Proof of The Trace of a Symmetric Matrix Is the Sum of Its Ordered Eigenvalues

Routine facts about the matrix product and the trace are used without comment (Toolkit for AL3).

Let $\lambda=(\lambda_{1}(X),\dots,\lambda_{n}(X))$. By The Spectral Theorem for Real Symmetric Matrices (i) there are $U\in\mathrm{O}(n)$ and $\mu\in\mathbb{R}^{n}$ with $\mu_{1}\ge\dots\ge\mu_{n}$ and $X=(U\operatorname{diag}(\mu))U^{\top}$. By The Ordered Eigenvalues of a Symmetric Matrix (i), $\lambda$ is the non-increasing tuple of any such decomposition; so $\mu=\lambda$ and

$$X=(U\operatorname{diag}(\lambda))U^{\top}.$$

We now compute the trace. The cyclicity of the trace (Basic Properties of the Trace (iii)) moves $U^{\top}$ to the front, the associativity of the matrix product (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (i)) regroups the factors, the orthogonality of $U$ gives $U^{\top}U=I_{n}$, and $I_{n}$ is neutral for the product by Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (iv):

$$\operatorname{tr}(X)=\operatorname{tr}\bigl(U^{\top}(U\operatorname{diag}(\lambda))\bigr)=\operatorname{tr}\bigl((U^{\top}U)\operatorname{diag}(\lambda)\bigr)=\operatorname{tr}\bigl(\operatorname{diag}(\lambda)\bigr).$$

Finally, the diagonal entries of $\operatorname{diag}(\lambda)$ are $\lambda_{1},\dots,\lambda_{n}$ (Diagonal Matrix with Prescribed Diagonal (i)), so the trace formula gives

$$\operatorname{tr}\bigl(\operatorname{diag}(\lambda)\bigr)=\sum_{k=1}^{n}\lambda_{k}(X).$$

Together with the previous display this is (i).