Finite-Dimensional Complex Inner-Product Spaces and Matrix Interfaces

The adjoint $T^{\ast }$ exists by [F3]. The definition [F1] gives $⟨x\mathrm{\mid }T\mathrm{\mid }y{⟩}_D=⟨Ty,x{⟩}_{K,\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}$.

Apply [F2] with $v=y$ and $w=x$, then apply [F1] again. This gives $⟨Ty,x{⟩}_{K,\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=⟨y,T^{\ast }x{⟩}_{H,\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=⟨T^{\ast }x\mathrm{\mid }y{⟩}_D$, proving the stated chain.

By [F1], $z={\sum }_{j<n}⟨z,e_j⟩e_j$. Applying $T$ and using linearity, then expanding each $T{e}_j$ by [F2], gives $Tz={\sum }_{j<n}{\sum }_{i<m}{T}_{ij}⟨z,e_j⟩f_i$.

Write $x_i=⟨x,f_i⟩$ and $y_j=⟨y,e_j⟩$, the coordinates from [F1]–[F2]. The image of $e_j$ under $\mathrm{\mid }x⟩⟨y\mathrm{\mid }$ is $⟨e_j,y⟩x=\overline{y_j}x$ by [F1] and [F7]. Its $i$-th coordinate is therefore $x_i\overline{y_j}$. By [F4]–[F6], these are exactly the entries of $[x{]}_F[y{]}_E^{†}$.

The amplitude identity is the defining Dirac conversion [F3].

By [F3], the operator sum in the Statement sends every $z$ to the right-hand side of step 1.1. It therefore agrees with $T$ on every vector.

If $n=0$, then $H=0$, $T=0$, and $y=0$; if $m=0$, then $K=0$, $T=0$, and $x=0$. In either case the displayed sums and rank-one map are zero, and the amplitude identity is $0=0$. Thus all formulas include zero dimensions.

Fix an ordered basis of $H$. For endomorphisms $R,S$ of $H$ and a scalar $\alpha$, [F9] shows that the diagonal entries of $\alpha R+S$ are $\alpha R_{jj}+S_{jj}$: the $j$-th coordinate column is obtained from $(\alpha R+S)e_j=\alpha R{e}_j+S{e}_j$. Summing these entries by [F4] gives ${\mathrm{tr}}_H(\alpha R+S)=\alpha {\mathrm{tr}}_H(R)+{\mathrm{tr}}_H(S)$. Thus trace homogeneity and additivity used below follow from the diagonal-sum definition. The same calculation covers $H=0$ by empty sums.

Let $C:H\to K$ be any finite-dimensional linear map. If $H=0$, its operator norm is $0$ by [F5]. If $H\mathrm{\ne }0$, choose an SVD as in [F2] with positive singular values $t_1\ge \cdots \ge t_q>0$ and orthonormal domain vectors $u_1,\dots ,u_{\dim H}$. If $q>0$, Bessel's inequality [F7] gives, for every unit vector $z$, $\mathrm{\parallel }Cz{\mathrm{\parallel }}^2={\sum }_{j\le q}{t}_j^2\mathrm{\mid }⟨z,u_j⟩{\mathrm{\mid }}^2\le t_1^2$, with equality at $z=u_1$. If $q=0$, the SVD says $C=0$, so every unit vector attains the value $0$. Thus the maximum in [F5] exists for every map used below, including the zero-rank case.

Choose an SVD as in [F2], with $r$ positive singular values. Then $\mathrm{\parallel }A{\mathrm{\parallel }}_1={\sum }_{i<r}{s}_i$ by [F1]. If $r=0$, [F2] gives $A=0$ and this sum is empty.

Write the SVD domain basis as $(e_j{)}_{j<n}$ and its positive singular values as $s_i$ for $i<r$. By [F4], [F7], and the fact that the $j$-th diagonal entry is the $e_j$-coordinate of the image of $e_j$, ${\mathrm{tr}}_H(B^{\ast }A)={\sum }_{j<n}⟨B^{\ast }A{e}_j,e_j{⟩}_H={\sum }_{i<r}{s}_i⟨B^{\ast }{f}_i,e_i{⟩}_H={\sum }_{i<r}{s}_i⟨f_i,B{e}_i{⟩}_K\mathrm{.}$ For the last equality, [F3] gives $⟨B{e}_i,f_i{⟩}_K=⟨e_i,B^{\ast }{f}_i{⟩}_H$, and conjugate symmetry [F10] gives the displayed order. This is a direct endomorphism-trace calculation and uses no cyclic trace identity for maps.

