Posts Tagged ‘real matrices’

If real matrices A, B satisfy A^2 + B^2 = AB, then det(BA – AB) ≥ 0

Posted: February 26, 2024 in Basic Algebra, Matrices
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Let M_n(\mathbb{R}) be the ring of n \times n matrices with real entries. A form of the following problem was posted on the Art of Problem Solving website a couple of days ago.

Problem. Show that if A,B \in M_n(\mathbb{R}) and A^2+B^2=AB, then \det(BA-AB) \ge 0.

Solution (Y. Sharifi). Let X=aA+B, \ Y=bA+B, where a=\frac{\sqrt{3}-1}{2}, \ b=\frac{-\sqrt{3}-1}{2}. Let i:=\sqrt{-1}. Then

(X+Yi)(X-Yi)=(aA+B+(bA+B)i)(aA+B-(bA+B)i)

=(a^2+b^2)A^2+2B^2+(a+b)(AB+BA)+(a-b)(BA-AB)i.

Thus, since a^2+b^2=2, \ a+b=-1, \ a-b=\sqrt{3}, and A^2+B^2=AB, we get that

\begin{aligned} (X+Yi)(X-Yi)=-(BA-AB)+\sqrt{3}(BA-AB)i=(-1+\sqrt{3}i)(BA-AB)=2e^{2\pi i/3}(BA-AB).\end{aligned}

Therefore

|\det(X+Yi)|^2=\det(X+Yi) \overline{\det(X+Yi)}=\det(X+Yi)\det(X-Yi)

=\det((X+Yi)(X-Yi))=\det(2e^{2\pi i/3}(BA-AB))=2^ne^{2n\pi i/3}\det(BA-AB),

which gives

\det(BA-AB)=2^{-n}e^{-2n\pi i/3}|\det(X+Yi)|^2. \ \ \ \ \ \ \ \ \ (*)

Now, if 3 \mid n, then e^{-2n\pi i/3}=1 and so (*) gives \det(BA-AB)=2^{-n}|\det(X+Yi)|^2 \ge 0. If 3 \nmid n, then e^{-2n\pi i/3} is not a real number. But both \det(BA-AB) and |\det(X+Yi)|^2 are real numbers, and hence, by (*), we must have \det(BA-AB)=0. So either way, \det(BA-AB) \ge 0. \ \Box