Review
This chapter is the core of advanced algebra,and it's also the most difficult one for most students,the most efficiency way to learn this chapter is understand the concept without ambiguous,have more practice,and review it regularly.

A special linear transformation

$A=(a_{ij})$ is a n-order square on the fileds of $\mathbb{P}$ ,we define a linear transformation $\mathscr{A}$ on $P^{n\times n}$ ,satisfied $\mathscr{A}X=AX$ ,please analyze the properties of $\mathscr{A}$

Minimum polynominal

Theorem if $A$ is a quasi-diagonal matrix,marked $A=diag\{A_1,A_2,\ldots,A_s\}$ which $A_i$ has a minimum polynominal $m_i(x)(i=1,2,\ldots,s)$ ,then the minimum polynominal of $A$ is the least common multiple of $m_1(x),m_2(x),\ldots,m_s(x)$ ,marked $[m_1(x),m_2(x),\ldots,m_s(x)]$

Example 1.if a matrix can be turned into a diagonal matrix,then it's minimum polynominal is the product of the one-time factor of mutual element.

Example 2.A real symmetry matrix $A$ has a characteristcs polynominal,marked $f(\lambda)=\lambda^5+3\lambda^4-6\lambda^3-10\lambda^2+21\lambda-9$ ,calculate the minimum polynominal of $A$

Invariant sub-spaces

Some simple but important theorem

proposition 1:Linear transformations $\mathscr{A}$ on the $V$ ,and it's kernel,range and chracteristcs sub-spaces are $\mathscr{A}-$ sub-spaces

proposition 2: $\mathscr{A,B}$ are two linear transformations,satisfied $\mathscr{AB=BA}$ ,then all of the characteristcs sub-spaces of $\mathscr{A}$ are the invariant sub-spaces of $\mathscr{B}$

proposition 3: $A$ is a linear transformations on fileds $\mathbb{P}$ of linear space $V$ , $\alpha\in W$ is a non-zero vector,then $L(\alpha)$ is a $\mathscr{A}$ -sub-spaces if and only if $\alpha$ is a chracteristics vector of $\mathscr{A}$ .

Theorem. $A$ is a transformations on fileds $\mathbb{P}$ of n-dimension linear space $V$ ,then $\mathscr{A}$ 'matrix under a set of bases is a diagonal matrix $diag\{A_1,A_2,\ldots,A_s\}$ if and only if $V$ can be decompositioned into the strict and of several non-trivial invariant subspace: $V=W_1\oplus W_2\oplus \ldots \oplus W_s$ ,and $A_i$ is the matrix under the base of $\mathscr{A}|W_i$

Important example
suppose $V$ is a n-dimension linear space of complex field,and the linear transformations $\mathscr{A}$ under the bases of $\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n$ 's matrix is a jordan block,marked $J=\left(\begin{array}{cccc} a&&&\\ 1&\ldots&&\\ &\ldots&a&\\ &&1&a\end{array} \right)$
proof:(1)there is only $V$ can be the $\mathscr{A}$ -sub-spaces who includes $\varepsilon_1$ .
(2)all of the non-zero $\mathscr{A}$ -sub-spaces includes $\varepsilon_n$
(3) $V$ can't be decompositioned into two non-trivial $\mathscr{A}$ -sub-spaces' strict and.
(4) $V$ have and only have $n+1$ $\mathscr{A}$ -sub-spaces,they are $\{0\},\{L(\varepsilon_n)\},\{L(\varepsilon_{n-1},\varepsilon_n)\},\ldots,\{L(\varepsilon_1,\varepsilon_2,\ldots,\varepsilon_n)\}$

Tips: let $\beta_1,\beta_2,\ldots,\beta_s$ is a group bases of $W$
suppose $\beta_i=a_{it}\varepsilon_t+\ldots+a_{1n}\varepsilon_n$ and,exists $a_{it}$ such that $a_{it}\not=0$ then,we can times $\mathscr{A}$ for several times.

The most important watershed(NB)

Lemma 1:if $V=V_1\oplus V_2,V_1=V_{11}\oplus \ldots\oplus V_{1s},V_2=V_{21}\oplus \ldots \oplus V_{2t}$ then $V=V_{11}\oplus \ldots\oplus V_{1s} \oplus V_{21} \oplus \ldots \oplus V_{2t}$
Lemma 2: $V$ us a linear space in filed $\mathbb{P}$ , $\mathscr{\sigma}$ is a zero-transformation on $V$ ,then $Ker \sigma=V$
Lemma 3:if a finite dimension linear space $V$ satisfied $V=V_1\oplus V_2\oplus \ldots \oplus V_s=W_1\oplus W_2\oplus\ldots\oplus W_s$ ,and $V_i\subseteq W_i(i=1,2,\ldots,s)$ then $V_i=W_i(i=1,2,\ldots,s)$

