# 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?

• 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? – Ben Grossmann 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

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.
• Actually the set $\{Ax:x\in V\}.$ – mfl Dec 1 '16 at 19:02