Question Number 201343 by gabi last updated on 05/Dec/23
Answered by aleks041103 last updated on 05/Dec/23
![A^3 +A^2 +A=0 λ is eigenvalue ⇒λ^3 +λ^2 +λ=λ(λ^2 +λ+1)=0 ⇒λ=0,w,w^∗ where w=e^(2πi/3) and w^∗ is the complex conjugate of w. let A=PLP^( −1) , where L is the Jordan normal form of A. ⇒P(L^3 +L^2 +L)P^( −1) =0 ⇒L^3 +L^2 +L=0 where L is block−diagonal with Jordan blocks i.e. L=diag(J_(11) ,..,J_(1a) ,J_(21) ,...,J_(2b) ,J_(31) ,...,J_(3c) ) where J_(ks) is a Jordan block with k−th eigenvalue ⇒J_(ks) ^( 3) +J_(ks) ^( 2) +J_(ks) =0 J_(ks) =λ_k E_s +S_s where E_s is the identity over the s×s square matrices and S_s is the s×s square matrix with 0s everywhere, except on the top off−diagonal. ex.: J_(14) = [(λ_1 ,1,0,0),(0,λ_1 ,1,0),(0,0,λ_1 ,1),(0,0,0,λ_1 ) ] J_(14) =λ_1 [(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1) ]+ [(0,1,0,0),(0,0,1,0),(0,0,0,1),(0,0,0,0) ]= =λ_1 E_4 +S_4 It is good to note that S_s ^k =0, if k≥s and {S_s ^k }_(k=0) ^(s−1) ={E_s }∪{S_s ^k }_(k=1) ^(s−1) is lin.indep. Then J_(ks) ^( 3) +J_(ks) ^( 2) +J_(ks) =0 ⇒(λE+S)^3 +(λE+S)^2 +λE+S=0 λ^3 E+3λ^2 S+3λS^2 +S^3 +λ^2 E+2λS+S^2 +λE+S=0 ⇒(λ^3 +λ^2 +λ)E+(3λ^2 +2λ+1)S+(3λ+1)S^2 +S^3 =0 (E,S∈M_(s×s) ) if s>3: {E,S,S^2 ,S^3 }⊆{S_s ^k }_(k=0) ^(s−1) is lin.ind. ⇒ { ((λ^3 +λ^2 +λ=0)),((3λ^2 +2λ+1=0)),((3λ+1=0)),((1=0)) :} which is imposs. if s=3: S^3 =0, {E,S,S^2 } are lin.indep. ⇒ { ((λ^3 +λ^2 +λ=0)),((3λ^2 +2λ+1=0)),((3λ+1=0)) :} which is imposs. if s=2: ⇒S^3 =S^2 =0 and {E,S} are lin.ind. ⇒ { ((λ^3 +λ^2 +λ=0)),((3λ^2 +2λ+1=0)) :} 1)⇒λ=0,w,w^∗ ⇒ λ^2 =−λ−1 or λ=0 replace in 2)⇒ −3(λ+1)+2λ+1=−λ−2⇒λ=−2 but (−2)^3 +(−2)^2 +(−2)≠0 or 3(0)^2 +2(0)+1≠0 ⇒imposs. ⇒s=1 ⇒ the jordan blocks are numbers ⇒A is diagonalisable. ⇒rg(A)=dim(ker(A−wE))+dim(ker(A−w^∗ E)) A is diagon. ⇒ker(A−wE)=span({v_1 ,...,v_k }) where {v_1 ,...,v_k } are lin.ind. and Av_m =wv_m ⇒(Av_m )^∗ =A^∗ v_m ^∗ =Av_m ^∗ =(wv_m )^∗ =w^∗ v_m ^∗ where it is used that A^∗ =A, because A∈M_(n×n) (R). ⇒{v_1 ^∗ ,...,v_k ^∗ } are eigenvectors corresponding to eigenvalue w^∗ ≠w. ⇒span({v_1 ^∗ ,...,v_k ^∗ })≤ker(A−w^∗ E) let u∈ker(A−w^∗ E)/span({v_1 ^∗ ,...,v_k ^∗ }) and u≠0 then, Au=w^∗ u ⇒Au^∗ =wu^∗ ⇒u^∗ ∈span({v_1 ,...,v_k }) ⇒u∈span({v_1 ^∗ ,...,v_k ^∗ }) ⇒u∈[span({v_1 ^∗ ,...,v_k ^∗ })]∩[ker(A−w^∗ E)/span({v_1 ^∗ ,...,v_k ^∗ })]={0} ⇒u=0 ⇒ contradiction ⇒ker(A−w^∗ E)=span({v_1 ^∗ ,...,v_k ^∗ }) ⇒rg(A)=dim(ker(A−wE))+dim(ker(A−w^∗ E))= =dim(span({v_1 ,...,v_k }))+dim(span({v_1 ^∗ ,...,v_k ^∗ }))= =k+k=2k ⇒2∣rg(A)](https://www.tinkutara.com/question/Q201404.png)
