The form is linear in the second argument because \(\omega \) is.
It is conjugate-linear in the first because
\((\lambda x)^*=\overline \lambda \,x^*\).
It is Hermitian because \(\langle y,x\rangle \) equals
\(\omega ((x^*y)^*)\), and Hermitian positivity makes that the
conjugate of \(\omega (x^*y)\). The square
\(\langle x,x\rangle =\omega (x^*x)\) is nonnegative by the state.
The inequality of Cauchy and Schwarz is
\(\lvert \langle x,y\rangle \rvert ^2\le \langle x,x\rangle \langle y,y\rangle \).
Over \(\mathbb C\), if \(\langle x,y\rangle \) is not zero, choose a
scalar \(u\) of modulus \(1\) with
\(u\langle x,y\rangle =\lvert \langle x,y\rangle \rvert \), and replace
\(y\) by \(uy\). The new square \(\langle y,y\rangle \) is unchanged, and
the new inner product is real and nonnegative. Over \(\mathbb R\) no
phase is required. For every real \(t\),
\[\langle x+ty,x+ty\rangle
=\langle x,x\rangle +2t\langle x,y\rangle +t^2\langle y,y\rangle \ge 0.\]
If \(\langle y,y\rangle =0\), the values \(2t\langle x,y\rangle \) cannot
stay above \(- \langle x,x\rangle \) for every real \(t\) unless
\(\langle x,y\rangle =0\). If \(\langle y,y\rangle >0\), complete the
square. The same quadratic is
\[\langle y,y\rangle \Bigl (t+\frac {\langle x,y\rangle }{\langle y,y\rangle }\Bigr )^2
+\langle x,x\rangle -\frac {\langle x,y\rangle ^2}{\langle y,y\rangle }.\]
The squared term can be made zero, so the remaining term is
nonnegative. Thus the inequality holds. Over \(\mathbb C\), putting
the unimodular factor back into \(y\) multiplies \(\langle x,y\rangle \)
by a scalar of modulus \(1\) and does not change \(\langle y,y\rangle \).
If \(\langle x,x\rangle =0\), the inequality forces
\(\langle x,y\rangle =0\) for every \(y\). Conversely the case \(y=x\)
shows that the radical is exactly \(N\). It is a subspace. If
\(x,y\in N\) and \(\lambda \in K\), then
\[\begin {aligned}
\langle x+\lambda y,x+\lambda y\rangle
={}&\langle x,x\rangle +\lambda \langle x,y\rangle \\
&+\overline {\lambda }\langle y,x\rangle
+\lvert \lambda \rvert ^2\langle y,y\rangle .
\end {aligned}\]
The two squares vanish because \(x,y\in N\). The two cross terms
vanish because an element of \(N\) is orthogonal to every vector.
It is a left ideal. For \(b\in A\) and \(x\in N\),
\(\langle bx,bx\rangle =\omega (x^*b^*bx)=\langle x,(b^*b)x\rangle =0\).
The form therefore descends to the quotient. If \(x-x'\) and
\(y-y'\) lie in \(N\), then
\(\langle x,y\rangle -\langle x',y'\rangle
=\langle x-x',y\rangle +\langle x',y-y'\rangle \).
The first summand vanishes because \(x-x'\) is in the radical. The
second is the conjugate of \(\langle y-y',x'\rangle \), which vanishes
for the same reason. The descended form is positive definite, and
finite dimensionality makes the quotient a Hilbert space with no
further completion. Its dimension is \(\dim A-\dim N\).
On a basis \(e_1,\ldots ,e_n\), the matrix
\(G_{ij}=\langle e_i,e_j\rangle =\omega (e_i^*e_j)\) is Hermitian. A
vector \(v\) represents \(x=\sum _i v_i e_i\). Hermitian symmetry gives
\(G_{ji}=\overline {G_{ij}}\). The number
\(\langle x,e_j\rangle =\sum _i\overline {v_i}G_{ij}\)
vanishes for every \(j\) if and only if \(x\in N\): the forward
direction was proved for every vector, and the converse is linearity
in the second argument. Conjugating those
sums and replacing \(\overline {G_{ij}}\) by \(G_{ji}\) produces
\((Gv)_j=0\). Rank-nullity
gives \(\operatorname {rank} G=n-\dim N=\dim H_\omega \). The class of
\(a\) has squared length \(\omega (a^*a)\). The operator
\(\pi _\omega (a)\) is well defined because \(N\) is a left ideal, and
\(\langle \pi _\omega (a)\Omega ,\pi _\omega (b)\Omega \rangle
=\langle a,b\rangle =\omega (a^*b)\).
