Matrix analysis notes (I)

Linear space

$V$ Nonempty set ,$F$ Number field , In collection $V$ Two algebraic operations addition are defined in $+$ Multiply with number $\cdot$ , And satisfy the operation law

• The law of addition exchange $\alpha +\beta =\beta +\alpha$
• The law of combination of addition $(\alpha +\beta )+\gamma =\alpha +(\beta +\gamma )$
• Zero element $\exists 0\in V,sendhave\; to\forall \alpha \in VYes\alpha +0=\alpha$
• Negative element $\forall \alpha \in V,allsavestay\beta sendhave\; to\alpha +\beta =0$
• $1\cdot \alpha =\alpha$
• The law of multiplication $k(l\alpha )=(kl)\alpha$
• Distributive law $(k+l)\alpha =k\alpha +l\alpha$
• Distributive law $k(\alpha +\beta )=k\alpha +k\beta$

$k,l\in F$

call $V$ It's a number field $F$ Colinear space

Common linear spaces

• Vector space
• Function space
• Matrix space
• Polynomial space

Relevant concepts

• The linear table out
  $$\alpha =k_1{\beta }_1+...+k_n{\beta }_n$$
• Linear correlation
  $$\begin{gathered} 0=k_1{\beta }_1+...+k_n{\beta }_n \\ {k}_1...k_n\ne 0 \end{gathered}$$
• Linearly independent
  Non linear correlation
• Maximum linear independent group
  There is a set of linear independent vectors in a vector group , And each vector can be linearly expressed by the vectors in the independent group
  Equivalent to a vector group
• Rank
  The number of maximal independent group vectors

Basic properties

• Vector groups with zero vectors are linearly correlated
• It's not about the whole $\Rightarrow$ Not part of it , Part of it $\Rightarrow$ It's all about
• A vector group with many vectors can be expressed linearly by a vector group with few vectors , Then a group of vectors with many vectors must be linearly correlated
• Rank unique but maximal independent group is not unique
• Vector group （Ⅰ） It can be a set of vectors （Ⅱ） The linear table out , Vector group （Ⅰ） The rank of $\le$ Vector group （Ⅱ） The rank of
• The equivalent vector group has the same rank

basal & coordinate & dimension

$V$ by $F$ Linear space , if $V$ in $n$ A linearly independent vector ${\alpha }_1,...,{\alpha }_n$, So that any vector can be represented by ${\alpha }_1,...,{\alpha }_n$ The linear table out , namely
$$\alpha =k_1{\alpha }_1+...+k_n{\alpha }_n$$
call ${\alpha }_1,...,{\alpha }_n$ For the base ,$(k_1,...,k_n{)}^T$ Is the coordinate under the base , call $V$ by $n$ One dimensional linear space , Write it down as $dimV=n$

The linear space we study is finite dimensional

Base transformation

set up ${\alpha }_1,...,{\alpha }_n$ and ${\beta }_1,...,{\beta }_n$ by $n$ One dimensional linear space $V$ Two groups of substrates , The relationship is
$${\beta }_i=[{\alpha }_1,...,{\alpha }_n]\left[\begin{array}{c}{\alpha }_{1i}\\ ...\\ {\alpha }_{ni}\end{array}\right],i=1,...,n$$
Matrixing has
$$[{\beta }_1,...,{\beta }_n]=[{\alpha }_1,...,{\alpha }_n]\left[\begin{array}{ccc}{a}_{11}& ...& a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& ...& a_{nn}\end{array}\right]$$
$n$ A square matrix of order is called a transition matrix $P$
Transition matrix $P$ reversible

Coordinate transformation

$\forall \xi \in V$, The coordinates under the two sets of bases are $[x_1,...,x_n{]}^T$ and $[y_1,...,y_n{]}^T$, Then there are
$$[x_1,...,x_n{]}^T=P[y_1,...,y_n{]}^T$$

Linear subspace

set up $V$ It's a number field $F$ On the one $n$ Dimensional subspace ,$W$ by $V$ A nonempty subset of , If yes $\forall \alpha ,\beta \in W$ as well as $\forall k,l\in F$ Yes
$$k\alpha +l\beta \in W$$
call $W$ by $V$ The subspace of

