Proof of The Ordered Eigenvalues of a Symmetric Matrix Are Determined by the Matrix

Fix $k\in[n]$. We show $\lambda_{k}\le\mu_{k}$; the reverse inequality then follows by symmetry.

First apply The Courant–Fischer Min–Max Theorem (i) to $A$ with the decomposition given by $Q$ and $\lambda$: the span $U_{k}=\operatorname{span}(Qe_{1},\dots,Qe_{k})$ is a subspace of $\mathbb{R}^{n}$ of dimension $k$, and $x\cdot(Ax)\ge\lambda_{k}$ for every unit vector $x\in U_{k}$.

Next apply The Courant–Fischer Min–Max Theorem (i) to $A$ with the decomposition given by $P$ and $\mu$, and to the subspace $U_{k}$ just obtained: since $\dim U_{k}=k$, it contains a unit vector $x$ with $x\cdot(Ax)\le\mu_{k}$. The previous paragraph applies to this $x$ as well, and so

$$\lambda_{k}\le x\cdot(Ax)\le\mu_{k}.$$

The hypotheses are symmetric in $(Q,\lambda)$ and $(P,\mu)$, so exchanging their roles in the two previous paragraphs gives $\mu_{k}\le\lambda_{k}$. Hence $\lambda_{k}=\mu_{k}$ for every $k\in[n]$, and $\lambda=\mu$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iii). This proves (i).