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Question Number 204957 by CrispyXYZ last updated on 03/Mar/24

2×2 matrix A and B satisfy that AB+A=BA+B. Prove that (A−B)^2 =O.

$$\mathrm{2}×\mathrm{2}\:\mathrm{matrix}\:\boldsymbol{\mathrm{A}}\:\mathrm{and}\:\boldsymbol{\mathrm{B}}\:\mathrm{satisfy}\:\mathrm{that} \\ $$$$\boldsymbol{\mathrm{AB}}+\boldsymbol{\mathrm{A}}=\boldsymbol{\mathrm{BA}}+\boldsymbol{\mathrm{B}}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\left(\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}\right)^{\mathrm{2}} =\boldsymbol{\mathrm{O}}. \\ $$

Answered by Rajpurohith last updated on 04/Mar/24

Need not hold true! take A=−I and B=I satisfy the given condition yet (A−B)^2 ≠0

$${Need}\:{not}\:{hold}\:{true}! \\ $$$${take}\:{A}=−{I}\:{and}\:{B}={I}\:{satisfy}\:{the}\:{given}\:{condition} \\ $$$${yet}\:\left({A}−{B}\right)^{\mathrm{2}} \neq\mathrm{0} \\ $$

Commented by MM42 last updated on 04/Mar/24

−2I≠0

$$−\mathrm{2}{I}\neq\mathrm{0}\: \\ $$

Answered by witcher3 last updated on 04/Mar/24

A= (((a b)),((c d)) );B= (((e f)),((g h)) ) AB+A= (((ae+bg+a af+bh+b)),((ce+dg+c cf +dh+d)) ) BA+B= (((ae+fc+e eb+fd+f)),((ga+hc+g gb+hd+h)) ) We know that Tr(AB)=Tr(BA) Tr(AB+A)=Tr(BA+B)⇒Tr(AB)+Tr(A)=Tr(BA)+Tr(B) ⇒Tr(A)=Tr(B) a+d=e+h det(AB+A)=(ae+bg+a)(cf+dh+d)−(af+bh+b)(ce+dg+c) =det(BA+B)=(ae+fc+e)(gb+hd+h)−(ga+hc+g)(eb+fd+f) (a−e)(d−h)−(b−f)(c−g)=0 ⇒det(A−B)=0 X=A−B⇒Tr(X)=0,det(X)=0 we knowX^2 −tr(X).X+det(X)=0⇒X^2 =0 (A−B)^2 =O_(2∗2)

$$\mathrm{A}=\begin{pmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}}\\{\mathrm{c}\:\:\:\:\:\mathrm{d}}\end{pmatrix};\mathrm{B}=\begin{pmatrix}{\mathrm{e}\:\:\:\:\mathrm{f}}\\{\mathrm{g}\:\:\:\:\mathrm{h}}\end{pmatrix} \\ $$$$\mathrm{AB}+\mathrm{A}=\begin{pmatrix}{\mathrm{ae}+\mathrm{bg}+\mathrm{a}\:\:\:\:\:\:\:\mathrm{af}+\mathrm{bh}+\mathrm{b}}\\{\mathrm{ce}+\mathrm{dg}+\mathrm{c}\:\:\:\:\:\:\:\mathrm{cf}\:+\mathrm{dh}+\mathrm{d}}\end{pmatrix} \\ $$$$\mathrm{BA}+\mathrm{B}=\begin{pmatrix}{\mathrm{ae}+\mathrm{fc}+\mathrm{e}\:\:\:\:\:\:\mathrm{eb}+\mathrm{fd}+\mathrm{f}}\\{\mathrm{ga}+\mathrm{hc}+\mathrm{g}\:\:\:\:\:\mathrm{gb}+\mathrm{hd}+\mathrm{h}}\end{pmatrix} \\ $$$$\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{Tr}\left(\mathrm{AB}\right)=\mathrm{Tr}\left(\mathrm{BA}\right) \\ $$$$\mathrm{Tr}\left(\mathrm{AB}+\mathrm{A}\right)=\mathrm{Tr}\left(\mathrm{BA}+\mathrm{B}\right)\Rightarrow\mathrm{Tr}\left(\mathrm{AB}\right)+\mathrm{Tr}\left(\mathrm{A}\right)=\mathrm{Tr}\left(\mathrm{BA}\right)+\mathrm{Tr}\left(\mathrm{B}\right) \\ $$$$\Rightarrow\mathrm{Tr}\left(\mathrm{A}\right)=\mathrm{Tr}\left(\mathrm{B}\right)\:\:\:\mathrm{a}+\mathrm{d}=\mathrm{e}+\mathrm{h} \\ $$$$\mathrm{det}\left(\mathrm{AB}+\mathrm{A}\right)=\left(\mathrm{ae}+\mathrm{bg}+\mathrm{a}\right)\left(\mathrm{cf}+\mathrm{dh}+\mathrm{d}\right)−\left(\mathrm{af}+\mathrm{bh}+\mathrm{b}\right)\left(\mathrm{ce}+\mathrm{dg}+\mathrm{c}\right) \\ $$$$=\mathrm{det}\left(\mathrm{BA}+\mathrm{B}\right)=\left(\mathrm{ae}+\mathrm{fc}+\mathrm{e}\right)\left(\mathrm{gb}+\mathrm{hd}+\mathrm{h}\right)−\left(\mathrm{ga}+\mathrm{hc}+\mathrm{g}\right)\left(\mathrm{eb}+\mathrm{fd}+\mathrm{f}\right) \\ $$$$\left(\mathrm{a}−\mathrm{e}\right)\left(\mathrm{d}−\mathrm{h}\right)−\left(\mathrm{b}−\mathrm{f}\right)\left(\mathrm{c}−\mathrm{g}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{det}\left(\mathrm{A}−\mathrm{B}\right)=\mathrm{0} \\ $$$$\mathrm{X}=\mathrm{A}−\mathrm{B}\Rightarrow\mathrm{Tr}\left(\mathrm{X}\right)=\mathrm{0},\mathrm{det}\left(\mathrm{X}\right)=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{knowX}^{\mathrm{2}} −\mathrm{tr}\left(\mathrm{X}\right).\mathrm{X}+\mathrm{det}\left(\mathrm{X}\right)=\mathrm{0}\Rightarrow\mathrm{X}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{A}−\mathrm{B}\right)^{\mathrm{2}} =\mathrm{O}_{\mathrm{2}\ast\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$

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