Show that $ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z) = ... $

Question

Let $X, Y, Z$ be $2 \times 2$ matrices. Show that these two matrix combinations are equal:

• $ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z) $

• $ \text{Tr}(X) \, \text{Tr}(YZ) + \text{Tr}(YX)\,\text{Tr}(Z)+ \text{Tr}(Z)\,\text{Tr}(XY) $

There's lots of identities for matrix trace, here's the only other one that I know, that we can switch the order of the matrix: $$ \text{Tr}(XYZ) = \text{Tr}(YZX) $$ I might specifically need to have $X,Y,Z \in \text{SL}_2(\mathbb{C})$. The right side looks symmetric under the cycle permutation of $(XYZ)$ (basically a triangle) while the left side does not.

Answer 1

(Edit: The result is true and can be derived from Cayley-Hamilton for $2\times 2$ matrices, and writing $det(A) = \frac{1}{2}[Tr(A)^2 - Tr(A^2)]$ for $A$ a $2\times 2$ matrix. See update 2 below.)

There is indeed a typo. We approach it brutally. Using what you have stated, SageMath (https://sagecell.sagemath.org/) gives the difference of your two expressions to be:

var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Z.trace()*(X*Y).trace())).expand()

---
output:
-x3*y2*z1 - x2*y3*z1 + x1*y4*z1 - x4*y4*z1 + x3*y1*z2 + x3*y4*z2 + x2*y1*z3 + x2*y4*z3 - x1*y1*z4 + x4*y1*z4 - x3*y2*z4 - x2*y3*z4

However, if we change the second expression to something more symmetric:

$$ \text{Tr}(X) \, \text{Tr}(YZ) + \text{Tr}(YX)\,\text{Tr}(Z)+ \color{blue}{ \text{Tr}(Y)\,\text{Tr}(XZ) }$$

Then the difference between this and $$ \text{Tr}(XYZ) + \text{Tr}(YXZ)+ \text{Tr}(X)\text{Tr}(Y)\text{Tr}(Z) $$ is

var('x1','x2','x3','x4','y1','y2','y3','y4','z1','z2','z3','z4')
X = matrix([[x1,x2],[x3,x4]])
Y = matrix([[y1,y2],[y3,y4]])
Z = matrix([[z1,z2],[z3,z4]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
0

(And thus a proof, albeit not enlightening.)

Remark. This identity is not true if $X,Y,Z$ are $3\times 3$ matrices:

var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
((X*Y*Z).trace() + (Y*X*Z).trace() + X.trace()*Y.trace()*Z.trace() - (X.trace()*(Y*Z).trace()+(Y*X).trace()*Z.trace()+Y.trace()*(X*Z).trace())).expand()

---
output:
x9*y5*z1 - x8*y6*z1 - x6*y8*z1 + x5*y9*z1 - x9*y4*z2 + x7*y6*z2 + x6*y7*z2 - x4*y9*z2 + x8*y4*z3 - x7*y5*z3 - x5*y7*z3 + x4*y8*z3 - x9*y2*z4 + x8*y3*z4 + x3*y8*z4 - x2*y9*z4 + x9*y1*z5 - x7*y3*z5 - x3*y7*z5 + x1*y9*z5 - x8*y1*z6 + x7*y2*z6 + x2*y7*z6 - x1*y8*z6 + x6*y2*z7 - x5*y3*z7 - x3*y5*z7 + x2*y6*z7 - x6*y1*z8 + x4*y3*z8 + x3*y4*z8 - x1*y6*z8 + x5*y1*z9 - x4*y2*z9 - x2*y4*z9 + x1*y5*z9

UPDATE.(1/20/2021) Upon some reflection, we see that there is some permutation cycle structure in these expressions. Let us denote $(M) := \text{Tr}(M)$ to simplify the notation. Then we do indeed have for four $3\times 3$ matrices $X,Y,Z,W$, the following equality $$ \sum_{\sigma\in \text{Perm}(X,Y,Z,W)}\text{sign}(\sigma) \sigma = 0 $$

where we interpret $\sigma$ first in cycle structure, and then interpret it as trace, (e.g., $\sigma = (X)(Y)(ZW) = \text{Tr}(X)\text{Tr}(Y)\text{Tr}(ZW)$). This is probably true for $n$ many matrices of size $k\times k$ where $k\le n-1$. The $n = 3$ case is this posted problem, and the $n=4$ case I verify it here:

var('x1','x2','x3','x4','x5','x6','x7','x8','x9','y1','y2','y3','y4','y5','y6','y7','y8','y9','z1','z2','z3','z4','z5','z6','z7','z8','z9','w1','w2','w3','w4','w5','w6','w7','w8','w9')
X = matrix([[x1,x2,x3],[x4,x5,x6],[x7,x8,x9]])
Y = matrix([[y1,y2,y3],[y4,y5,y6],[y7,y8,y9]])
Z = matrix([[z1,z2,z3],[z4,z5,z6],[z7,z8,z9]])
W = matrix([[w1,w2,w3],[w4,w5,w6],[w7,w8,w9]])

p4    = (X*Y*Z*W).trace() + (X*Y*W*Z).trace() + (X*Z*W*Y).trace()+ (X*Z*Y*W).trace()+ (X*W*Y*Z).trace() + (X*W*Z*Y).trace()
p1111 = X.trace()*Y.trace()*Z.trace()*W.trace()
p31   = (X.trace()*(Y*Z*W).trace() + X.trace()*(Y*W*Z).trace() 
       + Y.trace()*(X*Z*W).trace() + Y.trace()*(X*W*Z).trace()
       + Z.trace()*(X*Y*W).trace() + Z.trace()*(X*W*Y).trace()
       + W.trace()*(X*Y*Z).trace() + W.trace()*(X*Z*Y).trace())
p22   = (X*Y).trace()*(Z*W).trace() + (X*Z).trace()*(Y*W).trace() + (X*W).trace()*(Y*Z).trace()
p211  = ((X*Y).trace()*Z.trace()*W.trace() + (X*Z).trace()*Y.trace()*W.trace() + (X*W).trace()*Z.trace()*Y.trace() 
         +(Y*Z).trace()*X.trace()*W.trace() +(Y*W).trace()*X.trace()*Z.trace() + (W*Z).trace()*X.trace()*Y.trace())

(p1111 + p31 +p22 - p211-p4).expand()

---
output:
0

So a conjecture (at least to me): Let $X_1,\ldots,X_n$ be $n$ many $k\times k$ matrices over a commutative ring where $k\le n-1$, then $$ \sum_{\sigma\in \text{Perm}(X_1,\ldots,X_n)} \text{sign}(\sigma) \sigma = 0, $$

where we interpret $\sigma$ in cyclic decomposition and then as a product of traces as before. Spiritually this suggests there is a combinatorial sieve / inclusion-exclusion proof of this. It eludes me at this moment.

Update 2. Indeed this is true, and can be derived from Cayley-Hamilton. See Polynomial Identities and the Cayley-Hamilton Theorem by Formanek (he derives this particular one in Eqs (1)-(6)):

https://www.researchgate.net/publication/225640755_Polynomial_identities_and_the_Cayley-Hamilton_theorem