用户:CaffeineP/笔记/Linear Algebra Done Right (Sheldon Axler, 4th Edition)

• linear algebra

Chapter 1: Vector Spaces

1A Rⁿ and Cⁿ

Exercises 1A

8. Square roots of i

${(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)}^2={(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)}^2=i$

1B Definition of Vector Space

1.20 definition: vector space

$V$ is a set, $+$ an addition on $V$, $\ast$ a scalor multiplication on $V$, $ab:=a\ast b$ , and $𝐅$ a field with multiplication identity $1$, then

$\begin{array}{c}(V,+,\ast )\text{ is a}\\ \text{vector space}\\ \text{over }𝐅\end{array}:\iff \{\begin{array}{c}\underset{u\in V,v\in V}{\mathrm{\forall }}u+v=v+u\wedge \underset{u\in V,v\in V,w\in V}{\mathrm{\forall }}(u+v)+w=u+(v+w)\wedge \underset{a\in 𝐅,b\in 𝐅,v\in V}{\mathrm{\forall }}(ab)v=a(bv)\\ \underset{0\in V}{\mathrm{\exists }}\underset{v\in V}{\mathrm{\forall }}\left(v+0=v\wedge \underset{w\in V}{\mathrm{\exists }}v+w=0\right)\wedge \underset{v\in V}{\mathrm{\forall }}1v=v\\ \underset{a\in 𝐅,u\in V,v\in V}{\mathrm{\forall }}a(u+v)=au+av\wedge \underset{a\in 𝐅,b\in 𝐅,v\in V}{\mathrm{\forall }}(a+b)v=av+bv\end{array}$