Finite-dimensional duals and matrix transposes

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

From The transpose of a bounded operator, with its stated hypotheses: Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $T:X\to Y$ be bounded and linear between normed spaces. Its transpose, or Banach adjoint, is $T^{\ast }:Y^{\ast }⟶X^{\ast },(T^{\ast }g)(x)=g(Tx)\mathrm{.}$ The duals are def-dual-space-of-a-normed-space. Composition is bounded by lem-composition-operator-norm-inequality, so this has the displayed codomain. It is linear in $g$ over $\mathbb{K}$. No complex conjugation is inserted; a Hilbert adjoint uses a separate inner-product identification.

From The dual family of a finite basis is a basis of the dual space, with the same dimension, with its stated hypotheses: If $B=(b_1,\dots ,b_n)$ is a basis of a finite-dimensional $F$-vector space $V$, then its dual family $B^{\ast }=(b_1^{\ast },\dots ,b_n^{\ast })$ is a basis of $V^{\ast }$. Consequently $\dim V^{\ast }=\dim V=n$.

From A linear map from a finite-dimensional normed space is bounded, with its stated hypotheses: Let $X$ and $Y$ be normed spaces over the same scalar field, and assume $X$ admits an ordered basis of finite length. Then every linear map $S:X\to Y$ is a bounded linear operator in the sense of def-bounded-linear-operator.