4
$\begingroup$

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. :)

2 Answers 2

0

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.

  • 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) $? @DavideGiraudo2012-12-01
0

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.