Can Rank Multiplication Help Prove a Linear System Statement?

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Homework Help Overview

The discussion revolves around proving the inequality rank(A+B) ≤ rank(A) + rank(B) for matrices A and B in the context of linear algebra.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to connect the proof of the rank inequality with the known result rank(AB) ≤ rank(A) or rank(B). Some participants explore the implications of linear independence in the context of matrix addition.

Discussion Status

Participants are engaging with the concepts of rank and linear independence, with some expressing uncertainty about their proofs. There is an acknowledgment of the need for clarity and correction in the reasoning presented.

Contextual Notes

One participant notes a language barrier, indicating that their understanding of algebraic concepts in English may be limited. This may affect the clarity of their contributions.

annoymage
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Homework Statement



Show that rank(A+B) [tex]\leq[/tex] rank(A) + rank (B)

for every A,B [tex]\in[/tex] Mm,n (Real)

Homework Equations



N/A

The Attempt at a Solution



i only know how to proof this

rank(AB) [tex]\leq[/tex] rank(A) or rank(B),



and can this "rank(AB) [tex]\leq[/tex] rank(A) or rank(B)" help me to prove the above statement? can someone help me, to prove the above statement
 
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I'm not really sure whether it's a fine proof, so please correct me. And forgive my English, never read anything abt algebra in English :).

Let A and B be nxm matrices. Let's consider their columns as the columns of vector coordinates over the same vector space. Now rank(A) and rank(B) are the numbers of lineary independent vectors in each matrix respectively. Adding those two matrices, you are adding vectors. As there can be some linear dependencies between vectors in A and B, rk(A+B) can not overcome rk(A)+rk(B), as it still is nxm matrix. Basically speaking, by taking linear combinations of vectors, you can not get an independent vector, according to the very deffinition itself.

Huh, I hope one can understand what I wanted to write ;).
 
thanks i get it, and i will try convert that to mathematical form,

thank you
 
yep, good thinking
 

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