Suppose that for two $n \times n$ matrices $A,B$, $AB = A + B$. Prove that $$\text{rank}(A^2) + \text{rank} (B^2) \leq 2 \text{rank} (AB).$$

This reminds me of Sylvester's Rank Inequality theorem, but I'm not sure if that's really helpful here. I haven't really made significant progress on this beyond writing out a few matrix multiplication. Would appreciate any help at all! Thank you.


There's a generalisation of Sylvester's Rank Inequality Theorem attributed to Frobenius. It states for all matrices $X, Y, Z$ we have $$ rk(XY) + rk(YZ) \le rk(Y) + rk(XYZ). $$ Using $AB = A + B$ we get $(A-I)(B-I)=I$ hence $(A-I) = (B-I)^{-1}$ and so $(B-I)(A-I) = I$ which implies $BA = A + B = BA$. So the matrices commute. Then apply the above theorem with $Y = AB$, $X = A-I$ and $Z = B-I$ gives the desired inequality.


Step 1: $B-I$ is invertible

In fact, $$ Bv = v \implies Av = ABv = Av + Bv = Av + v\implies v=0 $$

Step 2: $AB=BA$

In fact, since $B-I$ is invertible, in particular $(B-I)^{-1}=P(B)$ where $P$ is a polynomial (obtained by euclidean division between the characteristic polynomial and $x-1$). So $$ AB = A+B \implies A(B-I) = B \implies A = BP(B) = P(B)B $$ and $A$ is thus a polynomial in $B$. In particular $AB=BA$.

Step 3: Profit

Notice now that $A^2 = P(B)BA$, so $rk(A^2)\le rk(BA)$ and the problem is symmetric in $B,A$, so $rk(B^2)\le rk(AB)$, leading to $$ rk(A^2) + rk(B^2)\le rk(BA)+ rk(AB) = 2rk(AB) $$

  • $\begingroup$ Thank you for the solution! I have one doubt, however, when you define the polynomial $P$, you say that it's obtained by Euclidean division between the characteristic polynomial of $B$ and $x-1$, but I'm not sure why this is true. I know that the polynomial of the inverse of a matrix is the reciprocal polynomial of the characteristic polynomial, but could you please explain why we perform Euclidean division with $x-1$? Thanks again! $\endgroup$ Apr 28 '20 at 15:26
  • $\begingroup$ One last thing, how do you go from $A^2 = P(B)BA$ implying $rk(A^2) \leq rk(BA)$? $\endgroup$ Apr 28 '20 at 15:43
  • 1
    $\begingroup$ @OmicronGamma in general, $rk(XY)\le rk(Y)$ for every couple of matrices, since the null space of $XY$ contains the one of $Y$. If $Q(x)$ is the characteristic polynomial of $B$, you have that $Q(B)=0$; moreover, $1$ is not an eigenvalue of $B$, so $Q(1)\ne 0$. As a consequence, by Euclidean division, $Q(x) = S(x)(x-1) + c$ for some nonzero constant $c$, so we divide by $-c$ and $-Q(x)/c = P(x)(x-1) - 1\implies -Q(B)/c = 0 = P(B)(B-I) - I\implies P(B) =(B-I)^{-1}$ $\endgroup$
    – Exodd
    Apr 28 '20 at 16:07

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