What information does the trace of a matrix give? I was recently thinking what information do we get from a matrix. So if we say the columns (or rows) of a matrix define the basis of a system, say vectors of 3 dimensional space. Then the determinant will tell about the volume of the space enclosed by those vectors. Does trace give any information about the space?
 A: @Qmechanic's answer is perfect as always :) I'd just like to give an (obviously) incomplete list of examples in which the trace of an operator is more manifestly related to some physical insight about the system.

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*Since the trace of an operator remains invariant under a change of basis, it gives you the sum of the eigenvalues as already pointed out. When the sum of the eigenvalues of an operator has direct physical significance, the trace of the operator becomes more manifestly physically significant. For example, the eigenvalues of a density matrix give you the probabilities of the system being in one of the eigenstates of the density operator. Since the summation of such probabilities should always be unity, the trace of a density matrix is always unity.


*The trace of the commutator of operators should always be zero for operators defined over a finite-dimensional vector space due to the famous property $\text{Tr}(AB)=\text{Tr}(BA)$. Thus, when such operators show up whose commutator manifestly cannot be traceless, you recognize that such operators cannot live in a finite-dimensional vector space. In particular, you can be in a situation where you know the commutation relations between two operators before knowing the vector space that they live in when symmetry considerations give you the commutation relations between the generators of a certain symmetry group. If the commutator of the generators cannot be traceless, you realize that you cannot obtain a finite-dimensional representation of that symmetry group. The famous example is the commutation relation $[\hat{x},\hat{p}]=i\hbar$ which tells you that the momentum and the position operators must live in an infinite-dimensional Hilbert space.


*Due to the property that the commutator of operators should always be zero for operators defined over a finite-dimensional vector space, and due to the property that an element of a Lie algebra of a symmetry group is proportional the commutator of two other elements of the Lie algebra, a generator of a compact symmetry group in its faithful representation is traceless. This means that the sum of the eigenvalues of such observables which correspond to the generators of some compact symmetry group add up to zero. For example, the eigenvalues of $L_z, S_z$, etc.


*The process of "tracing out" certain degrees of freedom of a system is of direct physical significance. In particular, when you don't care (or don't have access to) certain degrees of freedom of a system, you sum over all possible values of those degrees of freedom and the resultant model describes an effective theory for the degrees of freedom that you care about (or have access to). In this sense, taking the trace of a subsystem (i.e., certain inaccessible/uninteresting degrees of freedom) represents the procedure of obtaining an effective theory for the rest of the system. This is how you arrive at a density matrix to describe a system that is entangled to some other system that you don't have an access to.
A: I) More generally, 


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*Let $V$ be a (say, finite dimensional) vector space over a field $\mathbb{F}$. 

*Let $(e_i)_{i\in I}$ be a basis for $V$.

*Let $A\in {\rm End}(V)$ be an endomorphism in $V$, i.e. a $\mathbb{F}$-linear map $A:V\to V$.

*Let the matrix $(M^i{}_j)_{i,j\in I}$ be the unique $\mathbb{F}$-valued matrix that represents the linear map $A$ in the basis $(e_i)_{i\in I}$, which means that
$$\forall j\in I:~~ Ae_j ~=~ \sum_{i\in I} e_iM^i{}_j .$$
Then the trace ${\rm tr(A)}$ of the linear map $A$, defined as 
$${\rm tr(A)}~:=~\sum_{i\in I}M^i{}_i,$$ 
does not depend on the choice of basis $(e_i)_{i\in I}$. For this reason it is an important invariant. 
II) Another invariant is the determinant $\det (A)$ of $A$. The physical interpretation of the determinant is the (signed) quotient (often called the Jacobian $J$) between the (signed) volume of a region in $V$ before and after applying the map $A$.
III) A physical interpretation of the trace can be provided via i) the physical interpretation of the determinant and ii) the formula
$$\det (e^A) ~=~e^{\rm tr(A)}. $$ 
Therefore, for an infinitesimal map $A$, 
$$\det ({\bf 1}+A)~\approx~1+ {\rm tr(A)}.$$
So roughly speaking, in one interpretation, the trace provides information about infinitesimal change in the volume factor. 
