Why do I ask this question? Since I got into trouble with the problem below:

If $A,B$ are $n\times n$ matrices which satisfy $A^{2013}=0, AB=BA, B\neq 0$, then $$\mathrm{rank}(AB)\le \mathrm{rank}(B)-1.$$

I don't know how to solve it. Maybe it is not necessary to consider the equality case of the rank inequality to solve this problem, I still want to know more about rank inequalities and linear mappings.

In an algebraic inequality, we can (usually) find the equality case after solving it. But things are not the same with rank inequalities. For example:

If $f:V\rightarrow W$ is a linear mapping, and $a_1,a_2,\ldots,a_n$ are vectors in $V$. Then $$\mathrm{rank}(a_1,a_2\ldots,a_n)\ge \mathrm{rank}f(a_1),f(a_2),\ldots, f(a_n)).$$

Proof: Suppose $\mathrm{rank}(f(a_1),f(a_2),\ldots, f(a_n))=k$, and suppose $f(a_{i_1}),f(a_{i_2}),\ldots, f(a_{i_k})$ are linearly independent. Then $a_{i_1},a_{i_2},\ldots, a_{i_k}$ are also linearly independent. Thus $\mathrm{rank}(a_1,a_2,\ldots, a_n)\ge k$.

Or consider the Sylvester inequality:

If $A,B$ are $n\times n$ matrices, then $$\mathrm{rank}(A)+\mathrm{rank}(B)-n\le \mathrm{rank}(AB)\le \min (\mathrm{rank}(A),\mathrm{rank}(B)).$$

The left one is proved by using linear mappings, while the right one is by linear dependence.

As you can see, proving them is not so difficult, but the problem is when the equality occurs. Any help will be appreciated.

Thank you so much.

  • 1
    $\begingroup$ You only need to show that $rank(AB) \not = rank(B)$ Since it's obvious that $rank(AB) \leq rank(B)$ $\endgroup$
    – Brian
    Jan 22, 2015 at 15:44
  • $\begingroup$ @Syuizen Yes that's what I'm confusing at. Could you give me some more help? $\endgroup$ Jan 22, 2015 at 16:57

1 Answer 1


First if $A = 0$ the claim is true because $B \neq 0.$ Therefore wlog $A \neq 0.$ Since $A^{2003} = 0,$ there is smallest $k, 1 \le k \le 2003$ such that $$ A^k = 0, A^{k-1} \neq 0. $$

Now that the preliminaries are out of the way will show that $\mathrm{rank}(AB) \neq \mathrm{rank}(B)$ by contradiction. Assume the contrary $\mathrm{rank}(AB) = \mathrm{rank}(B).$ This implies that $\ker(AB) = \ker(B)$. In particular, $$ABx = 0 \implies Bx = 0. \tag 1$$

For an arbitrary $x,$ we have $ 0 = A^kBx = AB(A^{k-1}x)$ this implies $A^{k-1}x = 0$ by $(1).$ Since $x$ is arbitrary we have $A^{k-1} = 0$ which contradicts the definition of $k$ and concludes the proof.

  • $\begingroup$ Why "$rank(AB)=rank(A)$ implies $Ker(AB)=Ker(B)$"? $\endgroup$ Jan 22, 2015 at 17:11
  • 1
    $\begingroup$ nullity theorem says $rank(A) + dim(ker(A) = n$ apply that to $AB$ and subtract. that shows you $rank(A) = rank(AB)$ implies $dim(kerA)) = dim(ker(AB)).$ you always have $ker(B) \subset ker(AB)$ but the equality of dimensions gives you $ker(AB) = ker(B).$ $\endgroup$
    – abel
    Jan 22, 2015 at 17:15

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