Suppose $r>0$ and define $B_0:H\to K$ by $B_{0}z={\sum }_{i<r}⟨z,e_i⟩f_i$. By [F10], this is linear; it sends $e_i$ to $f_i$ for $i<r$ and sends the remaining basis vectors to $0$. By [F7], $\mathrm{\parallel }{B}_{0}z{\mathrm{\parallel }}^2={\sum }_{i<r}\mathrm{\mid }⟨z,e_i⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }z{\mathrm{\parallel }}^2$, with equality at $z=e_0$. The maximum property established in step 1.2 therefore gives $\mathrm{\parallel }{B}_0\mathrm{\parallel }=1$. Define $Cw={\sum }_{i<r}⟨w,f_i⟩e_i$. Expanding in the finite orthonormal bases using [F7] verifies $⟨B_{0}z,w{⟩}_K=⟨z,Cw{⟩}_H$, so [F3] gives $C=B_0^{\ast }$. Hence $B_0^{\ast }A{e}_i=s_i{e}_i$ for $i<r$ and $B_0^{\ast }A{e}_j=0$ for $j\ge r$. If $r=0$, set $B_0=0$.

Let $B:H\to K$ satisfy $\mathrm{\parallel }B\mathrm{\parallel }\le 1$. By [F5] and the maximum property established in step 1.2, $\mathrm{\parallel }B{e}_i\mathrm{\parallel }\le 1$ for every unit vector $e_i$; [F6] then gives $\mathrm{\mid }⟨f_i,B{e}_i⟩\mathrm{\mid }\le 1$. Applying [F8] to the sum in step 1.4 yields $\mathrm{\mid }{\mathrm{tr}}_H(B^{\ast }A)\mathrm{\mid }\le {\sum }_{i<r}{s}_i=\mathrm{\parallel }A{\mathrm{\parallel }}_1$. The same bound is immediate when $r=0$.

By [F4], [F7], and step 1.3, ${\mathrm{tr}}_H(B_0^{\ast }A)={\sum }_{i<r}{s}_i=\mathrm{\parallel }A{\mathrm{\parallel }}_1$; when $r=0$, both sides are $0$. Together with step 2.2 this proves the stated maximum, including zero-dimensional spaces.

For any scalar $\alpha$, apply the maximum formula to $\alpha A$. By step 1.1, ${\mathrm{tr}}_H(B^{\ast }(\alpha A))=\alpha {\mathrm{tr}}_H(B^{\ast }A)$, so taking absolute values and maxima gives $\mathrm{\parallel }\alpha A{\mathrm{\parallel }}_1=\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\parallel }A{\mathrm{\parallel }}_1$. The definition and [F2] give $\mathrm{\parallel }A{\mathrm{\parallel }}_1\ge 0$, with equality exactly when all singular values vanish, which by the SVD is exactly $A=0$.

For maps $A,D:H\to K$, every contraction $B$ satisfies $\mathrm{\mid }{\mathrm{tr}}_H(B^{\ast }(A+D))\mathrm{\mid }\le \mathrm{\mid }{\mathrm{tr}}_H(B^{\ast }A)\mathrm{\mid }+\mathrm{\mid }{\mathrm{tr}}_H(B^{\ast }D)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_1+\mathrm{\parallel }D{\mathrm{\parallel }}_1$ by the additivity from step 1.1, [F8], and step 3.1. Taking the maximum over $B$ proves $\mathrm{\parallel }A+D{\mathrm{\parallel }}_1\le \mathrm{\parallel }A{\mathrm{\parallel }}_1+\mathrm{\parallel }D{\mathrm{\parallel }}_1$. Thus the trace norm has all stated norm properties.

If $x=0$ or $y=0$, [F1] makes $A=\mathrm{\mid }x⟩⟨y\mathrm{\mid }$ the zero map. Then [F6] gives $\mathrm{\parallel }A{\mathrm{\parallel }}_1=0$, while [F2] gives $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }=0$.

Suppose $x,y\mathrm{\ne }0$. For $z\in H$ and $w\in K$, [F1]–[F2] give $⟨Az,w{⟩}_K=⟨z,y{⟩}_H⟨x,w{⟩}_K=⟨z,⟨w,x{⟩}_{K}y{⟩}_H\mathrm{.}$ Thus the adjoint identity [F3] gives $A^{\ast }w=⟨w,x{⟩}_{K}y$, and therefore $A^{\ast }Az=\mathrm{\parallel }x{\mathrm{\parallel }}^2⟨z,y{⟩}_{H}y$.

