Proof of Block Diagonal Symmetric Matrices and the Blocks of a Symmetric Matrix

Throughout we use the notation of the statement. We record first that for any real $p\times q$ matrix all of whose entries are $0$ and any $v\in\mathbb{R}^{q}$ one has $0_{p\times q}v=0_{\mathbb{R}^{p}}$, since by Matrix-Vector Product the $i$th coordinate of $0_{p\times q}v$ is $\sum_{j=1}^{q}0\cdot v_{j}$, a finite sum all of whose summands are $0$ and hence equal to $0$ by claim 7 of Properties of Finite Sums; and that $\zeta\cdot 0_{\mathbb{R}^{p}}=0$ for every $\zeta\in\mathbb{R}^{p}$, by the same argument applied to the coordinate description of the dot product.

Claim 1. By Block Matrix with Two Row Blocks and Two Column Blocks, the entries of $X\oplus Y$ are $(X\oplus Y)_{kl}=X_{kl}$ for $k,l\in[m]$, $(X\oplus Y)_{k,m+j}=0$ for $k\in[m]$ and $j\in[n]$, $(X\oplus Y)_{m+i,l}=0$ for $i\in[n]$ and $l\in[m]$, and $(X\oplus Y)_{m+i,m+j}=Y_{ij}$ for $i,j\in[n]$. Let $k,l\in[N]$. By the trichotomy recorded in the statement, each of $k$ and $l$ either lies in $[m]$ or has the form $m+i$ with $i\in[n]$, and exactly one of these holds; in each of the four resulting cases the entry in position $(k,l)$ equals the entry in position $(l,k)$, using $X_{kl}=X_{lk}$ in the first case, $Y_{ij}=Y_{ji}$ in the fourth, and $0=0$ in the two mixed cases. Hence $X\oplus Y$ is symmetric and lies in $\mathcal{S}(N)$.

The entries of $(X\oplus Y)-(X'\oplus Y')$ are the differences of the corresponding entries, by Difference of Real Matrices; comparing the four cases above with the corresponding entries of $(X-X')\oplus(Y-Y')$, and using $0-0=0$ in the mixed cases, gives equality of the two matrices. Similarly, the $(k,l)$ entry of $(aI_{m})\oplus(aI_{n})$ is, in the four cases, $a(I_{m})_{kl}$, $0$, $0$ and $a(I_{n})_{ij}$; by Identity Matrix and Standard Basis Vectors this equals $a$ when $k=l$ and $0$ otherwise, since in the two mixed cases $k$ and $l$ are distinct by the trichotomy, and since $k=l$ forces $k,l$ to lie in the same one of the two ranges. That is exactly the $(k,l)$ entry of $aI_{N}$.

Claim 2. Apply claim 2 of Action and Quadratic Form of a Block Matrix with $A=X$, $B=0_{m\times n}$, $C=0_{n\times m}$, $D=Y$ and with $\xi'=\xi$, $\eta'=\eta$:

$$\iota(\xi,\eta)\cdot\bigl((X\oplus Y)\iota(\xi,\eta)\bigr)=\xi\cdot(X\xi)+\xi\cdot(0_{m\times n}\eta)+\eta\cdot(0_{n\times m}\xi)+\eta\cdot(Y\eta),$$

and the two middle terms vanish by the facts recorded at the start.

Claim 3. Since $\le$ is a total order on $\mathbb{R}$, one of $\lVert X\rVert$ and $\lVert Y\rVert$ is at least the other; let $\lambda$ be the larger. Let $w\in\mathbb{R}^{N}$ with $\lVert w\rVert\le1$ and write $w=\iota(\xi,\eta)$, which is possible in exactly one way since $\iota$ is a bijection. By claim 3 of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space, $\lVert w\rVert^{2}=\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}$. By claim 2 above, claim 5 of Properties of the Absolute Value in an Ordered Field and claim 2 of Properties of the Norm of a Symmetric Real Matrix,

$$\bigl|w\cdot\bigl((X\oplus Y)w\bigr)\bigr|\le\lVert X\rVert\,\lVert\xi\rVert^{2}+\lVert Y\rVert\,\lVert\eta\rVert^{2}\le\lambda\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)=\lambda\lVert w\rVert^{2}\le\lambda,$$

using $\lVert w\rVert^{2}\le1$, which follows from $\lVert w\rVert\le1$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Hence $\lambda$ is an upper bound for the set of Norm of a Symmetric Real Matrix and $\lVert X\oplus Y\rVert\le\lambda$.

Conversely, let $\xi\in\mathbb{R}^{m}$ with $\lVert\xi\rVert\le1$ and put $w=\iota(\xi,0_{\mathbb{R}^{n}})$. Then $\lVert w\rVert^{2}=\lVert\xi\rVert^{2}\le1$, so $\lVert w\rVert\le1$, and claim 2 gives $w\cdot\bigl((X\oplus Y)w\bigr)=\xi\cdot(X\xi)+0_{\mathbb{R}^{n}}\cdot(Y0_{\mathbb{R}^{n}})=\xi\cdot(X\xi)$. Hence $|\xi\cdot(X\xi)|\le\lVert X\oplus Y\rVert$ for every such $\xi$, and therefore $\lVert X\rVert\le\lVert X\oplus Y\rVert$. The same argument with $w=\iota(0_{\mathbb{R}^{m}},\eta)$ gives $\lVert Y\rVert\le\lVert X\oplus Y\rVert$, so $\lambda\le\lVert X\oplus Y\rVert$ and the two bounds give $\lVert X\oplus Y\rVert=\lambda$.

