Trace From Linearity and Cyclicity

Assumptions¶

• $M_n(\mathbb C)$ is the algebra of complex $n\times n$ matrices: a vector space under addition and scalar multiplication, and a (noncommutative) ring under matrix multiplication.

• $e_1,\dots,e_n$ is the standard basis of $\mathbb C^n$. The matrix unit $E_{ij}$ is the $n\times n$ matrix with a 1 in row $i$, column $j$, and 0 everywhere else. The $E_{ij}$ form a basis of $M_n(\mathbb C)$, and they multiply by

  $$E_{ij}E_{kl}=\delta_{jk}\,E_{il}.$$

  (1)

• $\operatorname{diag}(d_1,\dots,d_n)$ is the diagonal matrix with the listed entries.

We are given a function $f:M_n(\mathbb C)\to\mathbb C$ such that:

(L) Linearity. $f(A+B)=f(A)+f(B)$ and $f(cA)=c\,f(A)$ for all $A,B\in M_n(\mathbb C)$ and $c\in\mathbb C$.

(C) Cyclicity. $f(AB)=f(BA)$ for all $A,B\in M_n(\mathbb C)$.

The case $n=1$ is tautological: $M_1(\mathbb C)=\mathbb C$, (C) is automatic by commutativity, and (L) gives $f(a)=f(1)\,a$, a scalar multiple of the identity map. Hence assume $n\ge 2$ below.

We use only elementary linear algebra in the standard basis. We do not assume any prior trace theory, the identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ as a known fact, eigenvalues, the spectral theorem, or that $f$ is continuous. The coordinate sum, when it appears below, is introduced as an explicit formula and its needed properties are proved directly. This note is the additive companion of Determinant From Homomorphism: there the universal object is the multiplicative determinant on the group $GL(n,\mathbb C)$; here it is the additive trace on the whole algebra $M_n(\mathbb C)$.

Step 1 — Conjugation invariance¶

For any invertible $P\in GL(n,\mathbb C)$ and any $A\in M_n(\mathbb C)$, apply (C) to the pair $(PA,\,P^{-1})$:

So $f$ is invariant under conjugation. Only conjugation invariance by invertible diagonal and permutation matrices is used below — exactly the two families that drive Steps 2 and 3 of the determinant note; see the remark at the end on how this weaker hypothesis already suffices.

Step 2 — $f$ vanishes on every off-diagonal matrix unit¶

Fix $i\neq j$. For an invertible diagonal $D=\operatorname{diag}(d_1,\dots,d_n)$ one computes entrywise

that is,

This is the same diagonal-conjugation formula used on transvections in Step 2 of the determinant note (there $T_{ij}(c)=I+c\,E_{ij}$ becomes $T_{ij}(c\,d_i/d_j)$). By conjugation invariance (2) and homogeneity (L),

Since $i\neq j$ and $n\ge 2$, we may choose $D$ with $d_i\neq d_j$, so $d_i/d_j\neq 1$ and the relation $\big(1-\tfrac{d_i}{d_j}\big)f(E_{ij})=0$ forces

Step 3 — $f$ is constant on the diagonal matrix units¶

For $i\neq j$ let $P_{ij}=I-E_{ii}-E_{jj}+E_{ij}+E_{ji}$ be the permutation matrix that swaps $e_i\leftrightarrow e_j$ and fixes every other basis vector. Then $P_{ij}^2=I$, so $P_{ij}\in GL(n,\mathbb C)$ with $P_{ij}^{-1}=P_{ij}$. Conjugating a diagonal matrix unit just relabels the two basis vectors, so it carries $E_{ii}=e_ie_i^{\mathsf T}$ to $e_je_j^{\mathsf T}=E_{jj}$:

By conjugation invariance (2), $f(E_{ii})=f(E_{jj})$. Hence all $n$ diagonal units share one common value

The component formula and the factorization $f=g\circ\operatorname{tr}$¶

Everything so far (Steps 1–3) used only that $f$ is linear (L) and cyclic (C). Expanding any $A=\sum_{i,j}a_{ij}E_{ij}$ by linearity and inserting (6) and (8),

So every linear cyclic functional is a fixed scalar times the sum of the diagonal entries. Introduce that sum as an explicit, everywhere-defined functional

single-valued by construction. Then (9) reads $f=\lambda\,\operatorname{tr}$. Three facts complete the picture, in parallel with the determinant note’s existence/uniqueness split.

