Problem 1 - Matrices

For an $n \times n$ square matrices $X$, $Y$, define $[X,Y]$ by

$$[X,Y] = XY -YX.$$

Prove the following:

1.

When $X$, $Y$ are alternating matrices, $[X,Y]$ is also an alternating matrix, where $X$ is an alternating matrix if $X^T + X = 0$ is satisfied.

2.

$$[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0$$

1.

$X$ and $Y$ are alternating matrices, that is:

$$X=-X^T,\quad Y=-Y^T.$$

We want to check whether we have $[X,Y]=-[X,Y]^T$ or not.

$[X,Y]^T = (XY-YX)^T,$
$=(XY)^T - (YX)^T,$
$=Y^TX^T - X^TY^,$
$=YX - XY,\quad \hbox{since $X$\ and $Y$\ are alternating matrices},$
$=-[X,Y]$

2.

The terms we eliminate at each step are in bold.
$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]$
$= X(YZ-ZY) - (YZ-ZY)X + Y(ZX-XZ) - (ZX-XZ)Y + Z(XY-YX) - (XY-YX)Z$
$=\boldsymbol{XYZ} - XZY - (YZ-ZY)X + Y(ZX-XZ) - (ZX-XZ)Y + Z(XY-YX) \boldsymbol{-XYZ} - YXZ,$
$= - XZY - (YZ - ZY)X + YZX \boldsymbol{-YXZ} - (ZX-XZ)Y + Z(XY-YX) \boldsymbol{-YXZ}$
$= - XZY \boldsymbol{- YZX} + ZYX \boldsymbol{+YZX} - (ZX-XZ)Y + Z(XY-YX)$
$= \boldsymbol{-XZY} + ZYX + - ZXY \boldsymbol{+XZY} + Z(XY-YX)$
$= \boldsymbol{ZYX} - ZXY + ZXY \boldsymbol{-ZYX}$
$= \boldsymbol{- ZXY + ZXY}$
$= 0$