ℝⁿ

${\mathbb{R}}^n$

Let $\mathbb{N}$ be the nonnegative integers and let $n\in \mathbb{N}$.

Define $[n]=\{i\in \mathbb{N}:i<n\}$. In particular, $[0]=\mathrm{\varnothing }$.

Let ${\mathbb{R}}^n={\mathbb{R}}^{[n]}$, the set of functions $[n]\to \mathbb{R}$.

$n=0$

${\mathbb{R}}^0={\mathbb{R}}^{\mathrm{\varnothing }}=\{\mathrm{\varnothing }\}$. ¹

${\mathbb{R}}^0$ is an $\mathbb{R}$-linear space, with one element, ∅.

$n\ge 1$

Let $n\in {\mathbb{N}}_{\ge 1}$. For $x,y\in {\mathbb{R}}^n$, we define $x+y\in {\mathbb{R}}^n$ by $(x+y)(i)=x(i)+y(i)$, $i\in [n]$.

For $x{\mathbb{R}}^n$ and $c\in \mathbb{R}$, we define $cx\in {\mathbb{R}}^n$ by $(cx)(i)=cx(i)$, $i\in [n]$. Then ${\mathbb{R}}^n$ is an $\mathbb{R}$-linear space.

For $i,j\in \mathbb{N}$, define

$${\delta }_{i,j}=\{\begin{array}{ll}1& i=j\\ 0& i\ne j\end{array}$$

For $k\in [n]$ define $e^k\in {\mathbb{R}}^n$ by

$$e^k(i)={\delta }_{i,k},i\in [n].$$

$\{e^k:k\in [n]\}$ is a basis for ${\mathbb{R}}^n$.

Dual space $({\mathbb{R}}^n{)}^{\ast }$

Let $({\mathbb{R}}^n{)}^{\ast }$ be the set of linear maps ${\mathbb{R}}^n\to \mathbb{R}$, the dual space of ${\mathbb{R}}^n$.

For $x\in {\mathbb{R}}^n$, define $x^T\in ({\mathbb{R}}^n{)}^{\ast }$ by

$$x^{T}y=\sum _{i\in [n]}x(i)y(i),y\in {\mathbb{R}}^n.$$

We call $x^T$ the transpose of $x$: $x$ is a vector/column vector and $x^T$ is a covector/row vector.

$\{(e^k{)}^T:k\in [n]\}$ is a basis for $({\mathbb{R}}^n{)}^{\ast }$.

Write $Tx=x^T$.Themap$T:x \mapsto x^T$isalinearisomorphism$\mathbb{R}^n \to (\mathbb{R}^n)^*$$.

Basis expansion

For $x\in {\mathbb{R}}^n$ and for $k\in [n]$, we have

$$x^T{e}^k=\sum _{i\in [n]}x(i)e^k(i)=\sum _{i\in [n]}x(i){\delta }_{i,k}=x(k)$$

and

$(e^k{)}^{T}x=\sum _{i\in [n]}{e}^k(i)x(i)=\sum _{i\in [n]}{\delta }_{i,k}x(i)=x(k)$.

Using this, one then works out that

$$x=\sum _{k\in [n]}((e^k{)}^{T}x)e^k.$$

Linear algebra

For real finite dimensional vector spaces $V$ and $W$, let $\mathcal{L}(V,W)$ be the set of linear transformations $V\to W$, which is itself a real finite dimensional vector space.

$$({\mathbb{R}}^n{)}^{\ast }=\mathcal{L}({\mathbb{R}}^n,\mathbb{R}).$$

An $m\times n$ matrix is an element of $\mathcal{L}({\mathbb{R}}^n,{\mathbb{R}}^m)$ and a choice of basis for ${\mathbb{R}}^n$.

$$\dim \mathcal{L}(V,W)=\dim V\cdot \dim W.$$

In particular,

$$\dim \mathcal{L}({\mathbb{R}}^m,{\mathbb{R}}^n)=\dim {\mathbb{R}}^m\cdot \dim {\mathbb{R}}^n=m\cdot n.$$

Transposes of linear maps

Let $A\in \mathcal{L}({\mathbb{R}}^n,{\mathbb{R}}^m)$.

Define $A^T\in \mathcal{L}(({\mathbb{R}}^m{)}^{\ast },({\mathbb{R}}^n{)}^{\ast })$ by

$$A^{T}fx=f(Ax),f\in ({\mathbb{R}}^m{)}^{\ast },x\in {\mathbb{R}}^n,$$

called the transpose of $A$.

Transposes of matrices

We remind ourselves that $\{(e^k{)}^T:k\in [m]\}$ is a basis for $({\mathbb{R}}^m{)}^{\ast }$ and $\text{Double exponent: use braces to clarify}$ is a basis for ${\mathbb{R}}^n$. Thus, $A^T\in \mathcal{L}(({\mathbb{R}}^m{)}^{\ast },({\mathbb{R}}^n{)}^{\ast })$ is also defined by

$$A^T(e^k{)}^T{e}^j=(e^k{)}^T(A{e}^j),j\in [n],k\in [m].$$

Bilinear maps

Tensor products

Determinants

1. empty function in nLab ↩