$${A}^{\mathrm{3}} +{A}^{\mathrm{2}} +{A}=\mathrm{0} \\ $$$$\lambda\:{is}\:{eigenvalue} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda=\lambda\left(\lambda^{\mathrm{2}} +\lambda+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\mathrm{0},{w},{w}^{\ast} \\ $$$${where}\:{w}={e}^{\mathrm{2}\pi{i}/\mathrm{3}} \:{and}\:{w}^{\ast} \:{is}\:{the}\:{complex} \\ $$$${conjugate}\:{of}\:{w}. \\ $$$${let}\:{A}={PLP}^{\:−\mathrm{1}} ,\:{where}\:{L}\:{is}\:{the}\:{Jordan}\:{normal} \\ $$$${form}\:{of}\:{A}. \\ $$$$\Rightarrow{P}\left({L}^{\mathrm{3}} +{L}^{\mathrm{2}} +{L}\right){P}^{\:−\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow{L}^{\mathrm{3}} +{L}^{\mathrm{2}} +{L}=\mathrm{0} \\ $$$${where}\:{L}\:{is}\:{block}−{diagonal}\:{with}\:{Jordan}\:{blocks} \\ $$$${i}.{e}. \\ $$$${L}={diag}\left({J}_{\mathrm{11}} ,..,{J}_{\mathrm{1}{a}} ,{J}_{\mathrm{21}} ,…,{J}_{\mathrm{2}{b}} ,{J}_{\mathrm{31}} ,…,{J}_{\mathrm{3}{c}} \right) \\ $$$${where}\:{J}_{{ks}} \:{is}\:{a}\:{Jordan}\:{block}\:{with}\:{k}−{th} \\ $$$${eigenvalue} \\ $$$$\Rightarrow{J}_{{ks}} ^{\:\mathrm{3}} +{J}_{{ks}} ^{\:\mathrm{2}} +{J}_{{ks}} =\mathrm{0} \\ $$$${J}_{{ks}} =\lambda_{{k}} {E}_{{s}} +{S}_{{s}} \\ $$$${where}\:{E}_{{s}} \:{is}\:{the}\:{identity}\:{over}\:{the}\:{s}×{s}\:{square} \\ $$$${matrices}\:{and}\:{S}_{{s}} \:{is}\:{the}\:{s}×{s}\:{square}\:{matrix} \\ $$$${with}\:\mathrm{0}{s}\:{everywhere},\:{except}\:{on}\:{the}\:{top} \\ $$$${off}−{diagonal}. \\ $$$${ex}.: \\ $$$${J}_{\mathrm{14}} =\begin{bmatrix}{\lambda_{\mathrm{1}} }&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\lambda_{\mathrm{1}} }&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\lambda_{\mathrm{1}} }&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\lambda_{\mathrm{1}} }\end{bmatrix} \\ $$$${J}_{\mathrm{14}} =\lambda_{\mathrm{1}} \begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}+\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix}= \\ $$$$=\lambda_{\mathrm{1}} {E}_{\mathrm{4}} +{S}_{\mathrm{4}} \\ $$$$ \\ $$$${It}\:{is}\:{good}\:{to}\:{note}\:{that} \\ $$$${S}_{{s}} ^{{k}} =\mathrm{0},\:{if}\:{k}\geqslant{s} \\ $$$${and}\:\left\{{S}_{{s}} ^{{k}} \right\}_{{k}=\mathrm{0}} ^{{s}−\mathrm{1}} =\left\{{E}_{{s}} \right\}\cup\left\{{S}_{{s}} ^{{k}} \right\}_{{k}=\mathrm{1}} ^{{s}−\mathrm{1}} \:{is}\:{lin}.{indep}. \\ $$$$ \\ $$$${Then} \\ $$$${J}_{{ks}} ^{\:\mathrm{3}} +{J}_{{ks}} ^{\:\mathrm{2}} +{J}_{{ks}} =\mathrm{0} \\ $$$$\Rightarrow\left(\lambda{E}+{S}\right)^{\mathrm{3}} +\left(\lambda{E}+{S}\right)^{\mathrm{2}} +\lambda{E}+{S}=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} {E}+\mathrm{3}\lambda^{\mathrm{2}} {S}+\mathrm{3}\lambda{S}^{\mathrm{2}} +{S}^{\mathrm{3}} +\lambda^{\mathrm{2}} {E}+\mathrm{2}\lambda{S}+{S}^{\mathrm{2}} +\lambda{E}+{S}=\mathrm{0} \\ $$$$\Rightarrow\left(\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda\right){E}+\left(\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}\right){S}+\left(\mathrm{3}\lambda+\mathrm{1}\right){S}^{\mathrm{2}} +{S}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({E},{S}\in{M}_{{s}×{s}} \right) \\ $$$${if}\:{s}>\mathrm{3}: \\ $$$$\left\{{E},{S},{S}^{\mathrm{2}} ,{S}^{\mathrm{3}} \right\}\subseteq\left\{{S}_{{s}} ^{{k}} \right\}_{{k}=\mathrm{0}} ^{{s}−\mathrm{1}} \:{is}\:{lin}.