Mathematics · Glossary

What is Transpose; trace?

Also known as: trace

Definition 21.5 University Mathematics — Year 1 · Chapter 21 — Matrices

The transpose of A=(aij)Mn,pA = (a_{ij}) \in \mathcal{M}_{n,p} is AT=(aji)Mp,nA^{\mathsf T} = (a_{ji}) \in \mathcal{M}_{p,n}; it satisfies (AB)T=BTAT(AB)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T} and (AT)T=A(A^{\mathsf T})^{\mathsf T} = A. The trace of a square matrix is trA=iaii\operatorname{tr} A = \sum_i a_{ii}; it is linear, and

tr(AB)=tr(BA)(AMn,p, BMp,n).\operatorname{tr}(AB) = \operatorname{tr}(BA) \qquad (A \in \mathcal{M}_{n,p},\ B \in \mathcal{M}_{p,n}).

Examples

Example 21.6 (The trace at work)

The projection of Chapter 20 onto Vect(1,1)\operatorname{Vect}(1,1) along Vect(0,1)\operatorname{Vect}(0,1), p(x,y)=(x,x)p(x, y) = (x, x), has matrix A=(1010)A = \begin{pmatrix} 1 & 0\\ 1 & 0\end{pmatrix} in the canonical basis: indeed A2=AA^2 = A, and

trA=1=rkA,\operatorname{tr} A = 1 = \operatorname{rk} A ,

illustrating Exercise 21.8: for idempotents the trace counts the dimension of the image, whatever slanted basis the matrix is written in. The invariance mechanism is the identity tr(AB)=tr(BA)\operatorname{tr}(AB) = \operatorname{tr}(BA):

tr(P1(AP))=tr((AP)P1)=trA,\operatorname{tr}\bigl(P^{-1}(AP)\bigr) = \operatorname{tr}\bigl((AP)P^{-1}\bigr) = \operatorname{tr} A ,

so all matrices similar to AA share its trace — the first numerical invariant of an endomorphism, to be joined by the determinant in Chapter 22 (the pair (s,p)(s, p) of the weekend problem below).

Example 21.7 (Symmetric plus antisymmetric)

Call AA symmetric when AT=AA^{\mathsf T} = A, antisymmetric when AT=AA^{\mathsf T} = -A. Every square matrix splits uniquely as one plus the other:

A=A+AT2symmetric+AAT2antisymmetric,A = \underbrace{\frac{A + A^{\mathsf T}}{2}}_{\text{symmetric}} + \underbrace{\frac{A - A^{\mathsf T}}{2}}_{\text{antisymmetric}},

and a matrix that is both is zero (A=AA = -A): the two sets are supplementary subspaces of Mn(K)\mathcal{M}_n(K) — the exact analogue of the even/odd split of functions (Example 18.11), with transposition playing the role of xxx \mapsto -x. Dimensions: a symmetric matrix is free on and above the diagonal, an antisymmetric one strictly above (zero diagonal):

n(n+1)2+n(n1)2=n2,\frac{n(n+1)}{2} + \frac{n(n-1)}{2} = n^2 ,

and the count balancing is Grassmann’s confirmation of directness. For n=2n = 2: (1512)=(1332)+(0220)\begin{pmatrix} 1 & 5\\ 1 & 2\end{pmatrix} = \begin{pmatrix} 1 & 3\\ 3 & 2\end{pmatrix} + \begin{pmatrix} 0 & 2\\ -2 & 0\end{pmatrix}. Symmetric matrices return as the second-derivative data of Chapter 25 (the Monge triple r,s,tr, s, t), and the symmetric-orthogonal ones are classified in Exercise 23.12.

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