1
$\begingroup$

Let:

  • $n \in \mathbb{N}$
  • $A$ a matrix of size $(n,n)$
  • $e_k$ any orthonormal vector then : $$ \operatorname{Tr}A=\sum_{1 \leq k \leq n} e_k'Ae_k, $$ The result is stated on this page and a closed result here.

My attempt :

  • Let $\mathcal{B}=(e_1 \dots e_n)$ be the initial basis (orthonormal) and $f$ the endomorphism represented by $A$ in $\mathcal{B}$.
  • $(v_1 \dots v_n)$ any orthonormal basis, there exists $P$ orthonormal : $Pe_i=v_i$
  • Let $1 \leq j \leq n$

$ \begin{align*} f(e_j)&= \sum_{i=1}^{n} a_{i,j} \\ \langle f(e_j),(e_j)\rangle &=a_{j,j} \\ \sum_{j=1}^{n} \langle f(e_j),(e_j)\rangle &=\operatorname{Tr}A \\ \sum_{i=1}^{n} e_i' A e_i &=\operatorname{Tr}A\\ \operatorname{Tr}A&=\operatorname{Tr}(P'AP)=\sum_{i=1}^{n} e_i' P'AP e_i = \sum_{i=1}^{n} (Pe_i)'A(P e_i) = \sum_{i=1}^{n} v_i'Av_i \\ \end{align*} $

$\endgroup$
1
$\begingroup$

You didn't really formulate a specific question, but I'm assuming you're asking for help to show the identity in question.

Let $\{e_k\}_{k=1}^n\subseteq\mathbb{R}^n$ be an orthonormal set of vectors, and let $A\in\mathbb{R}^n$. Define the matrix \begin{equation*} U = \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix}. \end{equation*} Note that $U$ is an orthogonal matrix, i.e., $UU^\top = U^\top U = I_n$. Therefore, \begin{equation*} \text{tr}(A) = \text{tr}(AI_n) = \text{tr}(AUU^\top) = \text{tr}(U^\top A U) = \text{tr}\begin{bmatrix} e_1^\top \\ e_2^\top \\ \vdots \\ e_n^\top \end{bmatrix} A \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix} = \text{tr}\begin{bmatrix} e_1^\top A e_1 & e_1^\top A e_2 & \cdots & e_1^\top Ae_n \\ e_2^\top Ae_1 & e_2^\top A e_2 & \cdots & e_2^\top A e_n \\ \vdots & \vdots & \ddots & \vdots \\ e_n^\top Ae_1 & e_n^\top A e_2 & \cdots & e_n^\top A e_n \end{bmatrix} = \sum_{k=1}^n e_k^\top A e_k. \end{equation*}

$\endgroup$
1
  • $\begingroup$ Thanks. Is my demonstration correct ? $\endgroup$ – zestiria Aug 23 '20 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.