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Let $A,B,C$ be $n \times n$ square real matrices with $ABC=0$. What is the maximum rank of $CBA$?

From an old written examination.

I've looked at the kernel and range of each matrix for simple cases (like if $n=2$), but how can we generalize to any $n$?

mathjacks
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3 Answers3

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First, note that $\dim(\ker(ABC)) \le \dim(\ker(A))+\dim(\ker(B))+\dim(\ker(C))$, so at least one of the three kernels is at least $\lceil n/3\rceil$. Next, note that $\textrm{rank}(CBA)\le \min\{\textrm{rank}(C),\textrm{rank}(B),\textrm{rank}(A)\}$, so we can conclude $\textrm{rank}(CBA) \le n-\lceil n/3\rceil = \lfloor 2n/3 \rfloor$.

We now demonstrate that this bound can actually be obtained. Write $n = 3k+l+m$, where $0\le l \le m\le 1$. Note that $2k+l = \lfloor 2n/3 \rfloor$. Now we label a basis of $\Bbb R^n$ as $x_1, x_2, \ldots, x_{k+m}, y_1, \ldots, y_{k+l}, z_1, \ldots, z_k$.

Define the maps $A,B,C$ on basis elements as follows:

  • $Ax_i = 0, Ay_i = y_i, Az_i = z_i$
  • $Bx_i = x_i, By_i = x_i, Bz_i = z_i$
  • $Cx_i = x_i, Cy_i = y_i, Cz_i = y_i$

Then we have $ABC = 0$, but $CBAy_i = x_i, CBAz_i = y_i$, so the rank of $CBA$ is equal to the number of $y_i$'s and $z_i$'s, ie $2k+l = \lfloor 2n/3 \rfloor$.

Rolf Hoyer
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  • Can you explain your notation for $k,l,m$ when you write $n=3k+l+m$? I understand it has to do with the ceiling/flooring function to get whole integer values but I don't understand how they were chosen. – mathjacks May 13 '16 at 16:49
  • @mathjacks Basically, $n$ has to be of the form $3k, 3k+1$, or $3k+2$. In the first case $l=m=0$, in the second $l = 0, m=1$, and in the third $l=1, m=1$. – Rolf Hoyer May 13 '16 at 19:04
  • Thank you. I marked your response as the answer because it is straightforward and self-contained. Thanks. – mathjacks May 13 '16 at 20:54
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The usual rank inequality gives $$ rk(CBA)\le {\rm min}\{rk(A),rk(B),rk(C)\} $$ for all $A,B,C$. Because in addition $ABC=0$ we can use the Frobenius rank inequality to obtain $$ rk(A)+rk(B)+rk(C)\le 2n. $$ Hence the upper bound for $rk(CBA)$ is at most $\lfloor 2n/3 \rfloor$, if $A$, $B$ and $C$ have the same rank. It is easy to see that this bound can be attained.

user26857
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Dietrich Burde
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  • From what I gather in the other answers, it is not actually that easy to see that the bound can be attained without some trial and error to guess the kernels and ranges. Do you have a better approach? – mathjacks May 13 '16 at 16:58
  • Well, the explicit construction of $A,B,C$ is quite easy with $Ax=0, Ay=y$ and $Az=z$, but I think you are not required to do this.The question in the "old written examination" asks: "What can be maximal possible rank of CBA?" This is different from "What is the maximal rank of $CBA$". – Dietrich Burde May 13 '16 at 17:37
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In general, the nullity of a product $\DeclareMathOperator{\rank}{rank}AB$ satisfies $$\max\{\dim \ker A, \dim \ker B\} \leq \dim \ker (AB) \leq \dim \ker A + \dim \ker B.$$ In our case we have $$n = \dim \ker (ABC) \leq \dim \ker A + \dim \ker B + \dim \ker C,$$ so at least one of $A, B, C$ has kernel of dimension $\geq \left\lceil \frac{n}{3} \right\rceil$. Thus, $$\ker (CBA) \geq \max\{\ker A, \ker B, \ker C\} \geq \left\lceil \frac{n}{3} \right\rceil$$ and hence by the Rank-Nullity Theorem, $$\rank (CBA) \leq n - \left\lceil \frac{n}{3} \right\rceil = \left\lfloor \frac{2n}{3} \right\rfloor.$$ We can prove that this bound is sharp by exhibiting a triple $(A, B, C)$ that achieves it. We can see that it's enough find matrices $\alpha, \beta, \gamma$ that respectively solve the problem for $n = 3$. Then, for any $n$, we can set $$A = \underbrace{\alpha \oplus \cdots \oplus \alpha}_{\left\lfloor\frac{n}{3}\right\rfloor} \oplus 0 ,$$ where $0$ is the $0 \times 0$ matrix of size $r \times r$, where $r$ is the remainder of $n$ modulo $3$, and define $B$ and $C$ similarly. I don't know a good way to find a $3 \times 3$ solution except by trial and error. A short script that tries random $(0, 1)$-matrices yields the pleasant-looking solution $$\alpha = \pmatrix{0&1&0\\0&0&1\\0&0&0}, \qquad \beta = \pmatrix{0&1&0\\1&0&0\\0&0&0}, \qquad \gamma = \pmatrix{0&0&0\\0&1&0\\1&0&0} .$$

Travis Willse
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  • Why is it sufficient to solve for $n=3$? I like this argument but it's unclear to me what setting $A = \alpha \oplus \cdots \oplus \alpha \oplus 0$ achieves. – mathjacks May 12 '16 at 20:17
  • The point is that multiplication of block diagonal matrices (with the same size blocks) amounts to multiplying corresponding blocks separately: Since $\gamma\beta\alpha$ has rank $2$, $CBA = \gamma\beta\alpha \oplus \cdots \gamma\beta\alpha \oplus 0$ has rank $\left\lfloor \frac{2n}{3} \right\rfloor$. On the other hand, computing $ABC$ we get $\alpha \beta \gamma \oplus \cdots \oplus \alpha\beta\gamma \oplus 0 = 0.$ – Travis Willse May 12 '16 at 20:34
  • Makes sense. Thanks! – mathjacks May 13 '16 at 05:15
  • Great, you're welcome. – Travis Willse May 13 '16 at 09:34