For the distance, claim 1 gives $(X\oplus Y)-(X'\oplus Y')=(X-X')\oplus(Y-Y')$, so by Distance Between Symmetric Real Matrices and what has just been proved, $d_{\mathcal{S}(N)}(X\oplus Y,X'\oplus Y')=\lVert(X-X')\oplus(Y-Y')\rVert$ is the larger of $\lVert X-X'\rVert=d_{\mathcal{S}(m)}(X,X')$ and $\lVert Y-Y'\rVert=d_{\mathcal{S}(n)}(Y,Y')$.

For the last assertion, note that a real number is smaller than $\varepsilon$ if the larger of it and a second number is, and that the larger of two numbers is smaller than $\varepsilon$ exactly when both are. Suppose $(X_{k}\oplus Y_{k})$ converges to $X\oplus Y$ and let $\varepsilon$ be positive; taking $N$ from Convergent Sequence in a Metric Space we get, for $k\ge N$, that the larger of $d_{\mathcal{S}(m)}(X_{k},X)$ and $d_{\mathcal{S}(n)}(Y_{k},Y)$ is smaller than $\varepsilon$, hence so is each of them; so both sequences converge. Conversely, if both converge, choose $N_{1}$ and $N_{2}$ for $\varepsilon$ and put $N=N_{1}+N_{2}$, which is at least each of them; for $k\ge N$ both distances, and hence their larger, are smaller than $\varepsilon$.

Claim 4. By The Positive Semidefinite Ordering on Symmetric Matrices, $X\oplus Y\preceq X'\oplus Y'$ means that

$$\xi\cdot(X\xi)+\eta\cdot(Y\eta)\le\xi\cdot(X'\xi)+\eta\cdot(Y'\eta)$$

for all $\xi\in\mathbb{R}^{m}$ and $\eta\in\mathbb{R}^{n}$, by claim 2 and the fact that every $w\in\mathbb{R}^{N}$ is $\iota(\xi,\eta)$ for exactly one pair. If $X\preceq X'$ and $Y\preceq Y'$, adding the two defining inequalities gives the display. Conversely, taking $\eta=0_{\mathbb{R}^{n}}$ in the display and using $0_{\mathbb{R}^{n}}\cdot(Y0_{\mathbb{R}^{n}})=0=0_{\mathbb{R}^{n}}\cdot(Y'0_{\mathbb{R}^{n}})$ gives $\xi\cdot(X\xi)\le\xi\cdot(X'\xi)$ for all $\xi$, that is $X\preceq X'$; taking $\xi=0_{\mathbb{R}^{m}}$ gives $Y\preceq Y'$.

Claim 5. For $i,j\in[m]$ we have $(Z^{11})_{ij}=Z_{ij}=Z_{ji}=(Z^{11})_{ji}$, so $Z^{11}\in\mathcal{S}(m)$; likewise $(Z^{22})_{ij}=Z_{m+i,m+j}=Z_{m+j,m+i}=(Z^{22})_{ji}$, so $Z^{22}\in\mathcal{S}(n)$.

Let $W$ be the block matrix determined by $Z^{11}$, $Z^{12}$, $(Z^{12})^{\top}$ and $Z^{22}$. By Block Matrix with Two Row Blocks and Two Column Blocks its entries are $W_{kl}=(Z^{11})_{kl}=Z_{kl}$ for $k,l\in[m]$; $W_{k,m+j}=(Z^{12})_{kj}=Z_{k,m+j}$; $W_{m+i,l}=\bigl((Z^{12})^{\top}\bigr)_{il}=(Z^{12})_{li}=Z_{l,m+i}=Z_{m+i,l}$, the last step by symmetry of $Z$ and the middle step by Transpose of a Real Matrix; and $W_{m+i,m+j}=(Z^{22})_{ij}=Z_{m+i,m+j}$. By the trichotomy every pair of indices in $[N]$ falls under exactly one of these four cases, so $W=Z$.

If every entry of $Z^{12}$ is $0$ then $Z^{12}=0_{m\times n}$ and, by Transpose of a Real Matrix, $(Z^{12})^{\top}=0_{n\times m}$, so $Z=W=Z^{11}\oplus Z^{22}$. Conversely if $Z=Z^{11}\oplus Z^{22}$ then, comparing entries in position $(i,m+j)$ using the entry description in the proof of claim 1, $(Z^{12})_{ij}=Z_{i,m+j}=0$ for all $i\in[m]$ and $j\in[n]$.

Finally, applying claim 2 of Action and Quadratic Form of a Block Matrix to $W=Z$ with $\xi'=\xi$ and $\eta'=\eta$,

$$\iota(\xi,\eta)\cdot\bigl(Z\iota(\xi,\eta)\bigr)=\xi\cdot(Z^{11}\xi)+\xi\cdot(Z^{12}\eta)+\eta\cdot\bigl((Z^{12})^{\top}\xi\bigr)+\eta\cdot(Z^{22}\eta).$$

By claim 5 of Elementary Properties of the Transpose of a Real Matrix together with claim 1 of that lemma and claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$\eta\cdot\bigl((Z^{12})^{\top}\xi\bigr)=\Bigl(\bigl((Z^{12})^{\top}\bigr)^{\top}\eta\Bigr)\cdot\xi=(Z^{12}\eta)\cdot\xi=\xi\cdot(Z^{12}\eta),$$

so the two middle terms are equal and their sum is $2\,\xi\cdot(Z^{12}\eta)$.