# $\operatorname{Tr}A=\sum_ke_k'Ae_k,$ where $e_k$ are any orthonormal vectors.

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*}

## 1 Answer

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, $$\begin{equation*} \text{tr}(A) = \text{tr}(AI_n) = \text{tr}(AUU^\top) = \text{tr}(U^\top A U) = \text{tr}\begin{bmatrix} e_1^\top \\ e_2^\top \\ \vdots \\ e_n^\top \end{bmatrix} A \begin{bmatrix} e_1 & e_2 & \cdots & e_n \end{bmatrix} = \text{tr}\begin{bmatrix} e_1^\top A e_1 & e_1^\top A e_2 & \cdots & e_1^\top Ae_n \\ e_2^\top Ae_1 & e_2^\top A e_2 & \cdots & e_2^\top A e_n \\ \vdots & \vdots & \ddots & \vdots \\ e_n^\top Ae_1 & e_n^\top A e_2 & \cdots & e_n^\top A e_n \end{bmatrix} = \sum_{k=1}^n e_k^\top A e_k. \end{equation*}$$

• Thanks. Is my demonstration correct ? – zestiria Aug 23 '20 at 7:28