On the direct sum, the product and the involution act blockwise, so
\(\omega (a^*a)=\sum _i\mathrm {Tr}(\rho _i a_i^*a_i)\). The trace is
invariant under a cyclic permutation, and
\(\mathrm {Tr}(b^*b)\) is the sum of the squared moduli of the entries
of \(b\). Hence
\[\mathrm {Tr}(\rho _i a_i^*a_i)
=\mathrm {Tr}(c_i a_i^*a_i c_i^*)
=\mathrm {Tr}\bigl ((a_i c_i^*)^*(a_i c_i^*)\bigr ),\]
which is nonnegative. The functional is therefore positive. It is
normalized because \(\omega (1)=\sum _i\mathrm {Tr}(\rho _i)=1\). Each
\(\rho _i=c_i^*c_i\) is self-adjoint. The trace of a conjugate
transpose is the conjugate of the trace, and
\((a_i\rho _i)^*=\rho _i a_i^*\), so
\(\mathrm {Tr}(\rho _i a_i^*)=\overline {\mathrm {Tr}(\rho _i a_i)}\).
Thus \(\omega \) is Hermitian.
The same identity shows that \(\omega (a^*a)=0\) if and only if
\(a_i c_i^*=0\) for every \(i\). Different blocks occupy
complementary summands, so
\(\dim N=\sum _i\dim \{a_i:a_i c_i^*=0\}\). A matrix vanishes on a
subspace precisely when it factors through the quotient by that
subspace. The image of \(c_i^*\) has dimension \(\operatorname {rank} c_i^*\).
The matrices vanishing on it are determined by their values on a
complement, so their space has dimension
\(m_i\bigl (m_i-\operatorname {rank} c_i^*\bigr )\). The quotient of that
block therefore has dimension \(m_i\operatorname {rank} c_i^*\). Summing
over the blocks gives
\(\dim H_\omega =\sum _i m_i\operatorname {rank} c_i^*\).
It remains to match that rank with \(\operatorname {rank}\rho _i\).
For the standard inner product,
\(\|c_i v\|^2=\langle c_i^*c_i v,v\rangle \).
If \(c_i^*c_i v=0\), the right-hand side vanishes, so \(c_i v=0\).
The converse is immediate, and rank-nullity gives
\(\operatorname {rank}\rho _i=\operatorname {rank} c_i\). A minor of a
transpose is the transpose of a minor, and the determinant is
unchanged by transposition. A minor of a conjugate transpose is the
conjugate of a minor of the transpose. The largest size of a nonzero
minor is therefore the same for \(c_i\), \(c_i^T\) and \(c_i^*\). Hence
\(\operatorname {rank} c_i^*=\operatorname {rank}\rho _i\), and
\(\dim H_\omega =\sum _i m_i\operatorname {rank}\rho _i\). On \(M_2\),
\(\mathrm {diag}(2/3,1/3)\) is the square of
\(\mathrm {diag}(\sqrt {2/3},\sqrt {1/3})\), so its rank is \(2\).
\(\mathrm {diag}(1,0)\) is the square of \(\mathrm {diag}(1,0)\), and
its only nonzero minor has size \(1\). The formula gives dimensions
\(4\) and \(2\).
On functions, the indicator basis is orthogonal for the Gram form.
Indeed \(\omega (1_x^*1_y)=\delta _{xy}w(x)\), so the Gram matrix is
diagonal with entries \(w(x)\). Its rank is the number of strictly
positive weights. A function lies in \(N\) exactly when it vanishes
on \(\operatorname {supp} w\), because
\(\omega (f^*f)=\sum _x w(x)\lvert f(x)\rvert ^2\). Two functions
determine the same class exactly when they agree on that support, and
every function on the support arises in this way. The descended inner
product is
\[\langle f,g\rangle
=\sum _{w(x)>0}w(x)\overline {f(x)}\,g(x).\]
Every function on a finite set is square-summable, so the space is
\(\ell ^2(\operatorname {supp} w)\).