# Rank of partitioned matrix

• May 8th 2010, 11:04 AM
phony
Rank of partitioned matrix
Given a 2n x 2n matrix M which can be decomposed into to matrices M = [A B] where A and B are 2n x n matrices. Rank(A) = n, Rank(B) = n and also ANY n rows of A are linearly independent and also any n rows of B. Also, any n rows of the 4n x n matrix
[A]
B]
are linearly independent.

Is it true that Rank(M) = 2n and how can it be shown?
• May 9th 2010, 05:21 AM
phony
first idea...
Here's now an idea how I could show it -- but actually I think, this is not totally correct :(

Let's look at the rows of (A B). Since all rows of A are linearly independent, we need a linear combination of n+1 rows of M such that we obtain a vector
(0 v_2).

v_2 is unequal 0 at all entries since the rows of B are linearly independent of the rows of A. (IS THIS TRUE?)

There exist further r-1 rows of B which are linearly independent of (0 v_2).

Hence rank(A B) = 2n.

Is this proof correct?? I am not happy with this proof at all...