Proof of Testing a Block Semidefinite Inequality on the Diagonal

Let $\xi$ be a point of Euclidean space $\mathbb{R}^n$, a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, with the dot product; write $P\zeta$ for the matrix-vector product and set $z=\iota(\xi,\xi)\in\mathbb{R}^{n+n}$, where $\iota$ is the concatenation map of Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space.

First, $0\,t=0$ for every $t\in\mathbb{R}$, since $0\,t=(0+0)\,t=0\,t+0\,t$ by distributivity and claim 2 of Additive Cancellation and Elementary Additive Identities in a Field applies; consequently $t+(-1)\,t=(1+(-1))\,t=0$, so $(-1)\,t=-t$ by claim 1 of that lemma. In particular $0_n=0\,I_n$ by Scalar Multiple of a Real Matrix and Identity Matrix.

By claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, $0_n\xi=(0\,I_n)\xi=0\,(I_n\xi)=0\,\xi$, and by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$, $\xi\cdot(0\,\xi)=0\,(\xi\cdot\xi)=0$. By the same two claims, $\xi\cdot\bigl((\beta I_n)\xi\bigr)=\beta\,(\xi\cdot\xi)$ for every $\beta\in\mathbb{R}$, and

$$\xi\cdot\bigl((-Y)\xi\bigr)=\xi\cdot\bigl((-1)(Y\xi)\bigr)=(-1)\bigl(\xi\cdot(Y\xi)\bigr)=-\,\xi\cdot(Y\xi).$$

Applying claim 2 of Action and Quadratic Form of a Block Matrix to $M$, with both arguments of $\iota$ equal to $\xi$ on each side,

$$z\cdot(Mz)=\xi\cdot(X\xi)+\xi\cdot(0_n\xi)+\xi\cdot(0_n\xi)+\xi\cdot\bigl((-Y)\xi\bigr)=\xi\cdot(X\xi)-\xi\cdot(Y\xi).$$

Applying the same claim to $N$, and writing $c=\xi\cdot\xi$,

$$z\cdot(Nz)=\alpha\,c+(-\alpha)\,c+(-\alpha)\,c+\alpha\,c=0,$$

since $(-\alpha)\,c=-(\alpha\,c)$ by the identity $(-1)\,t=-t$ together with associativity of multiplication.

The hypothesis $M\preceq N$, applied to the point $z$, gives $z\cdot(Mz)\le z\cdot(Nz)=0$, that is

$$\xi\cdot(X\xi)-\xi\cdot(Y\xi)\le0 .$$

By claim 3 of Elementary Arithmetic in an Ordered Field, together with claims 4 and 6 of Additive Cancellation and Elementary Additive Identities in a Field, this is equivalent to $0\le\xi\cdot(Y\xi)-\xi\cdot(X\xi)$ and hence, by claim 3 of Elementary Arithmetic in an Ordered Field again, to

$$\xi\cdot(X\xi)\le\xi\cdot(Y\xi).$$

Since $\xi\in\mathbb{R}^n$ was arbitrary, $X\preceq Y$ by The Positive Semidefinite Ordering on Symmetric Matrices.