1. Existence. The coordinate sum (10) is manifestly linear (each $a_{ii}$ is linear in $A$), and it is cyclic by a direct index computation valid for all matrices — no invertibility, no domain caveat:

Thus a linear cyclic functional with the normalization $\lambda=1$ exists; (11) is the concrete fact underlying the abstract hypothesis (C).

2. Uniqueness. Equation (9) pins the value of $f$ at every $A$ to $\lambda\sum_i a_{ii}$, where the single number $\lambda=f(E_{11})$ is read off from $f$ on one diagonal matrix unit (Step 3). So a linear cyclic functional has no freedom beyond that one value: two linear cyclic functionals $M_n(\mathbb C)\to\mathbb C$ that agree on a single diagonal matrix unit agree everywhere.

3. Factorization. Write $g:\mathbb C\to\mathbb C$ for the unique $\mathbb C$-linear map with $g(1)=\lambda$, namely $g(t)=\lambda t$. Since $\operatorname{tr}$ is surjective onto $\mathbb C$, equation (9) is exactly

In particular, suppose $f$ satisfies the normalization $f(E_{ii})=1$ for the diagonal units, i.e. $\lambda=1$ (equivalently $f(I)=n$). Then $g=\operatorname{id}$ and (9) reads $f=\operatorname{tr}$. Combined with existence, this characterizes the trace:

$\operatorname{tr}$ is the unique linear cyclic functional $M_n(\mathbb C)\to\mathbb C$ whose value on every diagonal matrix unit $E_{ii}$ is 1 (equivalently, with $\operatorname{tr}(I)=n$).

For a general $f$, this is the factorization theorem:

for a single free scalar $\lambda\in\mathbb C$ — derived from linearity and cyclicity alone, with no continuity, eigenvalues, or spectral theory.

Conversely, every $\lambda\,\operatorname{tr}$ is a linear cyclic functional: it is linear because $\operatorname{tr}$ is, and cyclic by (11). Hence these are exactly all of them: the trace is the universal linear cyclic functional, and every other is obtained by post-composing it with a $\mathbb C$-linear endomorphism $g$ of $\mathbb C$ — that is, multiplication by a scalar.

Remarks¶

Conjugation invariance already suffices. Steps 2 and 3 used only that $f$ is invariant under conjugation by invertible diagonal and permutation matrices, together with linearity. So the conclusion $f=\lambda\,\operatorname{tr}$ already follows from (L) plus conjugation invariance, a hypothesis weaker than full cyclicity (C). Since $\lambda\,\operatorname{tr}$ is in turn cyclic by (11), the two families

coincide. Step 1 is what bridges the stated hypothesis (C) to the conjugation invariance actually used, exactly as Step 1 of the determinant note bridges multiplicativity to conjugation invariance.

Direct route from cyclicity. One can also bypass conjugation entirely and read both vanishing facts straight off the matrix-unit products $E_{ij}E_{kl}=\delta_{jk}E_{il}$. For $i\neq j$, from $E_{ij}E_{jj}=E_{ij}$ and $E_{jj}E_{ij}=0$ (here $\delta_{ji}=0$), cyclicity (C) gives

For the diagonal, $E_{ij}E_{ji}=E_{ii}$ and $E_{ji}E_{ij}=E_{jj}$, so cyclicity gives

These reproduce (6) and (8) in two lines, after which (9) finishes as before. The conjugation argument in Steps 1–3 is kept as the spine because it parallels the determinant note tool for tool: diagonal conjugation (4) and permutation matrices.

Parallel with the determinant. The free factor $g$ here — a $\mathbb C$-linear map $\mathbb C\to\mathbb C$, i.e. multiplication by the scalar $\lambda$ — is the additive analog of the free homomorphism $g\in\operatorname{Hom}(\mathbb C^*,\mathbb C^*)$ in Determinant From Homomorphism. There, post-composition is with a multiplicative homomorphism of $\mathbb C^*$ and the universal object is $\det$ on $GL(n,\mathbb C)$; here, post-composition is with a linear endomorphism of $(\mathbb C,+)$ and the universal object is $\operatorname{tr}$ on $M_n(\mathbb C)$. In both cases the structural hypothesis (multiplicativity there, linearity-plus-cyclicity here) determines the invariant up to that single free post-composition. If the homogeneity half of (L) is dropped and only additivity is kept, $g$ may be any additive endomorphism of $(\mathbb C,+)$, a solution of Cauchy’s additive functional equation — pinned down to multiplication by a scalar once continuity or measurability is imposed, exactly as in Homomorphisms $\mathbb C^*\to\mathbb C^*$.