单阵和谱分解

矩阵可对角化的充要条件:n阶矩阵A相似于对角矩阵的充要条件是A有n个无关的特征向量。

证明:

必要性:证${P^{ - 1}}AP = D = left[ {egin{array}{*{20}{c}}{{lambda _1}}&0&0&0\0&{{lambda _2}}&0&0\0&0&{...}&0\0&0&0&{{lambda _n}}end{array}} ight] Rightarrow P = [egin{array}{*{20}{c}}{{X_1}}&{{X_2}}&{...}&{{X_n}}end{array}]$

[{P^{ - 1}}AP = D Rightarrow AP = PD Rightarrow Aleft[ {egin{array}{*{20}{c}}
{{alpha _1}}&{{alpha _2}}&{...}&{{alpha _n}}
end{array}} ight] = left[ {egin{array}{*{20}{c}}
{{alpha _1}}&{{alpha _2}}&{...}&{{alpha _n}}
end{array}} ight]D Rightarrow left[ {Aegin{array}{*{20}{c}}
{{alpha _1}}&{A{alpha _2}}&{...}&{A{alpha _n}}
end{array}} ight] = left[ {egin{array}{*{20}{c}}
{{lambda _1}{alpha _1}}&{{lambda _2}{alpha _2}}&{...}&{{lambda _n}{alpha _n}}
end{array}} ight]]

因为P可逆,则P的各列之间线性无关,并且$alpha_1,alpha_2,...,alpha_n$为特征向量,所以A有n个线性无关的特征向量。

单阵:$A = {A_{n imes n}}$为单阵$Leftrightarrow A sim D = left[ {egin{array}{*{20}{c}}{{lambda _1}}&0&0&0\0&{{lambda _2}}&0&0\0&0&{...}&0\0&0&0&{{lambda _n}}end{array}} ight]$$Leftrightarrow {P^{ - 1}}AP = D = left[ {egin{array}{*{20}{c}}{{lambda _1}}&0&0&0\0&{{lambda _2}}&0&0\0&0&{...}&0\0&0&0&{{lambda _n}}end{array}} ight]$$Leftrightarrow P = ({X_1},{X_2},...,{X_n})$,$X_k$均为无关特征向量

单阵判定定理:

(1)$A=A_{n imes n}$为单阵$ Leftrightarrow $A有n个无关的特征向量

(2)$A=A_{n imes n}$为单阵$ Leftrightarrow $A的每个k重根恰有k个无关特征向量

用法:A的任一一个k重根(k>1),$lambda in lambda (A)$:

(i)若$rank(A-lambda I)=n-k$,则A为单阵

(ii)若$rank(A-lambda I)  e n-k$,则A不为单阵

(3)若方阵A有n个互异根,则A必为单阵

(4)

(i)设A有k个互异根$ lambda_1, lambda _2,...,lambda _k (k le n)$,若$(A-lambda _1)(A-lambda _2)...(A-lambda _k) = 0 $,则A为单阵

(ii)设A有k个互异根$ lambda_1, lambda _2,...,lambda _k (k le n)$,若$(A-lambda _1)(A-lambda _2)...(A-lambda _k) e 0 $,则A不为单阵

(5)若$f(x)$无重根,且f(A)=0,则A为单阵

单阵谱分解:

A为单阵,则有

[{P^{ - 1}}AP = D = left[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

[D = {lambda _1}left[ {egin{array}{*{20}{c}}
1&0&0&0\
0&0&0&0\
0&0&{...}&0\
0&0&0&0
end{array}} ight] + {lambda _2}left[ {egin{array}{*{20}{c}}
0&0&0&0\
0&1&0&0\
0&0&{...}&0\
0&0&0&0
end{array}} ight] + ... + {lambda _n}left[ {egin{array}{*{20}{c}}
0&0&0&0\
0&0&0&0\
0&0&{...}&0\
0&0&0&1
end{array}} ight] = {lambda _1}{E_1} + {lambda _2}{E_2} + ... + {lambda _n}{E_n}]

