Determinant of a finite-dimensional matrix in terms of trace

Question

I have noticed that for the case of $1 \times 1$, $2 \times 2$ and $3 \times 3$ matrices $A$, $B$, I can write the determinant of their commutator $C=[A,B]$ in terms of traces:

$1 \times 1$ matrices $A$, $B$: $$\det(C)=\text{tr}(C)$$

$2 \times 2$ matrices $A$, $B$: $$\det(C)=-\frac{1}{2}\text{tr}(C^2)$$

$3 \times 3$ matrices $A$, $B$: $$\det(C)=\frac{1}{3}\text{tr}(C^3)$$

But I can't find a simple formula for $4 \times 4$ matrices--I have no idea how to generalize this general finite-dimensional matrices:

I can't quite understand the origin of this pattern for low-dimensional matrices.

So my question is can this be generalized to:

1. Larger $n\times n$ matrices, $n>3$?
2. Anticommutators $C=\{A,B\} = AB+BA$?

Answer 1

If you regard the determinant as a product of its eigenvalues, you can view this as the elementary symmetric function $e_n(x_1,x_2,\dots,x_n)$.

Your traces of powers of $C$ correspond to power symmetric functions $p_k=x_1^k+\dots+x_n^k$.

The condition on $C$ being a commutator gives you the condition that $C$ is traceless, that is, $p_1=e_1=0$.

So, what you want is to express an elementary symmetric function with power symmetric functions under the condition $p_1=0$.

Now, if you look at the first few expansions http://en.wikipedia.org/wiki/Newton%27s_identities#Expressing_elementary_symmetric_polynomials_in_terms_of_power_sums you see that the first few expansions only contain one term without $p_1$, so they are exceptionally simple.

However for $n=4$, you get:

$e_4= \frac 1{24} p_1^4 -\frac14 p_1^2p_2 +\frac 18 p_2^2+\frac13 p_1p_3 -\frac14 p_4$

The terms with $p_1$ disappear and we can conclude:

$$\det C = \frac 18 \operatorname{tr} (C^2)^2 -\frac14 \operatorname{tr}(C^4).$$

This kind of expansion exists for any matrix size, but it gets more and more complicated.

The general formula is:

$$e_n=\sum_k\sum_{\lambda_1\ge \lambda_2 \ge \dots \ge \lambda_k \text{ and } \lambda_1+\dots+\lambda_k=n} \hskip-1.5cm (-1)^{n-k} \frac{p_{\lambda_1}p_{\lambda_2} \dots p_{\lambda_k}}{\lambda_1 \lambda_2 \dots \lambda_k (\text{no. of $\lambda_i=1$})!\text{(no. of $\lambda_i=2$})! \dots}$$

Now eliminate all terms that contain a $p_1$, replace all $p_k$ with $\operatorname{tr}(C^k)$ and you get your formula for $n$.

Answer 2

The relations come from the fact that $\det(C)$ and ${\rm tr}(C^n)$ are symetric polynomials, so can be expressed with the elementary symetric polynomials. If $(x_1,\dots,x_n)$ are the eigen values of $C$ then $e_n(x_1,\dots,x_n)=\det(C)$, and $e_1(x_1,\dots,x_n)={\rm tr}(C)$.

For the other $e_k$, you can either write $e_k= \rm{tr}(\Lambda^k C)$, where $\Lambda^k C$ is the action of of $C$ on the exterior product. Or you can use $\rm{tr}(\otimes^\ell C)$ to express $e_k$.

For example if $n=4$, then one has $${\rm tr}(C^4) = e_1^4 -4e_1^2e_2 + 2e_2^2+4e_1e_3-4e_4.$$ Using the fact that $e_1={\rm tr}(C)=0$, $e_2=\rm{tr}(\Lambda^2 C)=\rm{tr}(\otimes^2 C)-\rm{tr}(C^2)$, one gets $${\rm tr}(C^4) = -4 \det(C) +2\rm{tr}(\Lambda^2 C)^2 = -4 \det(C) +2 \Big( {\rm tr}(\otimes^2 C)-\rm{tr}(C^2) \Big)^2.$$

EDIT: You can define $\Lambda^2 C$ in the following way: the matrix of $\Lambda^2 C$ is a $\binom{n}{2} \times \binom{n}{2}$-matrix whose coefficients are the determinants of all $2 \times 2$ submatrices. More precisely for $\Lambda^2 C$ has rows and column indexed by $(i_1,i_2)$ with $1 \leq i_1 < i_2 \leq n$, and $$[\Lambda^2 C]_{(i_1,i_2),(j_1,j_2)} := \det \begin{pmatrix} C_{i_1,j_1} & C_{i_1,j_2} \\ C_{i_2,j_1} & C_{i_2,j_2} \end{pmatrix}.$$ For $\otimes^2 C$, it is a $n^2 \times n^2$-matrix whose rows and columns are indexed by $(i_1,i_2)$ with $1 \leq i_1 \leq i_2 \leq n$ and $[\otimes^2 C]_{(i_1,i_2),(j_1,j_2)} = C_{i_1,j_1}.C_{i_2,j_2}$. Hence ${\rm tr}(\otimes^2 C) = \sum_{1\leq k,l \leq 2} C_{k,k} C_{l,l}$.