# Rank of the product of 3 matrices

Suppose I have 3 n by n matrices $A,B,C$ with $ABC=0$, what could be the maximal rank of $CBA$?

I guess the answer would be n but I failed to prove it( tried to use Rank-Nuillity Theorem but I don't know how to manipulate 3 matrices). Can anyone give me some hints? Thanks a lot!

• Of course the answer can't be $n$, that would force all three matrices to be invertible. Aug 20, 2014 at 21:26
• @ErickWong oh yes! thanks for pointing out! Aug 20, 2014 at 21:30
• Here's a hint to get you started: if $AB = 0$ then the columns of $B$ all lie in the nullspace of $A$. Is it possible for $\text{rank}(A)$ and $\text{rank}(B)$ to both be greater than $n/2$? Is it possible for $\text{rank}(AB)$ to be greater than $\text{rank}(A)$? Aug 20, 2014 at 21:42

In order to get $ABC = 0$, the sum of the dimensions of the kernels of $A,B,C$ must be at least $n$. Furthermore, the rank of $CBA$ is no more than the minimum of the ranks of $A,B,C$. So the best rank you could hope for for $CBA$ is $2n/3$ by the rank-nullility theorem. For simplicity let's say $n$ is divisible by $3$. Consider the following $3 \times 3$ matrices specified by columns, where $e_i$ denotes the $i$th basis vector (1 in position $i$ and 0 elsewhere), and $0$ denotes the $0$-vector.
$$A = (0,e_2,e_3), B = (0,e_1,e_2), C = (e_1, e_2, e_1)$$.
Then $ABC = 0$ but $CBA$ has rank 2. Now, you can replace each entry in the $3 \times 3$ matrices with a $d \times d$ identity matrix if the entry is $1$, and otherwise a $d \times d$ matrix of all zeros if the entry is $0$. Then you will get $3d \times 3d$ matrices such that $ABC = 0$ and $CBA$ has rank $2d$, which is the highest you could hope for. So $2n/3$ is the answer for the maximal rank of $CBA$.