由$P^{-1}AP=D Rightarrow  A=PDP^{-1}$

 [A = {P^{ - 1}}DP = {P^{ - 1}}({lambda _1}{E_1} + {lambda _2}{E_2} + ... + {lambda _n}{E_n})P = {lambda _1}{P^{ - 1}}{E_1}P + {lambda _2}{P^{ - 1}}{E_2}P + ... + {lambda _n}{P^{ - 1}}{E_n}P]

令$F_{i}=PE_{i}P^{-1}$

[A = {lambda _1}{F_1} + {lambda _2}{F_2} + ... + {lambda _n}{F_n}]

单阵谱分解公式:

若$A=A_{n imes n}$为单阵,全体不同根为$t_{1},t_{2},...,t_{k}$:

则有:

[A = {t_1}{G_1} + {t_2}{G_2} + ... + {t_k}{G_k}]

$G_{i}=lambda_i P sum {E_{k}}  P^{-1}=t_i sum{F_k}$

$G_i$的性质:

(1)$G_1+...+G_k=I$

(2)$G_{i}G_{j}=0$

(3)$G_i^{2}=G_{i}$,$G_{1},G_{2},...,G_{k}$称为“谱阵”

(4)$A^p=t_1^{p}G_{1}+t_2^{p}G_{2}+...+t_k^{p}G_{k}$

(5)任意多项式:$f(x)=c_0+c_1x+...=c_px^p$有:

[f(A) = f({t_1}){G_1} + f({t_2}){G_2} + ... + f({t_k}){G_k}]

求解$G_i$:

A为单阵:

[f(x) = (x - {t_1})(x - {t_2})...(x - {t_k}) Rightarrow f(A) = (A - {t_1})(A - {t_2})...(A - {t_k}) = 0]

[f(x) = (x - {t_2})...(x - {t_k}) Rightarrow f(A) = (A - {t_2})...(A - {t_k}) = f({t_1}){G_1} + f({t_2}){G_2} + ... + f({t_k}){G_k}]

其中$f(t_2)=...=f(t_k)=0$,所以:

[f(A) = (A - {t_2})...(A - {t_k}) = f({t_1}){G_1} Rightarrow {G_1} = frac{{f(A)}}{{f({t_1})}} = frac{{(A - {t_2})...(A - {t_k})}}{{f({t_1})}}]

同理可求得$G_2,...,G_k$

谱分解的平方根:若$A ge 0$且有谱公式$A = {lambda _1}{G_1} + {lambda _2}{G_2} + ... + {lambda _k}{G_k}$,则有平方根公式:

[sqrt A  = sqrt {{lambda _1}} {G_1} + sqrt {{lambda _2}} {G_2} + ... + sqrt {{lambda _k}} {G_k}]

逆谱公式:若$A为单阵,且全体不同根非0,谱公式$A = {lambda _1}{G_1} + {lambda _2}{G_2} + ... + {lambda _k}{G_k}$,则有逆谱根公式:

[{A^{ - 1}} = lambda _1^{ - 1}{G_1} + lambda _2^{ - 1}{G_2} + ... + lambda _k^{ - 1}{G_k}]

其他性质:

(1)$AG_i=lambda_iG_i$

(2)

引理:若$AP=tP$,则P中各列都是t的特征向量

证明:$A=[X_1,X_2,...,X_n]$

[AP = left[ {egin{array}{*{20}{c}}
{A{X_1}}&{A{X_2}}&{...}&{A{X_n}}
end{array}} ight] = left[ {egin{array}{*{20}{c}}
{t{X_1}}&{t{X_2}}&{...}&{t{X_n}}
end{array}} ight] = tP]

