Ivan Shishkin, Rye (1878)

Problems/Linear algebraExerciseUnreviewed

Trace d’une transposée

by darktoaster·
25
Difficulty scaleÉchelle de difficulté

This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 1–10First steps / middle schoolPremiers pas / collège
  2. 11–25Beginner / high schoolDébutant / lycée
  3. 26–50Intermediate / undergraduateIntermédiaire / licence
  4. 51–70Advanced / graduateAvancé / master
  5. 71–90Expert / specializedExpert / spécialisé
  6. 91–100Research levelNiveau recherche
These levels are approximate guides.Ces niveaux sont des repères approximatifs.
·
Français

Showing the Français version because no English translation exists yet. Add that translation.

Unreviewed. This problem has not been reviewed by trusted users yet.

Soit A∈Mn(R)A \in \mathcal{M}_{n}(\R). Montrer que la trace est invariante par transposition, c’est-à-dire : tr(AT)=tr(A)\mathrm{tr}(A^{T}) = \mathrm{tr}(A)

I solved itMark it doneAdd to my listKeep it in your list

Solutions

1
Reveal solutionsAre you sure? Give it a try first.

Solution by darktoaster

Discussions0 useful votes

En une phrase : on ne change pas la diagonale d’une matrice en la transposant, donc le calcul de la trace reste le même.
\newline
Soit i∈[ ⁣[1,n] ⁣]i \in [\![1,n]\!]
[AT]i,i=[A]i,i[A^{T}]_{i,i} = [A]_{i,i}
Les coefficients AA et ATA^{T} ont les mêmes coefficients diagonaux.
\newline
D’où tr(A)=tr(AT)\mathrm{tr}(A) = \mathrm{tr}(A^{T})

Report

For an unclear, ambiguous, or possibly incorrect statement, please use the Discussion tab on the right. Report content that needs moderator intervention, such as dangerous, clearly non-mathematical, or plagiarized content.