Proof of Action and Quadratic Form of a Block Matrix

Each result cited is universally quantified over the data in its own statement.

By Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space §index-split, $[m]\subseteq[m+n]$ and $m+i\in[m+n]$ for every $i\in[n]$, and every $k\in[m+n]$ satisfies exactly one of $k\in[m]$ and $k=m+i$ for some $i\in[n]$, this $i$ being unique. By Block Matrix with Two Row Blocks and Two Column Blocks §block, $M$ is the unique matrix of Existence and Uniqueness of the Block Matrix with Two Row Blocks and Two Column Blocks §existence; so $M\in\mathcal{M}_{m+n}(\mathbb{R})$ and, for $k,l\in[m]$ and $i,j\in[n]$,

$$M_{kl}=A_{kl},\qquad M_{k,\,m+j}=B_{kj},\qquad M_{m+i,\,l}=C_{il},\qquad M_{m+i,\,m+j}=D_{ij}.$$

Sums. Addition on $\mathbb{R}$ is associative and commutative, so Splitting an Operation over the Segment [m+n] into [m] and a Shifted Copy of [n] §interval-split, applied to $+$ on $\mathbb{R}$, gives for every $a:[m+n]\to\mathbb{R}$

$$\sum_{l=1}^{m+n}a_{l}=\sum_{l=1}^{m}a_{l}+\sum_{j=1}^{n}a_{m+j}, \tag{$\ast$}$$

in whichever reading the three sums from $1$ are taken, the first sum on the right being that of $a|_{[m]}$.

Claim 1 (action). Write $w=\iota(\xi,\eta)$. By Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space §concatenation, for every $u\in\mathbb{R}^m$ and $v\in\mathbb{R}^n$, $\iota(u,v)$ is the only point of $\mathbb{R}^{m+n}$ whose $k$th coordinate is $u_k$ for $k\in[m]$ and whose $(m+j)$th coordinate is $v_j$ for $j\in[n]$; in particular $w_l=\xi_l$ for $l\in[m]$ and $w_{m+j}=\eta_j$ for $j\in[n]$. Let $k\in[m+n]$. By Matrix-Vector Product §product and $(\ast)$,

$$(Mw)_k=\sum_{l=1}^{m+n}M_{kl}\,w_l=\sum_{l=1}^{m}M_{kl}\,\xi_l+\sum_{j=1}^{n}M_{k,\,m+j}\,\eta_j .$$

If $k\in[m]$, the block entries turn the right-hand side into $\sum_{l=1}^{m}A_{kl}\xi_l+\sum_{j=1}^{n}B_{kj}\eta_j=(A\xi)_k+(B\eta)_k$, again by Matrix-Vector Product §product, and this is $(A\xi+B\eta)_k$ by Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §operations. If $k=m+i$ with $i\in[n]$, they turn it into $\sum_{l=1}^{m}C_{il}\xi_l+\sum_{j=1}^{n}D_{ij}\eta_j=(C\xi)_i+(D\eta)_i=(C\xi+D\eta)_i$. Thus the point $Mw\in\mathbb{R}^{m+n}$ has $k$th coordinate $(A\xi+B\eta)_k$ for $k\in[m]$ and $(m+i)$th coordinate $(C\xi+D\eta)_i$ for $i\in[n]$. Since $A\xi+B\eta\in\mathbb{R}^m$ and $C\xi+D\eta\in\mathbb{R}^n$, the uniqueness just recalled gives $Mw=\iota(A\xi+B\eta,\ C\xi+D\eta)$.

Claim 2 (bilinear form). By claim 1 and Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space §dot-products,

$$\iota(\xi',\eta')\cdot\bigl(M\,\iota(\xi,\eta)\bigr)=\iota(\xi',\eta')\cdot\iota\bigl(A\xi+B\eta,\ C\xi+D\eta\bigr)=\xi'\cdot(A\xi+B\eta)+\eta'\cdot(C\xi+D\eta),$$

and expanding the two summands by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second gives $\bigl(\xi'\cdot(A\xi)+\xi'\cdot(B\eta)\bigr)+\bigl(\eta'\cdot(C\xi)+\eta'\cdot(D\eta)\bigr)$. By associativity of addition in $\mathbb{R}$, $(a+b)+(c+d)=((a+b)+c)+d$ for real $a,b,c,d$, so this equals $\xi'\cdot(A\xi)+\xi'\cdot(B\eta)+\eta'\cdot(C\xi)+\eta'\cdot(D\eta)$, read from the left.

Claim 3 (symmetry). By Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §membership, $M\in\mathcal{S}(m+n)$ if and only if $M_{kl}=M_{lk}$ for all $k,l\in[m+n]$. Splitting each of $k$ and $l$ as above, and using the block entries, the pairs $(k,l)$ with $k,l\in[m]$ give the equations $A_{kl}=A_{lk}$, the pairs $(m+i,m+j)$ with $i,j\in[n]$ give $D_{ij}=D_{ji}$, and the pairs $(k,m+j)$ and $(m+j,k)$ with $k\in[m]$, $j\in[n]$ both give $B_{kj}=C_{jk}$. Since these pairs are exactly the elements of $[m+n]\times[m+n]$, $M\in\mathcal{S}(m+n)$ holds if and only if

$$A_{kl}=A_{lk}\ (k,l\in[m]),\qquad D_{ij}=D_{ji}\ (i,j\in[n]),\qquad C_{jk}=B_{kj}\ (j\in[n],\ k\in[m]).$$

By Symmetric Real Matrices and the Set $\mathcal{S}(n)$ §membership the first family says $A\in\mathcal{S}(m)$ and the second says $D\in\mathcal{S}(n)$. By Transpose of a Real Matrix §transpose, $B^{\top}\in\mathcal{M}_{n\times m}(\mathbb{R})$ has entries $(B^{\top})_{jk}=B_{kj}$, so the third family says that $C$ and $B^{\top}$, both in $\mathcal{M}_{n\times m}(\mathbb{R})$, have the same entries, which is $C=B^{\top}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality.