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I was trying to solve the following problem (SEEMOUS 2019 Problem 3):

Let $A,B$ be complex-valued $n \times n$ matrices such that $B^2 = B$. Show that $$\text{rank}(AB - BA) \leq \text{rank}(AB + BA)$$

$B$ is a projection matrix; My idea was to diagonalise $B$. It turns out that $B$ is similar to a matrix of the form $$\begin{bmatrix} I_k & 0 \\ 0 & 0 \end{bmatrix}, 0 \leq k \leq n. $$

Now my question is, can I just suppose without loss of generality that $B$ is of that form? I would think yes, because $B$ is of that form under a change of basis, which is actually an isomorphism, so for any $A$, if we solve the problem for the diagonal form of $B$ and the matrix $A'$ representing $A$ under the change of basis, we can revert through the isomorphism to the original form.

Anyways, then my proof continues as follows: Write $A = \begin{bmatrix} X & Y \\ Z & T \end{bmatrix}$ such that the block sizes allow multiplication with $B$. Then $AB - BA = \begin{bmatrix} 0 & -Y \\ Z & 0 \end{bmatrix}$ and $AB + BA = \begin{bmatrix} 2X & Y \\ Z & 0 \end{bmatrix}.$ Then the rank inequality is obvious.

Tanny Sieben
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1 Answers1

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Alternatively, instead of the change of basis you can consider the matrix B to be Diagonal with eigenvalues 0, 1, because rank is invariant under similar matrices.

Then $AB - BA = \begin{bmatrix} 0 & -Y \\ Z & 0 \end{bmatrix}$ and $AB + BA = \begin{bmatrix} 2X & Y \\ Z & 0 \end{bmatrix}.$ Then the rank inequality is obvious.

Can you please explain that last step? It doesn't seem trivial to me.

Summand
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  • Having the matrix $X$ instead of the $0$ matrix in the top left could only potentially increase the rank of the matrix. A row in $Y$ that is linearly dependent with the rows above it might form a linearly independent row in $AB+BA$ now that the left part isn't always going to be all $0$'s – Ryan Feb 29 '20 at 01:04