Ordinary subspace

Linear space $V$ and $0$ Called a trivial subspace

Generate subspace

$span\{{\alpha }_1,...,{\alpha }_s\}$

Intersection of subspaces & and

• $V_1\cap V_2$
• $V_1+V_2$
  The intersection space and the sum space of the subspace constitute $V$ The subspace of
  $dim{V}_1+dim{V}_2=dim(V_1+V_2)+dim(V_1\cap V_2)$

Linear mapping

set up $V_1,V_2$ It's a number field $F$ On two linear spaces , mapping $\varphi :V_1\to V_2$, about $V_1$ Any two vectors ${\alpha }_1,{\alpha }_2$ And any number $\lambda \in F$ There are
$$\begin{gathered} \varphi ({\alpha }_1+{\alpha }_2)=\varphi ({\alpha }_1)+\varphi ({\alpha }_1) \\ \varphi (\lambda {\alpha }_1)=\lambda \varphi ({\alpha }_1) \end{gathered}$$
Called mapping $\varphi$ By $V_1$ To $V_2$ The linear mapping of . call $\alpha$ by $\varphi (\alpha )$ The original ,$\varphi (\alpha )$ by $\alpha$ like

nature

• $\varphi (0)=0$
• $\varphi (\sum k_i{\alpha }_i)=\sum k_i\varphi ({\alpha }_i)$
• set up ${\alpha }_1,...,{\alpha }_s\in V$ And linear correlation , be $\varphi ({\alpha }_1),...,\varphi ({\alpha }_s)$ Linear correlation

range

set up $\varphi$ by $V_1$ To $V_2$ The linear mapping of
Make $\varphi (V_1)=\{\beta =\varphi (\alpha )\in V_2,\forall \alpha \in V_1\}$
be $\varphi (V_1)$ by $V_2$ Linear subspace , Called linear mapping $\varphi$ Range of values , Write it down as $R(\varphi )$

Nucleon space

Make $N(\varphi )={\varphi }^{-1}(0)=\{\alpha \in V_1|\varphi (\alpha )=0\}$
be $N(\varphi )$ by $V_1$ Linear subspace , Called linear mapping $\varphi$ The nucleon space of ,$dimN(\varphi )$ be called $\varphi$ Zero degree of

Rank and zero degree theorem

set up $\varphi$ yes $n$ One dimensional linear space $V_1$ To $m$ One dimensional linear space $V_2$ The linear mapping of , that
$$dimR(\varphi )+dimN(\varphi )=n$$

linear transformation

set up $\varphi$ by $V$ To $V$ A linear mapping of , call $\varphi$ It's a linear space $V$ Linear transformation of

For example, differential transformation

be similar

set up $\varphi$ by $V$ Linear transformation of ,${\alpha }_1,...,{\alpha }_n$ And ${\beta }_1,...,{\beta }_n$ yes $V$ Two groups of bases . from $\{{\alpha }_i\}$ To $\{{\beta }_i\}$ The transition matrix is $P$,$\varphi$ At base ${\alpha }_1,...,{\alpha }_n$ The matrix under is $A$, At base ${\beta }_1,...,{\beta }_n$ The matrix under is $B$, be
$$B=P^{-1}AP$$
And $A$ And $B$ be similar , Write it down as $BA$

The eigenvalue & Eigenvector

set up $A$ by $n$ Square matrix . If there is a number $\lambda$ And $n$ Nonzero column vector $X$, bring
$$AX=\lambda Xor(\lambda I-A)X=0$$
said $\lambda$ For matrix $A$ The eigenvalue ,$X$ For matrix $A$ Of are characteristic values $\lambda$ Eigenvector of

Related definitions

• set up $A$ It's a number field $F$ Of $n$ Order matrix , matrix $\lambda I-A$ be called $A$ Characteristic matrix
• determinant
  $$|\lambda I-A|=\left|\begin{array}{cccc}\lambda -a_{11}& -a_{12}& ...& -a_{1n}\\ -a_{21}& \lambda -a_{22}& ...& -a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{n1}& -a_{n2}& ...& \lambda -a_m\end{array}\right|$$
  be called $A$ Characteristic polynomial
• $n$ Subalgebral equation $|\lambda I-A|=0$ be called $A$ Characteristic equation
  The root is called $A$ The characteristic root of the （ The eigenvalue ）
• All eigenvalues of a matrix are called $A$ Spectrum of , Write it down as $\sigma (A)$
• $(\lambda I-A)X=0$ Called characteristic equations

nature

• $|A|={\lambda }_1...{\lambda }_n$

The necessary and sufficient condition for a matrix to be invertible is that all eigenvalues are not zero

