Rank of matrix products is never greater than what

Question

If I have a product of matrices of rank 1, the product is not going to have rank greater than 1. why?

Hint: Think about it in terms of compositions of linear transformations. — , Dec 01 '16 at 18:39
@Crostul : Proper notation is $\operatorname{rank} AB \le \operatorname{rank} A,$ coded with \operatorname{rank}. That does not only de-italicize the letters, but also results in proper spacing in things like $a\operatorname{rank} b$ and $a\operatorname{rank}(b)$ (where you'll notice that the space to the right of "rank" in the second example is smaller than it is in the first, i.e. it depends on the context). (With some standard operatornames like \det, \sin, \max, \lim etc., just use the backslash.) — Michael Hardy, Dec 01 '16 at 18:42

score 1 · Answer 1 · answered Dec 01 '16 at 18:39

1

Consider two matrices $A,B$ with rank one. Then the image of $B$ has rank one. Now $A$ (when you do the product $AB$) can be considered as acting on an $1$-dimensional subspace. So its image has at most rank one.

answered Dec 01 '16 at 18:39

mfl

29,399

Can you please explain what the image of matrix means. – Abhishek Bhatia Dec 01 '16 at 18:56
Are you familiar with linear transformations? – mfl Dec 01 '16 at 18:58
Yeah a bit of. Do you mean Ax is the image of A? – Abhishek Bhatia Dec 01 '16 at 19:01
Actually the set ${Ax:x\in V}.$ – mfl Dec 01 '16 at 19:02
Why is the image of B have rank one? – Abhishek Bhatia Dec 01 '16 at 19:03
Because if the matrix $B$ has rank one then the image of the linear map associated to $B$ has at most dimension one. – mfl Dec 01 '16 at 19:05

Rank of matrix products is never greater than what

1 Answers1