[proofplan] We prove both identities directly from the definition $[A,B]=AB-BA$. Antisymmetry follows by expanding $[Y,X]$ and factoring out a minus sign. For the Jacobi identity, we expand each double commutator using distributivity and associativity of matrix multiplication, then check that each resulting triple product occurs once with coefficient $1$ and once with coefficient $-1$. [/proofplan] [step:Expand the commutator with reversed arguments] Let $X,Y\in M(n,\mathbb C)$. By the definition of the commutator bracket, \begin{align*} [Y,X]=YX-XY. \end{align*} Factoring out $-1$ gives \begin{align*} YX-XY=-(XY-YX). \end{align*} Using again the definition $[X,Y]=XY-YX$, we obtain \begin{align*} [Y,X]=-[X,Y]. \end{align*} Equivalently, \begin{align*} [X,Y]=-[Y,X]. \end{align*} [/step] [step:Expand the three double commutators] Let $X,Y,Z\in M(n,\mathbb C)$. First, \begin{align*} [Y,Z]=YZ-ZY. \end{align*} Using the commutator definition, distributivity of matrix multiplication over matrix addition, and associativity of matrix multiplication, \begin{align*} [X,[Y,Z]]=X(YZ-ZY)-(YZ-ZY)X. \end{align*} Therefore \begin{align*} [X,[Y,Z]]=XYZ-XZY-YZX+ZYX. \end{align*} Similarly, since \begin{align*} [Z,X]=ZX-XZ, \end{align*} we have \begin{align*} [Y,[Z,X]]=Y(ZX-XZ)-(ZX-XZ)Y, \end{align*} and hence \begin{align*} [Y,[Z,X]]=YZX-YXZ-ZXY+XZY. \end{align*} Finally, since \begin{align*} [X,Y]=XY-YX, \end{align*} we have \begin{align*} [Z,[X,Y]]=Z(XY-YX)-(XY-YX)Z, \end{align*} and hence \begin{align*} [Z,[X,Y]]=ZXY-ZYX-XYZ+YXZ. \end{align*} [guided] We now expand each term in the Jacobi expression. The only operations being used are the definition of the commutator, distributivity, and associativity of matrix multiplication. Associativity is what allows us to write products such as $X(YZ)$ and $(XY)Z$ both as $XYZ$. Start with the first term. Since the inner commutator is \begin{align*} [Y,Z]=YZ-ZY, \end{align*} the outer commutator gives \begin{align*} [X,[Y,Z]]=X(YZ-ZY)-(YZ-ZY)X. \end{align*} Distributing the products across the differences gives \begin{align*} [X,[Y,Z]]=XYZ-XZY-YZX+ZYX. \end{align*} For the second term, the inner commutator is \begin{align*} [Z,X]=ZX-XZ. \end{align*} Thus \begin{align*} [Y,[Z,X]]=Y(ZX-XZ)-(ZX-XZ)Y. \end{align*} After distributing and using associativity to remove parentheses from triple products, this becomes \begin{align*} [Y,[Z,X]]=YZX-YXZ-ZXY+XZY. \end{align*} For the third term, the inner commutator is \begin{align*} [X,Y]=XY-YX. \end{align*} Therefore \begin{align*} [Z,[X,Y]]=Z(XY-YX)-(XY-YX)Z. \end{align*} Distributivity and associativity give \begin{align*} [Z,[X,Y]]=ZXY-ZYX-XYZ+YXZ. \end{align*} These expansions are the whole mechanism behind the Jacobi identity: once every term is written as a signed triple product, the cancellation becomes a finite algebraic check. [/guided] [/step] [step:Cancel the signed triple products in the Jacobi sum] Adding the three expansions obtained above gives \begin{align*} [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]] \end{align*} equal to \begin{align*} XYZ-XZY-YZX+ZYX+YZX-YXZ-ZXY+XZY+ZXY-ZYX-XYZ+YXZ. \end{align*} Group the terms by their ordered triple products: \begin{align*} XYZ-XYZ=0. \end{align*} Also, \begin{align*} -XZY+XZY=0. \end{align*} Similarly, \begin{align*} -YZX+YZX=0. \end{align*} Moreover, \begin{align*} ZYX-ZYX=0. \end{align*} The remaining two cancellations are \begin{align*} -YXZ+YXZ=0 \end{align*} and \begin{align*} -ZXY+ZXY=0. \end{align*} Hence the whole sum is the zero matrix: \begin{align*} [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0. \end{align*} This proves the Jacobi identity and completes the proof. [/step]