# Rank proof

• Jan 8th 2012, 03:38 PM
itpro
Rank proof
Would someone please show me how to show that:

$\displaystyle rank(A) + rank(B) \leq rank(C)$

Thank you.
• Jan 8th 2012, 04:02 PM
alexmahone
Re: Rank proof
Quote:

Originally Posted by itpro
Would someone please show me how to show that:

$\displaystyle rank(A) + rank(B) \leq rank(C)$

Thank you.

Define A, B and C.
• Jan 8th 2012, 04:07 PM
itpro
Re: Rank proof
Quote:

Originally Posted by alexmahone
Define A, B and C.

They are $\displaystyle nxn$ matrices.
• Jan 8th 2012, 04:08 PM
alexmahone
Re: Rank proof
Quote:

Originally Posted by itpro
They are $\displaystyle nxn$ matrices.

They can't just be any n x n matrices, can they? What is the relationship among them?
• Jan 8th 2012, 04:13 PM
itpro
Re: Rank proof
Quote:

Originally Posted by alexmahone
They can't just be any n x n matrices, can they? What is the relationship among them?

Sorry, missed that: $\displaystyle C = A + B$

My intuition for the proof is that rank(A) and rank(b) is at most n and since addition preserves the dimension they will be at most of the rank of a C
as C rank is less or equal to n.
• Jan 9th 2012, 03:30 AM
Deveno
Re: Rank proof
what you want to show is that:

rank(A) + rank(B) ≤ rank(A+B).

but this is not true. suppose that A =

[1 0]
[0 0] and that B =

[-1 0]
[ 0 0].

clearly, A and B both have rank 1, so rank(A) + rank(B) = 1 + 1 = 2.

however, A+B is the 0-matrix, which has rank 0, and 2 is not less than or equal to 0.