若$(A-t)P=0$,则P中各列都是t的特征向量

(3)谱公式的全体谱阵的列都是特征向量,分别是$lambda_1,lambda_2,...,lambda_k$的特征向量

(4)A为单阵,谱阵$G_1,G_2,...,G_k$中恰有n个无关的特征向量。

证明:即证明$rank(G_1,G_2,...,G_k)=n$即可

$rank(mn) le min(rank(m),rank(n))$

$rank(A_{m imes n}) le min(m,n)$

[left[ {egin{array}{*{20}{c}}
{{G_1}}&{{G_2}}&{...}&{{G_k}}
end{array}} ight]left[ {egin{array}{*{20}{c}}
{{I_1}}\
{egin{array}{*{20}{c}}
{{I_2}}\
{...}
end{array}}\
{{I_k}}
end{array}} ight] = {G_1} + ... + {G_k} = {I_{n imes n}}]

[rank({I_{n imes n}}) = n le rank(left[ {egin{array}{*{20}{c}}
{{G_1}}&{{G_2}}&{...}&{{G_k}}
end{array}} ight]) le min (n,nk) = n Rightarrow rank(left[ {egin{array}{*{20}{c}}
{{G_1}}&{{G_2}}&{...}&{{G_k}}
end{array}} ight] = n]

遗传公式:

(1)$f(x)$为任意多项式

[lambda (A) = { {lambda _1},{lambda _2},...,{lambda _n}}  Rightarrow lambda [f(A)] = { f({lambda _1}),f({lambda _2}),...,f({lambda _n})} ]

(2)若A有n个特征向量$x_1,x_2,...,x_n$,则f(A)也有相同的特征向量$x_1,x_2,...,x_n$,且$f(A)x_1=f(lambda_1)x_1,f(A)x_1=f(lambda_1)x_2,...,f(A)x_1=f(lambda_1)x_n$

Cayley-Hamilton定理:$A=A_{n imes n}$,A的特征多项式为$f(lambda ) = left| {lambda I - A} ight|$,必有$f(A)=0$

零化式:根据Cayley-Hamilton定理,矩阵A必然存在函数$f(x)$,使得f(A)=0,称f(x)为A的零化式,也称多项式f(x)以A为根。

最小零化式(极小式或者最小多项式):固定$A=A_{n imes n}$,次数最低的首项系数为1的那个多项式$f(A)$称为A的最小零化式,记作$m_A(x)$。

极小式性质:

(1)矩阵A的极小式是唯一的

(2)A的极小式为$m_A(lambda)$,且$f(A)=0$(即f(x)以A为根),则$m_A(x)|f(x)$,这里的"|"符号为整除符号。

(3)A的极小式是A的特征多项式的一个因子,也可记为“${m_A}(x)|left| {lambda I - A} ight|$”。

(4)A的极小式和A的特征多项式$f(x)$有相同的根。

证明:

  由(3)得到,$m_A(x)$的根一定是$f(x)$的根。

  下面证明$f(x)$的根一定是$m_A(x)$的根,也就是证明A的特征根都是$m_A(x)$的根:

  任意A的特征根$lambda_i$和$x_i$,根据$lambda [f(A)] = { f({lambda _1}),...,f({lambda _n})} $有

  ${m_A}(A){X_i} = {m_A}({lambda _i}){X_i}$,因为$m_A(A)=0$则有:

  ${m_A}(A){X_i} = {m_A}({lambda _i}){X_i} = 0$

  因为$X_i$不为0,所以${m_A}({lambda _i})=0$,也就是任意A的根$lambda_i$为最小式的根,得证。

(5)相似的方阵具有相同的最小多项式。

(6)A为分块矩阵:

[A = left[ {egin{array}{*{20}{c}}
{{A_1}}&0\
0&{{A_2}}
end{array}} ight]]

则A的极小式为$A_1$的极小式和$A_2$的极小式的最小公倍数。

原文地址:https://www.cnblogs.com/codeDog123/p/10210875.html