Let $e=y\mathrm{/}\mathrm{\parallel }y\mathrm{\parallel }$ and put $\lambda =\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2$. Step 1.2 gives $A^{\ast }Ae=\lambda e$ and $A^{\ast }Az=0$ whenever $z\perp e$. Every $z\in H$ has the decomposition $z=⟨z,e⟩e+(z-⟨z,e⟩e)$, whose second term is orthogonal to $e$; the two summands intersect only at $0$. Therefore $H=\mathrm{span}\{e\}\oplus e^{\perp }$, and $A^{\ast }A$ acts by $\lambda$ on the first summand and by $0$ on the second.

Define $R$ on the direct sum $H=\mathrm{span}\{e\}\oplus e^{\perp }$ by $Re=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }e$ and $Rz=0$ for $z\perp e$, extending linearly. For $z,w\in H$, put $c=⟨z,e⟩$ and $d=⟨w,e⟩$. Then $⟨Rz,w⟩=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }c\bar{d}=⟨z,Rw⟩$, so $R$ is self-adjoint; its eigenvalues are $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ on $\mathrm{span}\{e\}$ and $0$ on $e^{\perp }$. These eigenvalues are nonnegative, so [F5] makes $R$ non-negative. It satisfies $R^2=A^{\ast }A$ by step 2.1. Uniqueness in [F5] gives $R=\mathrm{\mid }A\mathrm{\mid }=\sqrt{A^{\ast }A}$. Its eigenvalues are $\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ once and $0$ on $e^{\perp }$, so [F4] and [F6] give $\mathrm{\parallel }A{\mathrm{\parallel }}_1=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$. Together with step 1.1 this proves the formula, including zero vectors and zero-dimensional spaces.

The operator $\mathrm{\mid }A\mathrm{\mid }=\sqrt{A^{\ast }A}$ in [F1] is non-negative. By [F2], choose an orthonormal eigenbasis of $H$; its eigenvalues, with multiplicity, are exactly the singular values of $A$ by [F1]. In that basis [F3] gives ${\mathrm{tr}}_H(\mathrm{\mid }A\mathrm{\mid })={\sum }_{j<\dim H}{s}_j(A)$.

The sum in step 1.1 is $\mathrm{\parallel }A{\mathrm{\parallel }}_1$ by [F4]. If $H=0$, both are the empty sum $0$ by [F3]–[F4]. This proves the formula for rectangular maps and all finite dimensions.

Choose an SVD of $A$ as in [F1], with domain basis $(e_j)$ and codomain basis $(f_i)$. By [F2], $(V^{\ast }{e}_j)$ and $(U{f}_i)$ are orthonormal bases. For every $z\in H$, [F3] gives $UAVz={\sum }_{j<r}{s}_j⟨Vz,e_j⟩U{f}_j={\sum }_{j<r}{s}_j⟨z,V^{\ast }{e}_j⟩U{f}_j\mathrm{.}$ This is an SVD of $UAV$ with the same positive singular values, so [F1] gives $\mathrm{\parallel }UAV{\mathrm{\parallel }}_1=\mathrm{\parallel }A{\mathrm{\parallel }}_1$.

Let $P$ be non-negative. By [F4], choose an orthonormal eigenbasis $(e_j)$ with $P{e}_j={\lambda }_j{e}_j$ and ${\lambda }_j\ge 0$. Since $P^{\ast }=P$, $P^{\ast }P=P^2$. The operator $P$ is a non-negative square root of $P^2$, so uniqueness [F5] gives $\sqrt{P^{\ast }P}=P$. Its singular values are therefore the ${\lambda }_j$; by [F1] the trace norm is ${\sum }_j{\lambda }_j$.

Computing the trace in the same basis by [F6] gives ${\mathrm{tr}}_H(P)={\sum }_j{\lambda }_j=\mathrm{\parallel }P{\mathrm{\parallel }}_1$. If $H=0$, both sums are empty and equal $0$. This proves both assertions.

Put $\beta =\mathrm{\parallel }B\mathrm{\parallel }$. If $H=0$, [F2] defines $\beta =0$. If $H\mathrm{\ne }0$, choose an SVD of $B$ from [F7], with positive singular values $t_1\ge \cdots \ge t_q>0$ and orthonormal domain vectors $(u_j)$. If $q>0$, [F8] gives $\mathrm{\parallel }Bv{\mathrm{\parallel }}^2={\sum }_{j\le q}{t}_j^2\mathrm{\mid }⟨v,u_j⟩{\mathrm{\mid }}^2\le t_1^2$ for every unit vector $v$, with equality at $u_1$. If $q=0$, the SVD gives $B=0$, and every unit vector attains $0$. Thus the maximum in [F2] exists in all cases. When $\beta =0$, every unit vector has $Bv=0$; scaling and linearity give $B=0$ (also when $H=0$). By [F3], $B^{\ast }=0$, and then $B^{\ast }A=0$, whose trace is $0$ by [F4]; both sides of the inequality are $0$.

