Proof of Compression of a Bounded Symmetric Bilinear Form along an Orthonormal Tuple

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement of the lemma. The properties of $\Lambda$ and $\Lambda^{\sharp}$ recorded in Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (i) are used throughout: $\Lambda\Lambda^{\sharp}\zeta=\zeta$, $\zeta\cdot\Lambda x=\langle\Lambda^{\sharp}\zeta,x\rangle$ and $|\Lambda^{\sharp}\zeta|=\lVert\zeta\rVert$ for all $\zeta\in\mathbb{R}^{m}$ and $x\in H$, and $\Lambda^{\sharp}\zeta$ is the finite sum $\sum_{i=1}^{m}\zeta_{i}e_{i}$ in $H$ fixed in that lemma. Two matrices of the same size are equal when their entries agree, a matrix being a family of entries by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations.

Claim 1 (clause 1). Let $b\in\mathrm{Sym}(H)$. Its values satisfy $b(e_{i},e_{j})=b(e_{j},e_{i})$ for all $i,j\in[m]$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i), so the entries of $b^{\flat}$ are unchanged when the two indices are interchanged, and $b^{\flat}$ is therefore symmetric, that is $b^{\flat}\in\mathcal{S}(m)$.

Let $T_{b}:H\to H$ be the operator representing $b$, so that $b(x,y)=\langle T_{b}x,y\rangle$ for all $x,y\in H$. Fix $\zeta,\eta\in\mathbb{R}^{m}$. By Matrix-Vector Product the $i$th coordinate of $b^{\flat}\eta$ is $\sum_{j=1}^{m}b(e_{i},e_{j})\,\eta_{j}$, so by Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$

$$\zeta\cdot\bigl(b^{\flat}\eta\bigr)=\sum_{i=1}^{m}\zeta_{i}\Bigl(\sum_{j=1}^{m}b(e_{i},e_{j})\,\eta_{j}\Bigr).$$

Fix $i\in[m]$. The terms of the inner sum are $b(e_{i},e_{j})\eta_{j}=\eta_{j}\langle T_{b}e_{i},e_{j}\rangle$, by the representation and the commutativity of multiplication in $\mathbb{R}$, so Inner Products Against Finite Sums, and Orthonormal Families, in a Real Inner Product Space (ii), read from right to left with $T_{b}e_{i}$ in the first argument, gives

$$\sum_{j=1}^{m}b(e_{i},e_{j})\,\eta_{j} =\Bigl\langle T_{b}e_{i},\ \sum_{j=1}^{m}\eta_{j}e_{j}\Bigr\rangle =\bigl\langle T_{b}e_{i},\,\Lambda^{\sharp}\eta\bigr\rangle =b\bigl(e_{i},\Lambda^{\sharp}\eta\bigr) =b\bigl(\Lambda^{\sharp}\eta,e_{i}\bigr) =\bigl\langle T_{b}\Lambda^{\sharp}\eta,\,e_{i}\bigr\rangle ,$$

the last three steps by the representation, by the symmetry of $b$ (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (i)) and by the representation again. Substituting, and applying Inner Products Against Finite Sums, and Orthonormal Families, in a Real Inner Product Space (ii) once more in the same way, now with $T_{b}\Lambda^{\sharp}\eta$ in the first argument and the coefficients $\zeta_{i}$,

$$\zeta\cdot\bigl(b^{\flat}\eta\bigr) =\sum_{i=1}^{m}\zeta_{i}\bigl\langle T_{b}\Lambda^{\sharp}\eta,\,e_{i}\bigr\rangle =\Bigl\langle T_{b}\Lambda^{\sharp}\eta,\ \sum_{i=1}^{m}\zeta_{i}e_{i}\Bigr\rangle =\bigl\langle T_{b}\Lambda^{\sharp}\eta,\,\Lambda^{\sharp}\zeta\bigr\rangle =b\bigl(\Lambda^{\sharp}\eta,\Lambda^{\sharp}\zeta\bigr) =b\bigl(\Lambda^{\sharp}\zeta,\Lambda^{\sharp}\eta\bigr),$$

the last step again by the symmetry of $b$. This proves clause 1.

Claim 2 (clause 2). Let $b,b'\in\mathrm{Sym}(H)$ and $\lambda\in\mathbb{R}$, and let $i,j\in[m]$. By Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv) the form $b+b'$ has the value $b(e_{i},e_{j})+b'(e_{i},e_{j})$ at $(e_{i},e_{j})$, which by Sum of Real Matrices is the $(i,j)$ entry of $b^{\flat}+(b')^{\flat}$; hence $(b+b')^{\flat}=b^{\flat}+(b')^{\flat}$. Likewise $(\lambda b)(e_{i},e_{j})=\lambda\,b(e_{i},e_{j})$ is the $(i,j)$ entry of $\lambda\,b^{\flat}$ by Scalar Multiple of a Real Matrix, so $(\lambda b)^{\flat}=\lambda\,b^{\flat}$.

