implications of these ranks in the context of linear algebra.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the potential ranks of A+B, suggesting possibilities of 0, 1, or 2. There is an exploration of dimensionality and the relationship between the spans of the matrices' columns.

Discussion Status

Some participants have provided examples to illustrate the different cases of rank, while others are questioning the validity of these examples and the generalizability of the findings. There is ongoing exploration of the definitions and implications of matrix rank.

Contextual Notes

Participants note the challenge of demonstrating the rank possibilities in a general sense, with some expressing uncertainty about the implications of rank 0, 1, or 2 for square matrices. There is mention of constraints related to homework rules and the need for clarity in notation and definitions.

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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