Operator-Algebraic States and GNS Representations

By distributing finite sums the displayed form is linear in its first variable, and conjugating the sum gives $⟨x,y⟩=\overline{⟨y,x⟩}$. Also $⟨x,x⟩={\sum }_j\mathrm{\mid }{x}_j{\mathrm{\mid }}^2$ is nonnegative and zero only for $x=0$, by [F6]. Thus it satisfies the complex inner-product conditions [F1]. The matrix units $E_{ij}$ form a basis of the complex vector space $M_n(\mathbb{C})$. Matrix multiplication is bilinear by its coordinate formula, has unit $I_n$, and is associative because both $(a(bc){)}_{ij}$ and $((ab)c{)}_{ij}$ equal ${\sum }_{k,l}{a}_{ik}{b}_{kl}{c}_{lj}$. Entrywise conjugation and transposition make $a\mapsto a^{\ast }$ conjugate-linear and give $(a^{\ast }{)}^{\ast }=a$ and $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$: the $(i,j)$ entry of $(ab{)}^{\ast }$ is $\overline{{\sum }_k{a}_{jk}{b}_{ki}}={\sum }_k\overline{b_{ki}}\overline{a_{jk}}$, which is the $(i,j)$ entry of $b^{\ast }{a}^{\ast }$ by [F6]. Also $⟨ax,y⟩=⟨x,a^{\ast }y⟩$ by expanding the finite coordinate sums using [F1] and [F6]. The coordinate formula makes $x\mapsto ax$ complex-linear. For each row $i$, apply [F3] to $(a_{i1},\dots ,a_{in})$ and $(\overline{x_1},\dots ,\overline{x_n})$ to get $\mathrm{\mid }{\sum }_j{a}_{ij}{x}_j{\mathrm{\mid }}^2\le ({\sum }_j\mathrm{\mid }{a}_{ij}{\mathrm{\mid }}^2)\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, so $\mathrm{\parallel }ax{\mathrm{\parallel }}_2^2\le ({\sum }_{i,j}\mathrm{\mid }{a}_{ij}{\mathrm{\mid }}^2)\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and every matrix action is bounded. Conversely, every linear map on $H_n$ is represented by its columns in this basis. Hence $M_n(\mathbb{C})=\mathcal{B}(H_n)$.

For $a,b\in M_n(\mathbb{C})$ and $\mathrm{\parallel }x{\mathrm{\parallel }}_2\le 1$, rescaling the unit-ball definition [F2] gives $\mathrm{\parallel }abx{\mathrm{\parallel }}_2\le \mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }bx{\mathrm{\parallel }}_2\le \mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }$, so taking the supremum yields $\mathrm{\parallel }ab\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }$. For any matrix $a$, [F3] and the adjoint identity in step 1.1 show ${\sup }_{\mathrm{\parallel }x{\mathrm{\parallel }}_2,\mathrm{\parallel }y{\mathrm{\parallel }}_2\le 1}\mathrm{\mid }⟨ax,y⟩\mathrm{\mid }=\mathrm{\parallel }a\mathrm{\parallel }=\mathrm{\parallel }{a}^{\ast }\mathrm{\parallel }$: the inner supremum over $y$ equals $\mathrm{\parallel }ax{\mathrm{\parallel }}_2$ (choose $y=ax\mathrm{/}\mathrm{\parallel }ax{\mathrm{\parallel }}_2$ if $ax\mathrm{\ne }0$, which has unit norm by [F7]), and the adjoint identity swaps the two unit-ball suprema. Therefore $\mathrm{\parallel }{a}^{\ast }a\mathrm{\parallel }\le \mathrm{\parallel }{a}^{\ast }\mathrm{\parallel }\mathrm{\parallel }a\mathrm{\parallel }=\mathrm{\parallel }a{\mathrm{\parallel }}^2$. For every unit vector $x$, $\mathrm{\parallel }ax{\mathrm{\parallel }}_2^2=⟨ax,ax⟩=⟨x,a^{\ast }ax⟩\le \mathrm{\parallel }{a}^{\ast }ax{\mathrm{\parallel }}_2\le \mathrm{\parallel }{a}^{\ast }a\mathrm{\parallel }$ by [F1] and [F3]; taking the supremum over the nonempty unit sphere gives the reverse inequality $\mathrm{\parallel }a{\mathrm{\parallel }}^2\le \mathrm{\parallel }{a}^{\ast }a\mathrm{\parallel }$. Thus the C*-identity holds.

The operator norm is nonnegative by [F2]. The vector norm is absolutely homogeneous by [F7], so the unit-ball formula gives $\mathrm{\parallel }\lambda a\mathrm{\parallel }=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }a\mathrm{\parallel }$; its triangle inequality [F7] gives $\mathrm{\parallel }(a+b)x{\mathrm{\parallel }}_2\le \mathrm{\parallel }ax{\mathrm{\parallel }}_2+\mathrm{\parallel }bx{\mathrm{\parallel }}_2$ for every unit-ball vector $x$, hence $\mathrm{\parallel }a+b\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }+\mathrm{\parallel }b\mathrm{\parallel }$. If $\mathrm{\parallel }a\mathrm{\parallel }=0$, then each basis vector $e_j$ has $\mathrm{\parallel }a{e}_j{\mathrm{\parallel }}_2=0$, so $a{e}_j=0$ by [F7]; all columns vanish and $a=0$. Conversely, $a=0$ has norm zero by the definition. Thus the operator norm is a norm. The basis $(E_{ij}{)}_{1\le i,j\le n}$ is finite, so [F4] makes $M_n(\mathbb{C})$ complete for this norm. Steps 1.1 and 2.1 supply associativity, the unit, the involution, submultiplicativity, and the C*-identity; these are exactly the clauses in [F5]. Since $n\ge 1$, $I_n\mathrm{\ne }0$, so the algebra is unital under the library's convention.