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Question from homework in Linear Algebra:

Let $A,B$ be two matrices of size $n \times n$ such that $AB=0$.

Show that: $rank(A) + rank(B) \le n$ .

It probably has something to do with the dim of the null space or column space but I can't put things together from what we've learned...

Please help.. Thanks. :)

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2 Answers 2

up vote 0 down vote accepted

Hint: show that $\operatorname{Im}(B)\subset \ker A$ and a well-known formula linking the rank and the dimension of the kernel of a matrix with the dimension of the underlying space.

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when you say Ker(A) you mean Null Space ? when you say Image sapce(B) you mean Column Space? we haven't learned the term Image space... –  Dor Shalom Dec 1 '12 at 12:52
    
Yes, that's what I mean. What is the notation used? –  Davide Giraudo Dec 1 '12 at 12:56
1  
Okay, thank you very much. Now I understand the way to the solution. lest say that column space of B is C(B) and null space of A is N(A) .For showing that $C(B) \subset N(A)$, Is it enough to just say that because $ AB=0 $, then for each vector $ v \in C(B) : A∗v=0 $, therefor foreach $ v \in C(B) : v \in N(A) $, therefor $ C(B) \subset N(A) $? @DavideGiraudo –  Dor Shalom Dec 1 '12 at 13:47

One can also use the inequality $rank(AB) \geq rank(B)-nul(A)$ which is true for all square matrices A nad B. By the rank-nullity theorem, we have $nul(A)= n - rank(A)$, so

$$ \begin{align} 0=rank(AB) &\geq rank(B)-nul(A)\\ &\geq rank(B)-n+rank(A) \end{align} $$ Which after rearrangement gives the desired inequality.

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