Let $A,B,C$ be $n\times n$ matrices. It is always true that $rank(ABC)\leq rank(AC)$ ?
It is always true that $rank(ABC)\leq rank(AC)$
#1
Đã gửi 04-07-2014 - 09:41
#2
Đã gửi 17-07-2014 - 12:52
Let $A,B,C$ be $n\times n$ matrices. It is always true that $rank(ABC)\leq rank(AC)$ ?
It's true!
I have not written up the proof but here is the idea. Think about the rank of a matrix as the dimension of the range space of the linear transformation represented by the matrix. In fact, this is the most natural definition for rank of a matrix.
I will write up the proof if you need it, but if you already solved this problem then I will not spoil other people's joy of discovery.
- maitram yêu thích
#3
Đã gửi 28-05-2015 - 00:26
#4
Đã gửi 28-05-2015 - 00:34
And lots of courterexamples.
#5
Đã gửi 10-10-2015 - 21:34
I don't think so
What is your method which you used to find the counterexample?
Đời người là một hành trình...
#6
Đã gửi 31-12-2015 - 15:07
Try for 2x2 matrices first and be lucky!. ..
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