# How to prove this rank inequality? [duplicate]

Let $$n\geq2$$ and $$A,B\in M_{n}(\mathbb{C})$$ such that $$B^2=B$$. Prove that $$\mbox{ rank }(AB-BA)\leq\mbox{ rank }(AB+BA).$$

If $$B$$ is zero or the identity matrix, we are done. But $$B$$ will always be diagonalisable. From this, how we can proceed?

• If we could somehow replace the matrix B with the diagonal matrix, that should make it way easier May 2, 2019 at 18:48
• Try the case for diagonalised matrix for n=2, after considering identity and zero May 2, 2019 at 19:03

Disclaimers: First of all, I will provide a simple explanation why $$B$$ is diagonalizable (for those who are interested to see how this can be justified) and then I will explain how one can get the full solution. However, at some point there are "miserable" and simple calculations, that should be done, and I omit them intentionally. I hope this won't be a problem.

$$B$$ is diagonalizable:

As you observed, $$B$$ is diagonalizable. Let's see how one can justify this. The minimal polynomial of $$B$$ should divide the polynomial $$x^2-x$$, which is vanished by $$B$$ (recall that the minimal polynomial of $$B$$ divides every polynomial that vanishes at $$B$$). Then, the minimal polynomial has only distinct roots (since possible minimal polynomials are $$x,\ x-1$$ and $$x^2-x$$) and this means that $$B$$ is diagonalizable.

The full solution:

Now let's move on to what you need to see in this answer. There exists an invertible matrix $$U$$ such that $$B=UDU^{-1},$$ where $$D$$ is the diagonal matrix with $$1$$'s at the first diagonal entries (if $$1$$ is eigenvalue of $$B$$) and $$0$$'s at the rest of them (if $$0$$ is eigenvalue of $$B$$). Since rank is invariant for similar matrices, we have that

$$\mbox{rank}(AB-BA)=\mbox{rank}(U(AB-BA)U^{-1})=\mbox{rank}(CD-DC)\\ \mbox{rank}(AB+BA)=\mbox{rank}(U(AB+BA)U^{-1})=\mbox{rank}(CD+DC),$$

where $$C=UAU^{-1}.$$ Thus, we only need to prove that

$$\mbox{rank}(CD-DC)\leq \mbox{rank}(CD+DC)$$

for any $$C\in M_n(\mathbb{C}).$$ Now write $$C=(c_{ij})_{1\leq i,j\leq n}$$ and compute the matrices $$CD-DC$$ and $$CD+DC.$$ Doing so (and it's extremely easy, because $$D$$ is a diagonal matrix), you will find that $$CD-DC$$ and $$CD+DC$$ are equal to $$2\times 2$$ block matrices (the blocks are not necessarily square blocks), where the matrices at the blocks $$(1,2)$$ of $$CD-DC$$ and $$CD+DC$$ are opposite, the matrices at the blocks $$(2,1)$$ and $$(2,2)$$ are equal (equal matrices at the blocks $$(2,1)$$ in $$CD-DC$$ and $$CD+DC$$ and equal matrices at the $$(2,2)$$ blocks of those too), while the matrix at the block $$(1,1)$$ of $$CD-DC$$ is a zero matrix. Moreover, the $$(2,2)$$ blocks are the zero matrices. Finding that, it is very easy to prove that whenever $$k\in \{1,\ldots,n\}$$ rows of $$CD-DC$$ are linearly independent, the same rows of $$CD+DC$$ will also be linearly independent (for this we can deduce linear independency by using the definition, which says that the rows $$u_1,\ldots ,u_k$$ are linearly independent iff $${\lambda}_1u_1+\cdots+{\lambda}_ku_k=0$$ implies $${\lambda}_1=\cdots ={\lambda}_n=0$$). Of course, the rank of the matrix is equal to the number of its linearly independent rows. Consequently, the aforementioned observation implies that

$$\mbox{rank}(CD-DC)\leq \mbox{rank}(CD+DC)$$

and we are done.

• Would you please explain why the blocks (1,2),(2,1) and (2,2) are equal. May 4, 2019 at 6:42
• Hmm, they are not all equal. I will edit that. The matrices at the blocks $(1,2)$ of $CD-DC$ and $CD+DC$ are opposite, but the next argument about linear independency still works. Anyway, this claim and the fact that the matrices at the blocks $(2,1)$ and $(2,2)$ are equal (i.e. the matrices at the $(2,1)$ blocks of $CD-DC$ and $CD+DC$ are the same and the matrices at the $(2,2)$ blocks of these matrices are also the same) follow by a simple calculation of multiplying, subtracting and adding matrices. May 4, 2019 at 9:22
• To be more clear, I don't say that the matrices at the $(2,1)$ and $(2,2)$ blocks of $CD-DC$ or $CD+DC$ will be the same. I am just saying that the $(2,1)$ blocks of $CD-DC$ and $CD+DC$ are equal in those two matrices. The same is true for their $(2,2)$ blocks. May 4, 2019 at 9:39
• Note also that you should conclude - as I added to my answer - that the $(2,2)$ blocks are zero matrices in both $CD-DC$ and $DC+CD$, when at least one diagonal entry of $D$ is $0.$ Next, you can work the rest of the proof out. May 4, 2019 at 9:51