[Mathematics for postgraduate entrance examination] Linear Algebra Chapter 3 - Vector | 3) The nature of the rank of vector groups, vector spaces, transition matrices

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introduction

Immediately after learning the basic concept of vector group rank in the previous article, continue to learn the content of vectors.

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3. Vector group equivalence, maximum linearly independent group and rank of vector group

3.2 Properties of rank of vector group

Property 1 (three ranks are equal) —— Let $A=(b_1,b_2,\dots ,b_n)=(a_1,a_2,\dots ,a_n{)}^T$，among them$a_1,a_2,\dots ,a_n$与$b_1,b_2,\dots ,b_n$Respectively matrix $The\; row\; vector\; group\; and\; column\; vector\; group\; of\; A$ , then the matrixAAChichi$A$ , AAThe rank of the row vector group of$A$ , AASets of column vectors of$A\; are\; of\; equal\; rank.$

Property 2 —— Suppose $A:a_1,a_2,\dots ,a_n$与$B:b_1,b_2,\dots ,b_n$are two vector groups with the same dimension, if the vector group $A$ can be composed of vector group$B$ linear representation, then the vector group$The\; rank\; of\; A$ does not exceed the vector group$B$ 's rank.

Property 3 - The ranks of equivalent vector groups are equal, and vice versa.

1. Let vector group $A:a_1,a_2,\dots ,a_n$与$B:b_1,b_2,\dots ,b_n$The ranks of are equal, and the vector group $A$ can be composed of vector group$B$ linear representation, then the vector group$A$ with vector group$B$ is equivalent.
2. Let vector group$A:a_1,a_2,\dots ,a_n$可由$B:b_1,b_2,\dots ,b_n$linear representation, but the vector set $A$ is not available from the vector group$B$ linear representation, then the vector group$The\; rank\; of\; A$ is less than the vector group$B.$ _
3. For two equivalent vector groups, the matrices they form are also equivalent, but not necessarily vice versa.

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Four, $n-$ dimensional vector space

4.1 Basic concepts

$n-$ dimensional vector space- all$The\; addition\; of\; n-$ dimensional vectors together with vectors and the multiplication of numbers and vectors is called$n-$ dimensional vector space, denoted as$R^n.$

Base—— Let $R^n$ is$n-$ dimensional vector space, let$a_1,a_2,\dots ,a_n$is nn in the vector space$n$ vectors, if satisfy:
(1)$a_1,a_2,\dots ,a_n$Linearly independent;
(2) For any $b\in R^n,\beta$ can be composed of vector groups$a_1,a_2,\dots ,a_n$Linear representation,
called $a_1,a_2,\dots ,a_n$for $n-$ dimensional vector space$R^{base\; of\; n}$ .
In particular, if$a_1,a_2,\dots ,a_n$Two pairs of orthogonal, and are unit vectors, called orthonormal basis.

The coordinates of the vector under the base - Let $a_1,a_2,\dots ,a_n$for $R^{The\; basis\; of\; n}$ ,$b\in R^n$，若$b=k_1{a}_1+k_2{a}_2+\cdots +k_n{a}_n$, say $(k_1,k_2,\dots ,k_n)$ is the vector$\beta$在基$a_1,a_2,\dots ,a_n$coordinates below.

Transition matrix - Transform from one set of bases to another set of bases, which can be multiplied by a matrix, which is called a transition matrix.

Some intuition is needed to better understand vector spaces. The first thing to understand is that a matrix represents a transformation.

4.2 Basic properties

Theorem 1 Suppose $a_1,a_2,\dots ,a_n$for $n-$ dimensional vector space$R^{The\; basis\; of\; n}$ ,$b\in R^n$，令$A=(a_1,a_2,\dots ,a_n)$ , then the vector$\beta$在基$a_1,a_2,\dots ,a_n$The coordinates below are $X=A^{-1}b.$

Theorem 2 —— Suppose $a_1,a_2,\dots ,a_n$与$b_1,b_2,\dots ,b_n$is the vector space $R^{Two\; bases\; of\; n}$ , let$A=(a_1,a_2,\dots ,a_n),B=(b_1,b_2,\dots ,b_n)$ , then from the basis$a_1,a_2,\dots ,a_n$到基$b_1,b_2,\dots ,b_n$The transition matrix of is $Q=A^{-1}B.$

Theorem 3 —— From the basis $a_1,a_2,\dots ,a_n$到基$b_1,b_2,\dots ,b_n$The transition matrix and from the basis $b_1,b_2,\dots ,b_n$To the basis $a_1,a_2,\dots ,a_n$The resulting transition matrices are the inverses of each other.

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write at the end

This concludes the theoretical part of vectors. The connection among matrix, vector, and equation system will be summarized and issued recently.