Proposition 1
(Extremely important) $V$ is a linear space on field $\mathbb{P}$ , $\mathscr{A}$ is a linear transformations on space $V$ , $f(x),f_1(x),f_2(x),\ldots,f_s(x)\in \mathbb{P}[x]$ , $f(x)=f_1(x)f_2(x)\ldots f_s(x)$ and $f_1(x),f_2(x),\ldots,f_s(x)$ ,any of two are mutal vegetarian.then $Kerf(\mathscr{A})=Kerf_1(\mathscr{A})\oplus \ldots\oplus Kerf_s(\mathscr{A})$

Proposition 2
$V$ is a linear space on field $\mathbb{P}$ , $\mathscr{A}$ is a transformations on space $V$ , $f(x)\in \mathbb{P}[x]$ is a zero-polunominal.(then $f(\mathscr{A})=\mathscr{0}$ ),and $f(x)=f_1(x)f_2(x)\ldots f_s(x)$ ,which $f_1(x),f_2(x),\ldots,f_s(x)\in \mathbb{P}$ and any of two are mutal vegetarian,then $V=Kerf_1(\mathscr{A})\oplus Kerf_2(\mathscr{A})\oplus\ldots \oplus Kerf_s(\mathscr{A})$

Proposition 3
$V$ is a linear space in $\mathbb{C}$ , $\mathscr{A}$ is a linear transformation on $V$ , $m(\lambda)$ is the minimum polynominal of $\mathscr{A}$ ,and $m(\lambda)=(\lambda-\lambda_1)^{t_1}(\lambda-\lambda_2)^{t_2}\ldots (\lambda-\lambda_s)^{t_s}$ which $\lambda_1,\lambda_2,\ldots,\lambda_s$ are multal differentiation characteristc values.then,any $k_i \geq t_i(i=1,2,\ldots,s)$ we have $Ker(\mathscr{A}-\lambda_1\epsilon)^{t_i}=Ker(\mathscr{A}-\lambda_i\epsilon)^{k_i}$

Proposition 4
$V$ is a linear space on the field $\mathbb{P}$ , $\mathscr{A}$ is a transformation on $V$ ,and $f(x),f_1(x),\ldots,f_s(x)\in \mathbb{P}[x],f(x)=f_1(x)f_2(x)\ldots f_s(x)$ is a zero-polynominal of $\mathscr{A}$ ,and $f_1(x),f_2(x),\ldots,f_s(x)$ any of two are mutal vegetarian.we marked $F_i(x)=\frac{f(x)}{f_i(x)}(i=1,2,\ldots,s)$ then for any $i=1,2,\ldots,s$ ,we have $Kerf_i(\mathscr{A})=ImF_i(\mathscr{A})$

Example 1:
$f(x),g(x)$ are two non-zero polynominal on field $\mathbb{P}$ , $d(x),r(x)$ are respectively the max common divisor and the min male-double,both of their first item are 1, $A$ is a n-order square matrix on field $\mathbb{P}$ ,we marked $f(A)X=0,g(A)X=0,d(A)X=0,r(A)X=0$ solving spaces respectively are $V_1,V_2,V_3,V_4$ ,proof: $V_3=V_1\cap V_2,V_4=V_1+V_2$

Example 2:if $V$ is a finite dimension linear space, $\mathscr{A},\mathscr{B}$ are the linear transformations on $V$ ,satisfied $\mathscr{A}^2=\mathscr{B}^2=0,\mathscr{AB+BA=\varepsilon}$
(1)proof: $Ker\mathscr{A}=\mathscr{A}Ker\mathscr{B},Ker\mathscr{B}=\mathscr{B}Ker\mathscr{A}$ and $V=Ker\mathscr{A}\oplus Ker\mathscr{B}$
(2)proof:the dimension of $V$ is even
(3)proof:if $dimV=2$ ,then $V$ has a group of bases such that $\mathscr{A,B}$ matrix respectively are $\left( \begin{array}{cc} 0&1\\ 0&0 \end{array} \right)$ , $\left( \begin{array}{cc} 0&0\\ 1&0 \end{array} \right)$ under this group of bases.

Tips: (2)suppose $\alpha_1,\alpha_2,\ldots,\alpha_s$ are the bases of $Ker\mathscr{A}$ ,and $\beta_1,\beta_2,\ldots,\beta_t$ are the bases of $Ker\mathscr{B}$ ,from(1),we know that eixist a group of vectors $\gamma_1,\gamma_2,\ldots,\gamma_s\in Ker\mathscr{B}$ such that $\alpha_i=\mathscr{A}\gamma_i(i=1,2,\ldots,s)$ ,this means $s\geq t$ and,we can do it in a similar way,and we can get $s=t$ ,or we can proof that $\gamma_i$ are linear independent.
(3)there is only one vector in each space.