Suppose $\beta >0$ and put $C=B\mathrm{/}\beta$. For every unit vector $v$, $\mathrm{\parallel }Cv{\mathrm{\parallel }}^2=⟨{\beta }^{-1}Bv,{\beta }^{-1}Bv⟩={\beta }^{-2}\mathrm{\parallel }Bv{\mathrm{\parallel }}^2,$ so nonnegativity gives $\mathrm{\parallel }Cv\mathrm{\parallel }=\mathrm{\parallel }Bv\mathrm{\parallel }\mathrm{/}\beta \le 1$ by [F2]. Applying the argument in step 1.1 with $C$ in place of $B$ shows its defining maximum exists and is at most $1$, hence $\mathrm{\parallel }C\mathrm{\parallel }\le 1$. Trace-norm duality [F1] gives $\mathrm{\mid }{\mathrm{tr}}_H(C^{\ast }A)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_1$.

Since $\beta$ is a positive real, [F3] gives $C^{\ast }={\beta }^{-1}{B}^{\ast }$. Fix an ordered basis of $H$. The coordinate column of ${\beta }^{-1}(B^{\ast }A)e_j$ is ${\beta }^{-1}$ times the coordinate column of $(B^{\ast }A)e_j$, by uniqueness of coordinates and [F9]; hence their diagonal entries have the same scaling. Summing them by [F4] gives ${\mathrm{tr}}_H(C^{\ast }A)={\beta }^{-1}{\mathrm{tr}}_H(B^{\ast }A)$. Thus step 2.1 and modulus homogeneity [F6] yield ${\beta }^{-1}\mathrm{\mid }{\mathrm{tr}}_H(B^{\ast }A)\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_1$. Multiplying by $\beta$ proves the inequality when $\beta >0$; step 1.1 proves the $\beta =0$ case.

By [F1], choose an orthonormal eigenbasis $(e_j{)}_{j<n}$ with $X{e}_j={\lambda }_j{e}_j$. Self-adjointness and conjugate symmetry give ${\lambda }_j=⟨X{e}_j,e_j⟩=⟨e_j,X^{\ast }{e}_j⟩=⟨e_j,X{e}_j⟩=\overline{⟨X{e}_j,e_j⟩}$, so every ${\lambda }_j$ is real.

For $v\in \mathcal{H}$, define $X_{+}v={\sum }_{j<n}\max ({\lambda }_j,0)⟨v,e_j⟩e_j$ and $X_{-}v={\sum }_{j<n}\max (-{\lambda }_j,0)⟨v,e_j⟩e_j$. These are linear by the first-slot rule [F2] and the finite-sum definition [F9]. Their diagonal entries are real and nonnegative. If $v={\sum }_j{c}_j{e}_j$, then [F2] gives $⟨X_{+}v,v⟩={\sum }_j\max ({\lambda }_j,0)\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge 0$, and the same calculation applies to $X_{-}$. Thus [F3] makes both operators non-negative. Every $X_{+}u$ is a linear combination of basis vectors with ${\lambda }_j>0$, while every $X_{-}v$ is a linear combination of basis vectors with ${\lambda }_j<0$; these subsets of the orthonormal basis are orthogonal. Thus $⟨X_{+}u,X_{-}v⟩=0$ for all $u,v$. On every $e_j$, $(X_{+}-X_{-})e_j={\lambda }_j{e}_j=X{e}_j$, hence $X=X_{+}-X_{-}$.

In this basis $X^{\ast }X=X^2$ has eigenvalues ${\lambda }_j^2$. Define $R$ by $R{e}_j=\mathrm{\mid }{\lambda }_j\mathrm{\mid }{e}_j$. Its diagonal entries are nonnegative, so [F3] makes it non-negative, and $R^2=X^{\ast }X$; uniqueness [F7] therefore gives $R=\sqrt{X^{\ast }X}$. Thus [F4]–[F5] give $\mathrm{\parallel }X{\mathrm{\parallel }}_1={\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$, $\mathrm{tr}(X_{+})={\sum }_{{\lambda }_j>0}{\lambda }_j$, and $\mathrm{tr}(X_{-})={\sum }_{{\lambda }_j<0}(-{\lambda }_j)$. Their latter two sums add to ${\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$, proving the trace-norm identity.

If $\mathrm{tr}(X)=0$, [F5] gives ${\sum }_j{\lambda }_j=0$. Hence the sum of positive eigenvalues equals the sum of absolute values of negative eigenvalues. Step 3.1 then yields $\mathrm{\parallel }X{\mathrm{\parallel }}_1=2\mathrm{tr}(X_{+})=2\mathrm{tr}(X_{-})$. If $n=0$, all sums are empty and the identities still hold.