• ${\sum }_{i=1}^n{a}_{ii}={\sum }_{i=1}^n{\lambda }_i$
• if $A$ The characteristic value is $\lambda$,$X$ yes $A$ Corresponding to $\lambda$ Eigenvector of , be
  • $kA$ The characteristic value is $k\lambda$
  • $A^m$ The characteristic value is ${\lambda }^m$
  • if $A$ reversible , be $A^{-1}$ The characteristic value is ${\lambda }^{-1}$
  • $f(x)$ by $x$ polynomial , be $f(x)$ The characteristic value is $f(\lambda )$
• $X$ Still matrix $kA,A^m,A^{-1},f(A)$ Corresponding to $k\lambda ,{\lambda }^m,{\lambda }^{-1},f(\lambda )$ Eigenvector of

Feature subspace

$n$ Order matrix $A$ Corresponding eigenvalue of ${\lambda }_0$ Add zero vector to all eigenvectors of , It can make up $R^n$ The subspace of , It's called a matrix $A$ Belong to characteristic value ${\lambda }_0$ The characteristic subspace of , Write it down as $V_{{\lambda }_0}$, It's not hard to see. $V_{{\lambda }_0}$ Is a set of characteristic equations
$$({\lambda }_{0}I-A)X=0$$
The solution space of

Algebraic repetition

set up ${\lambda }_1,...,{\lambda }_r$ by $r$ Two different eigenvalues , The corresponding multiplicity is $p_1,...,p_r$, call $p_i$ by ${\lambda }_i$ Algebraic sufficient repetition
$$|\lambda I-A|=(\lambda -{\lambda }_1{)}^{p_1}...(\lambda -{\lambda }_r{)}^{p_r}$$

Geometric repeatability

Feature subspace $V_{{\lambda }_0}$ Dimension of $q_i$ by ${\lambda }_i$ Geometric repeatability of
$$q_i=n-rank({\lambda }_{i}I-A)$$

nature

• Eigenvectors belonging to different eigenvalues are linearly independent
• set up ${\lambda }_1,...,{\lambda }_r$ by $A$ Of $r$ Two different eigenvalues .$q_i$ yes ${\lambda }_i$ Geometric repeatability ,${\alpha }_{i1},...,{\alpha }_{i{q}_i}$ Is corresponding to ${\lambda }_i$ Of $q_i$ A linearly independent eigenvector , be $A$ All eigenvectors ${\alpha }_{11},...,{\alpha }_{1q_1}$,…,${\alpha }_{r1};...;{\alpha }_{r{q}_r}$ Are linearly independent
• An eigenvector does not belong to different eigenvalues
• $A$ Any eigenvalue ${\lambda }_i$ Geometric repeatability $q_i$ Not greater than algebraic repetition $p_i$

Similarity matrix

nature

Similar matrices have the same

• Characteristic polynomial
• The eigenvalue
• Determinant value
• Rank
• trace
• Spectrum

Diagonalizable condition

Diagonalize

$n$ Order matrix $A$ Similar to diagonal matrix , said $A$ Diagonalize , Also known as $A$ Is a simple matrix

Conditions

• $n$ Order matrix $A$ The necessary and sufficient condition for diagonalization is $n$ A linearly independent eigenvector
• $n$ Order matrix $A$ A necessary and sufficient condition for diagonalization is that the geometric repetition of each eigenvalue is equal to the algebraic repetition
• if $n$ The order matrix has $n$ Different eigenvalues （ Characteristic polynomials have no multiple roots ）, be $A$ Similar diagonalization

At the same time diagonalize

• set up $A\in C^{n\times n},B\in C^{m\times m}$, And $C=\left[\begin{array}{cc}A& 0\\ 0& B\end{array}\right]$, be $C$ The necessary and sufficient condition for diagonalization is $A,B$ Can be diagonalized
• set up $A,B\in C^{n\times n}$ Can be diagonalized , be $A,B$ At the same time, the necessary and sufficient condition for diagonalization is $AB=BA$