Proof of Action and Quadratic Form of a Block Matrix

Write $w=\iota(\xi,\eta)$, so that, by the description of the concatenation map in Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space, $w_l=\xi_l$ for $l\in[m]$ and $w_{m+j}=\eta_j$ for $j\in[n]$. Throughout, an index $k\in[m+n]$ is treated according to the alternative recorded in Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space: either $k\in[m]$, or $k=m+i$ for a unique $i\in[n]$.

Claim 1. Let $k\in[m+n]$. By Matrix-Vector Product and Splitting a Finite Sum at an Index,

If $k\in[m]$, the entries of the block matrix are $M_{kl}=A_{kl}$ and $M_{k,m+j}=B_{kj}$, so the right-hand side equals $(A\xi)_k+(B\eta)_k$, which is the $k$th coordinate of $A\xi+B\eta$ by Sum of Points of $\mathbb{R}^n$. If $k=m+i$ with $i\in[n]$, those entries are $M_{m+i,l}=C_{il}$ and $M_{m+i,m+j}=D_{ij}$, so the right-hand side equals $(C\xi)_i+(D\eta)_i$, the $i$th coordinate of $C\xi+D\eta$. Thus $Mw$ has $k$th coordinate $(A\xi+B\eta)_k$ for $k\in[m]$ and $(m+i)$th coordinate $(C\xi+D\eta)_i$ for $i\in[n]$, which is exactly the description of $\iota(A\xi+B\eta,\ C\xi+D\eta)$ in Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space.

Claim 2. By claim 1 and claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space,

and expanding each of the two summands by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ gives the stated identity.

Claim 3. By Transpose of a Real Matrix and Symmetric, Positive Semidefinite, and Positive Definite Real Matrices, $M$ is symmetric exactly when $M_{kl}=M_{lk}$ for all $k,l\in[m+n]$. Splitting each of $k$ and $l$ according to the alternative above, this system of equations is the conjunction of

the pairs $(k,m+j)$ and $(m+j,k)$ yielding the same third condition. The first family of equations says $A=A^{\top}$, the second says $D=D^{\top}$, and the third says that $C_{jk}=(B^{\top})_{jk}$ for all $j\in[n]$ and $k\in[m]$, that is, $C=B^{\top}$.