Example 3: $\mathscr{A,B}$ are the linear transformations on linear space $V$ ,satisfied $A^2=A,B^2=B$
proof:(1) $\mathscr{A,B}$ have a same value range if and only if $\mathscr{AB=B,BA=A}$
(2) $A,B$ have the same kernel if and only if $\mathscr{AB=A,BA=B}$

Tips: $\mathscr{AB=B}\Leftarrow\Rightarrow\mathscr{(A-\varepsilon)B=0}\Leftarrow\Rightarrow\forall\alpha\in V,\mathscr{B\alpha}\in Ker\mathscr{(A-\varepsilon)=A}V,then ,\mathscr{B}V \subseteq \mathscr{A}V$

Example 4:if $\mathscr{A,B}$ are the linear transformations on linear space $V$ of field $\mathbb{P}$ ,and $\mathscr{A}$ satisfied $\mathscr{A^2=A}$ ,proof:
(1) $\mathscr{A^{-1}(0)=\{\alpha-A\alpha|\alpha \in }V\}$
(2) $V=\mathscr{A^{-1}(0)\oplus A}V$
(3) $\mathscr{A^{-1}(0),A}V$ are the $\mathscr{B}$ -sub-space if and only if $\mathscr{AB=BA}$

Tips: we need $\forall \alpha\in V,\mathscr{AB}\alpha=\mathscr{BA}\alpha$ ,we decompositioned $\alpha=\alpha_1+\alpha_2$ ,and then we consider about $\mathscr{BA}\alpha=\mathscr{BA}\alpha_2=\mathscr{BA^{2}}\beta_2=\mathscr{BA}\beta_2=\mathscr{B}\alpha_2$ and $\mathscr{AB}\alpha=\mathscr{AB}\alpha_1+\mathscr{AB}\alpha_2=\mathscr{A^2}\beta_2=\mathscr{A}\beta_2=\mathscr{B}\alpha_2$

Proposition
Extrmely important:if $V$ is a n-dimension linear space on field $\mathbb{P}$ , $\mathscr{A}$ is a linear transformations on $V$ ,then the $\mathscr{A}$ can be diagonalized if and only if the minimum polynominal of $\mathscr{A}$ is a product of one-time-dependent on field $\mathbb{P}$ .

Example 5:if $V$ is a 6-dimension linear space on complex field, $\mathscr{A}$ is a transformations on $V$ ,the characteristic polynominal is $f(\lambda)=(\lambda-1)(\lambda-2)^2(\lambda-3)^3$ ,then $V$ can be decomposition into the strict and of 3 $\mathscr{A}$ -sub-space,and their dimension respectively are1,2,3.

Tips: use the knowledge of jordan type.

Example 6:if $V$ is a n-dimension linear space on filed $\mathbb{P}$ , $\mathscr{A}$ is a transformation on $V$ ,and $m(\lambda)$ is the minimum polynominal of $\mathscr{A}$ ,we know that $m(\lambda)=g(\lambda)h(\lambda)$ ,which $(g(\lambda),h(\lambda))=1$ ,proof:exists $\mathscr{A}$ -sub-space $V_1,V_2$ ,such that $V=V_1\oplus V_2$ ,and $\mathscr{A}|V_1$ has minimum polynominal $g(\lambda)$ , $\mathscr{A}|V_2$ has a minimum polynominal $h(\lambda)$

Tips: $\mathscr{A}(\alpha_1,\alpha_2,\ldots,\alpha_r,\alpha_{r+1},\ldots,\alpha_n)=(\alpha_1,\ldots,\alpha_n)\left(\begin{array}{cc} A_1&\\ &A_2\end{array}\right)$ and,we marked the minimum polynominal of $A_1,A_2$ respectively as $m_1(\lambda),m_2(\lambda)$ ,then $m_1(\lambda)|g(\lambda),m_2(\lambda)|h(\lambda)$ ,and we know that the minimum polynominal of $A$ is $m(\lambda)=g(\lambda)h(\lambda)=[m_1(\lambda),m_2(\lambda)]\Rightarrow m_1(\lambda)=g(\lambda),m_2(\lambda)=h(\lambda)$

The further discussing about whether a matrix can be diagonal.

