Gate-ECE- properties of trace

In linear algebra, the trace of an n-by-n square matrix A is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right) of A, i.e.,
where a_ii denotes the entry on the ith row and ith column of A. The trace of a matrix is the sum of the (complex) eigenvalues, and it is invariant with respect to a change of basis. This characterization can be used to define the trace of a linear operator in general. Note that the trace is only defined for a square matrix (i.e., n × n).

Properties

Basic properties

The trace is a linear mapping. That is,

${\displaystyle {\begin{aligned}\operatorname {tr} (A+B)&=\operatorname {tr} (A)+\operatorname {tr} (B)\\\operatorname {tr} (cA)&=c\operatorname {tr} (A)\end{aligned}}}$.

for all square matrices A and B, and all scalars c.

A matrix and its transpose have the same trace:

${\displaystyle \operatorname {tr} (A)=\operatorname {tr} \left(A^{\mathrm {T} }\right)}$.

This follows immediately from the fact that transposing a square matrix does not affect elements along the main diagonal.

Trace of a product

The trace of a product can be rewritten as the sum of entry-wise products of elements:

${\displaystyle \operatorname {tr} \left(X^{\mathrm {T} }Y\right)=\operatorname {tr} \left(XY^{\mathrm {T} }\right)=\operatorname {tr} \left(Y^{\mathrm {T} }X\right)=\operatorname {tr} \left(YX^{\mathrm {T} }\right)=\sum _{i,j}X_{ij}Y_{ij}}$.

This means that the trace of a product of matrices functions similarly to a dot product of vectors. For this reason, generalizations of vector operations to matrices (e.g. in matrix calculus and statistics) often involve a trace of matrix products.

For real matrices, the trace of a product can also be written in the following forms:

${\displaystyle \operatorname {tr} \left(X^{\mathrm {T} }Y\right)=\sum _{ij}(X\circ Y)_{ij}}$	(using the Hadamard product, i.e. entry-wise product).
${\displaystyle \operatorname {tr} \left(X^{\mathrm {T} }Y\right)=\operatorname {vec} (Y)^{\mathrm {T} }\operatorname {vec} (X)=\operatorname {vec} (X)^{\mathrm {T} }\operatorname {vec} (Y)}$	(using the vectorization operator).

The matrices in a trace of a product can be switched without changing the result: If A is an m × n matrix and B is an n × m matrix, then

${\displaystyle \operatorname {tr} (AB)=\operatorname {tr} (BA)}$.^[1]

More generally, the trace is invariant under cyclic permutations, i.e.,

${\displaystyle \operatorname {tr} (ABCD)=\operatorname {tr} (BCDA)=\operatorname {tr} (CDAB)=\operatorname {tr} (DABC)}$.

This is known as the cyclic property.

Note that arbitrary permutations are not allowed: in general,

${\displaystyle \operatorname {tr} (ABC)\neq \operatorname {tr} (ACB)}$.

However, if products of three symmetric matrices are considered, any permutation is allowed. (Proof: tr(ABC) = tr(A^T B^T C^T) = tr(A^T(CB)^T) = tr((CB)^TA^T) = tr((ACB)^T) = tr(ACB), where the last equality is because the traces of a matrix and its transpose are equal.) For more than three factors this is not true.

Unlike the determinant, the trace of the product is not the product of traces, that is:

${\displaystyle \operatorname {tr} (XY)\neq \operatorname {tr} (X)\operatorname {tr} (Y)}$

What is true is that the trace of the Kronecker product of two matrices is the product of their traces:

${\displaystyle \operatorname {tr} (X\otimes Y)=\operatorname {tr} (X)\operatorname {tr} (Y)}$.

Other properties

The following three properties:

${\displaystyle {\begin{aligned}\operatorname {tr} (A+B)&=\operatorname {tr} (A)+\operatorname {tr} (B)\\\operatorname {tr} (cA)&=c\cdot \operatorname {tr} (A)\\\operatorname {tr} (AB)&=\operatorname {tr} (BA)\end{aligned}}}$,

characterize the trace completely in the sense that follows. Let f be a linear functional on the space of square matrices satisfying f(x y) = f(y x). Then f and tr are proportional.^[2]

The trace is similarity-invariant, which means that A and P⁻¹AP have the same trace. This is because

${\displaystyle \operatorname {tr} \left(P^{-1}AP\right)=\operatorname {tr} \left(P^{-1}(AP)\right)=\operatorname {tr} \left((AP)P^{-1}\right)=\operatorname {tr} \left(A\left(PP^{-1}\right)\right)=\operatorname {tr} (A)}$.

If A is symmetric and B is antisymmetric, then

${\displaystyle \operatorname {tr} (AB)=0}$.

The trace of the identity matrix is the dimension of the space; this leads to generalizations of dimension using trace. The trace of an idempotent matrix A (for which A² = A) is the rank of A. The trace of a nilpotent matrix is zero.

More generally, if f(x) = (x − λ₁)^d₁···(x − λ_k)^d_k is the characteristic polynomial of a matrix A, then

${\displaystyle \operatorname {tr} (A)=d_{1}\lambda _{1}+\cdots +d_{k}\lambda _{k}}$.

When both A and B are n-by-n, the trace of the (ring-theoretic) commutator of A and B vanishes: tr([A, B]) = 0; one can state this as "the trace is a map of Lie algebras ${\displaystyle gl_{n}\to k}$ from operators to scalars", as the commutator of scalars is trivial (it is an abelian Lie algebra). In particular, using similarity invariance, it follows that the identity matrix is never similar to the commutator of any pair of matrices.

Conversely, any square matrix with zero trace is a linear combinations of the commutators of pairs of matrices.^[3] Moreover, any square matrix with zero trace is unitarily equivalent to a square matrix with diagonal consisting of all zeros.

The trace of any power of a nilpotent matrix is zero. When the characteristic of the base field is zero, the converse also holds: if ${\displaystyle \operatorname {tr} \left(x^{k}\right)=0}$ for all ${\displaystyle k}$, then ${\displaystyle x}$ is nilpotent.

The trace of a Hermitian matrix is real, because the elements on the diagonal are real.

The trace of a projection matrix is the dimension of the target space.

${\displaystyle {\begin{aligned}P_{X}&=X\left(X^{\mathrm {T} }X\right)^{-1}X^{\mathrm {T} }\\\Rightarrow \operatorname {tr} \left(P_{X}\right)&=\operatorname {rank} \left(X\right)\end{aligned}}}$.

Note that ${\displaystyle P_{X}}$ is idempotent, and more generally the trace of any idempotent matrix equals its rank.