rank of the matrix product

Asked Jan 22 '19 at 09:16

Active Jan 22 '19 at 09:36

Viewed 56 times

So I have matrices $A \in \mathbb{R}^{m\times n}$ and $B \in \mathbb{R}^{n\times k}$, I should proof these two inequalities: $rank(AB) \leq \max(rank(A),rank(B))$ and $rank(AB) \leq \min(rank(A),rank(B))$.

So I understand that $A: \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ $\Rightarrow$ $\max(rank(A)) \leq \max(\dim(Im(A))) = n$ and $B: \mathbb{R}^{n} \rightarrow \mathbb{R}^{k}$ $\Rightarrow$ $\max(rank(B)) \leq \max(\dim(Im(B))) = k$ and $\min(rank(A)) = \min(rank(B)) = \min (rank(AB)) = 0$.

But what after?

edited Jan 22 '19 at 09:36

asked Jan 22 '19 at 09:16

Just do it

What does rank$(A,B)$ mean? – Jose Brox Jan 22 '19 at 09:30
@JoseBrox $rank(A)$ and $rank(B)$ – Just do it Jan 22 '19 at 09:31
A suggestion: In order to get people interested in answering your questions, it is best if you stick to usual conventions! Otherwise they can read your statement and don't bother to try and understand it... – Jose Brox Jan 22 '19 at 09:33
@JoseBrox You are right, this is my mistake, I'll fix it now) – Just do it Jan 22 '19 at 09:35
Ok! Now, what do max$($rank$(A))$ and max$(\dim(Im(A)))$ mean? – Jose Brox Jan 22 '19 at 09:46
@JoseBrox so $\max(rank(A))$ -- maximum possible rank of matrix $A$, $\max(dim(Im(A)))$ -- maximum dimension of the image of a linear map corresponding to matrix $A$, $Im(A) \subset \mathbb{R}^{n}$
may be I am incerrect writing $Im(A)$ and I should use $ImA$?
– Just do it Jan 22 '19 at 09:50
You mean the maximum rank when $A$ ranges in all $m\times n$ matrices? That's min$(m,n)$, since the rank is the maximum number of linearly independent vectors in the column or row space – Jose Brox Jan 22 '19 at 10:03
@JoseBrox I think that this is equal to $\max \dim (ImA)$, no? – Just do it Jan 22 '19 at 10:07
@vermator okey, thx, I will read it – Just do it Jan 22 '19 at 10:08

rank of the matrix product

0 Answers0