Theorem.if $V$ is a n-dimension linear space on field $\mathbb{P}$ , $\mathscr{A}$ is a transformation on $V$ , $\lambda_1,\lambda_2,\ldots,\lambda_s\in \mathbb{P}$ are all mutal different characteristic values,then these conditions are equal:
(1) $\mathscr{A}$ can be diagonal.
(2) $\mathscr{A}$ have $n$ linear independent characteristic vectors,they build a group of bases of $V$ .
(3) $V=V_{\lambda_1}\oplus V_{\lambda_2}\oplus \ldots\oplus V_{\lambda_s}$
(4) $\mathscr{A}$ dimension of the summary of different characteristic values' space are $n$ ,other words $dimV_1+\ldots+dimV_s=n$
(5)The algebra repeat times of each characteristic equals the geometric repeat times.
(6) $\mathscr{A}$ has a minimum polynominal $m(\lambda)$ is the product of multal vegetarian one-times factors on field $\mathbb{P}$ ,other words $m(\lambda)=(\lambda-\lambda_1)\ldots(\lambda-\lambda_s)$
(6)' $\mathscr{A}$ exists a zero-polynominal which can be decompositioned into the product ofmultal vegetarian one-times factor.

Example 7.if $V$ is an-dimension linear space on complex field, $\mathscr{A}$ is a linear transformation on $V$ ,satisfied $\mathscr{A^4=4A^3+2A}$ ,then $\mathscr{A}$ can be diagonal.

Example 8.if $\mathscr{A}$ is a linear transformation on field $\mathbb{P}$ of n-dimension linear space,it has a group of bases $\alpha_1,\ldots,\alpha_n$ ,and it's matrix under this base is $A=\left(\begin{array}{ccc} &&1\\ &\ldots&\\ 1&&\end{array}\right)$ ,whether $\mathscr{A}$ can be diagonal?if it can,please calculate a group of bases of $V$ ,such that $\mathscr{A}$ become a diagonal matrix under this group of base,and write down this diagonal matrix.

Exchangeble problems and simultaneous upper triangular problem.

Example 1. $V$ is a n-dimension linear space on the complex field, $\mathscr{A,B}$ are linear transformations on $V$ ,satisfied $\mathscr{AB=BA}$ ,then $\mathscr{A,B}$ have common characteristic vectors.
Tips: $\mathscr{A,B}$ $\Rightarrow$ all the characteristic sub-space of $\mathscr{A}$ are $\mathscr{B}$ -sub-space.
we let $V_{\lambda}$ as the characteristic space of $\mathscr{A}$ $\Rightarrow$ $V_{\lambda}$ is $\mathscr{B}$ -sub-space $\Rightarrow$ $\mathscr{B}_{|V_{\lambda}}$ is a linear transformation on $V_{\lambda}$ $\Rightarrow$ $\mathscr{B}_{|V_{\lambda}}$ has a characteristic space $V_u \subseteq V_{\lambda}$ ,then for all the $\alpha\in V_u,\alpha\not=0\Rightarrow$ $\left\{ \begin{array}{c} \mathscr{B}\alpha=\mu\alpha\\ \mathscr{A}\beta=\lambda\beta \end{array}\right.$

Example 2. $V$ is a n-dimension linear space on complex field, $\mathscr{A,B}$ are linear transformations on $V$ ,satisfied $\mathscr{AB=BA}$ ,if $\mathscr{A}$ has $s$ different characteristic values,then $\mathscr{A,B}$ at least have $s$ common and linear independent characteristic vectors.

Proposition
if $A,B$ are two n-order matrix on complex field,and $AB=BA$ ,then exists a reversible matrix $P$ ,such that $P^{-1}AP$ and $P^{-1}BP$ can become the upper triangle at the same time.

Example 3.if $A,B$ are two n-order matrix on complex field,and $A$ is power zero matrix(exist a positive integer m,such that $A^m=0$ ),and $AB=BA$ ,then $|A+B|=|B|$

Example 4.if $A,B$ are n-order square, $AB=BA$ ,and $A,B$ can be diagonal,please proof $A,B$ can be diagonal at the same time.
Tips: first,we suppose exists $P$ such that $P_1^{-1}AP_1=diag\{\lambda _1E_1,\lambda_2 E_2,\ldots,\lambda_sE_s\}$ ,and from $P_1^{-1}ABP_1=P_1{-1}BAP_1$ we know $P_1^{-1}BP_1=diag\{B_1,B_2,\ldots,B_s\}$ ,and $B$ can be diagonal too,so exists a $P_2$ such that $P_2^{-1}P_1^{-1}BP_1P_2=\{\mu_1,\mu_2,\ldots,\mu_n\}$ ,and the $P$ we are looking for is $P_1P_2$

Example 5. $A,B$ are the n-order real symmetry matrix,proof:exist orthogonal matrix $T$ such that $T'AT$ and $T'BT$ are diagonal matrix at the same time if and only if $AB=BA$

Example 6. $A,B$ are n-order real symmetry matrix,then $AB=BA$ if and only if they have common vectors as a standard orthogonal base on the European space $\mathbb{R}^n$ .

5.Linear space