# Rank of matrix products is never greater than what [duplicate]

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

## marked as duplicate by Dietrich Burde, Namaste, Martin Sleziak, R_D, user91500Dec 2 '16 at 7:02

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• Hint: Think about it in terms of compositions of linear transformations. – user137731 Dec 1 '16 at 18:39
• What's your preferred definition of rank? – Omnomnomnom Dec 1 '16 at 18:40
• @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 1 '16 at 18:42

## 1 Answer

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.

• Can you please explain what the image of matrix means. – Abhishek Bhatia Dec 1 '16 at 18:56
• Are you familiar with linear transformations? – mfl Dec 1 '16 at 18:58
• Yeah a bit of. Do you mean Ax is the image of A? – Abhishek Bhatia Dec 1 '16 at 19:01
• Actually the set $\{Ax:x\in V\}.$ – mfl Dec 1 '16 at 19:02
• Why is the image of B have rank one? – Abhishek Bhatia Dec 1 '16 at 19:03