For the norm, let $\xi\in\mathbb{R}^{m}$ satisfy $\lVert\xi\rVert\le1$. The norm $\lVert b\rVert$ is a greatest lower bound of a set of nonnegative reals bounded below by $0$ (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (ii)), hence $0\le\lVert b\rVert$ by Lower Bound and Greatest Lower Bound in a Totally Ordered Set. By clause 1 and Elementary Properties of Bounded Symmetric Bilinear Forms: Norm, Quadratic Form, Order and Continuity (ii),

$$\bigl|\xi\cdot(b^{\flat}\xi)\bigr|=\bigl|b(\Lambda^{\sharp}\xi,\Lambda^{\sharp}\xi)\bigr|\le\lVert b\rVert\,\bigl|\Lambda^{\sharp}\xi\bigr|\,\bigl|\Lambda^{\sharp}\xi\bigr|=\lVert b\rVert\,\lVert\xi\rVert\,\lVert\xi\rVert\le\lVert b\rVert ,$$

the last step by claim 5 of Elementary Arithmetic in an Ordered Field applied with the nonnegative multiplier $\lVert b\rVert$ to the inequality $\lVert\xi\rVert\lVert\xi\rVert\le1$, which holds by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field since $\lVert\xi\rVert$ and $1$ are nonnegative and $\lVert\xi\rVert\le1$. Thus $\lVert b\rVert$ is an upper bound of the set whose least upper bound is $\lVert b^{\flat}\rVert$ by Norm of a Symmetric Real Matrix; a least upper bound is at most every upper bound of its set by Upper Bound and Least Upper Bound, so $\lVert b^{\flat}\rVert\le\lVert b\rVert$.

For the order, suppose $b\preceq b'$, that is $b(x,x)\le b'(x,x)$ for every $x\in H$ (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iii)). For every $\zeta\in\mathbb{R}^{m}$, clause 1 applied twice gives

$$\zeta\cdot\bigl(b^{\flat}\zeta\bigr)=b\bigl(\Lambda^{\sharp}\zeta,\Lambda^{\sharp}\zeta\bigr)\le b'\bigl(\Lambda^{\sharp}\zeta,\Lambda^{\sharp}\zeta\bigr)=\zeta\cdot\bigl((b')^{\flat}\zeta\bigr),$$

which is $b^{\flat}\preceq(b')^{\flat}$ by The Positive Semidefinite Ordering on Symmetric Matrices. This proves clause 2.

Claim 3 (clause 3). Let $\zeta\in\mathbb{R}^{m}$. The identity form has $I(x,y)=\langle x,y\rangle$ (Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv)), so clause 1 gives

$$\zeta\cdot\bigl(I^{\flat}\zeta\bigr)=\bigl\langle\Lambda^{\sharp}\zeta,\Lambda^{\sharp}\zeta\bigr\rangle=\zeta\cdot\Lambda\Lambda^{\sharp}\zeta=\zeta\cdot\zeta ,$$

the second equality by $\zeta\cdot\Lambda x=\langle\Lambda^{\sharp}\zeta,x\rangle$ with $x=\Lambda^{\sharp}\zeta$ and the third by $\Lambda\Lambda^{\sharp}\zeta=\zeta$. Also $\zeta\cdot(I_{m}\zeta)=\zeta\cdot\zeta$ by claim 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum. Hence $\zeta\cdot(I^{\flat}\zeta)\le\zeta\cdot(I_{m}\zeta)$ and $\zeta\cdot(I_{m}\zeta)\le\zeta\cdot(I^{\flat}\zeta)$ for every $\zeta$, so $I^{\flat}\preceq I_{m}$ and $I_{m}\preceq I^{\flat}$ by The Positive Semidefinite Ordering on Symmetric Matrices, and $I^{\flat}=I_{m}$ by claim 6 of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure.

By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (iii) the projection form satisfies $\Pi(z,w)=\Lambda z\cdot\Lambda w$, so clause 1 gives $\zeta\cdot(\Pi^{\flat}\zeta)=\Lambda\Lambda^{\sharp}\zeta\cdot\Lambda\Lambda^{\sharp}\zeta=\zeta\cdot\zeta$, and the argument just given yields $\Pi^{\flat}=I_{m}$.

By the same clause $N=I-\Pi$, that is $N=I+(-1)\Pi$ by Bounded Symmetric Bilinear Forms on a Real Inner Product Space: Norm, Order, Identity Form and Restriction (iv), so clause 2 gives $N^{\flat}=I^{\flat}+(-1)\Pi^{\flat}=I_{m}-I_{m}$, whose entries are $\delta_{ij}-\delta_{ij}=0$ by Identity Matrix and Standard Basis Vectors and Difference of Real Matrices; hence $N^{\flat}=0_{m}$. Finally, for $b\in\mathrm{Sym}(H)$ and $t\in\mathbb{R}$, clause 2 gives $(b+t\,N)^{\flat}=b^{\flat}+t\,N^{\flat}=b^{\flat}+t\,0_{m}$, and the entries of $t\,0_{m}$ are $t\cdot0=0$ by Scalar Multiple of a Real Matrix and claim 1 of Zero Products and Elementary Identities in a Field, so adding it leaves every entry of $b^{\flat}$ unchanged and $(b+t\,N)^{\flat}=b^{\flat}$. This proves clause 3.

Claim 4 (clause 4). Let $M\in\mathcal{S}(m)$ and $\zeta\in\mathbb{R}^{m}$. By Coordinate Maps of a Finite Orthonormal Tuple: Forms, the Tail Form, and Quadratic Test Functions (ii) the form $M^{\Lambda}$ has the value $\Lambda z\cdot(M\,\Lambda w)$ at $(z,w)$, so clause 1 gives

$$\zeta\cdot\bigl((M^{\Lambda})^{\flat}\zeta\bigr)=M^{\Lambda}\bigl(\Lambda^{\sharp}\zeta,\Lambda^{\sharp}\zeta\bigr)=\Lambda\Lambda^{\sharp}\zeta\cdot\bigl(M\,\Lambda\Lambda^{\sharp}\zeta\bigr)=\zeta\cdot(M\zeta).$$

As in Claim 3 this gives $(M^{\Lambda})^{\flat}\preceq M$ and $M\preceq(M^{\Lambda})^{\flat}$ by The Positive Semidefinite Ordering on Symmetric Matrices, hence $(M^{\Lambda})^{\flat}=M$ by claim 6 of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure. This proves clause 4 and completes the proof of the lemma.