# Square Matrix Inequality

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 $$rk(A^2) + rk(B^2)\le rk(BA)+ rk(AB) = 2rk(AB)$$

• 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! Apr 28 '20 at 15:26
• One last thing, how do you go from $A^2 = P(B)BA$ implying $rk(A^2) \leq rk(BA)$? Apr 28 '20 at 15:43
• @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 $-Q(x)/c = P(x)(x-1) - 1\implies -Q(B)/c = 0 = P(B)(B-I) - I\implies P(B) =(B-I)^{-1}$ Apr 28 '20 at 16:07