{ind}. \\ $$$$\Rightarrow\begin{cases}{\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda=\mathrm{0}}\\{\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}=\mathrm{0}}\\{\mathrm{3}\lambda+\mathrm{1}=\mathrm{0}}\\{\mathrm{1}=\mathrm{0}}\end{cases}\:{which}\:{is}\:{imposs}. \\ $$$${if}\:{s}=\mathrm{3}: \\ $$$${S}^{\mathrm{3}} =\mathrm{0},\:\left\{{E},{S},{S}^{\mathrm{2}} \right\}\:{are}\:{lin}.{indep}. \\ $$$$\Rightarrow\begin{cases}{\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda=\mathrm{0}}\\{\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}=\mathrm{0}}\\{\mathrm{3}\lambda+\mathrm{1}=\mathrm{0}}\end{cases}\:{which}\:{is}\:{imposs}. \\ $$$${if}\:{s}=\mathrm{2}: \\ $$$$\Rightarrow{S}^{\mathrm{3}} ={S}^{\mathrm{2}} =\mathrm{0}\:{and}\:\left\{{E},{S}\right\}\:{are}\:{lin}.{ind}. \\ $$$$\Rightarrow\begin{cases}{\lambda^{\mathrm{3}} +\lambda^{\mathrm{2}} +\lambda=\mathrm{0}}\\{\mathrm{3}\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$\left.\:\:\:\:\:\:\:\mathrm{1}\right)\Rightarrow\lambda=\mathrm{0},{w},{w}^{\ast} \:\Rightarrow\:\lambda^{\mathrm{2}} =−\lambda−\mathrm{1}\:{or}\:\lambda=\mathrm{0} \\ $$$$\left.\:\:\:\:\:\:\:{replace}\:{in}\:\mathrm{2}\right)\Rightarrow\:−\mathrm{3}\left(\lambda+\mathrm{1}\right)+\mathrm{2}\lambda+\mathrm{1}=−\lambda−\mathrm{2}\Rightarrow\lambda=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:{but}\:\left(−\mathrm{2}\right)^{\mathrm{3}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\left(−\mathrm{2}\right)\neq\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{or}\:\mathrm{3}\left(\mathrm{0}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{0}\right)+\mathrm{1}\neq\mathrm{0} \\ $$$$\Rightarrow{imposs}. \\ $$$$ \\ $$$$\Rightarrow{s}=\mathrm{1}\:\Rightarrow\:{the}\:{jordan}\:{blocks}\:{are}\:{numbers} \\ $$$$\Rightarrow{A}\:{is}\:{diagonalisable}. \\ $$$$ \\ $$$$\Rightarrow{rg}\left({A}\right)={dim}\left({ker}\left({A}−{wE}\right)\right)+{dim}\left({ker}\left({A}−{w}^{\ast} {E}\right)\right) \\ $$$${A}\:{is}\:{diagon}. \\ $$$$\Rightarrow{ker}\left({A}−{wE}\right)={span}\left(\left\{{v}_{\mathrm{1}} ,…,{v}_{{k}} \right\}\right) \\ $$$${where}\:\left\{{v}_{\mathrm{1}} ,…,{v}_{{k}} \right\}\:{are}\:{lin}.{ind}.\:{and}\: \\ $$$${Av}_{{m}} ={wv}_{{m}} \\ $$$$\Rightarrow\left({Av}_{{m}} \right)^{\ast} ={A}^{\ast} {v}_{{m}} ^{\ast} ={Av}_{{m}} ^{\ast} =\left({wv}_{{m}} \right)^{\ast} ={w}^{\ast} {v}_{{m}} ^{\ast} \\ $$$${where}\:{it}\:{is}\:{used}\:{that}\:{A}^{\ast} ={A},\:{because} \\ $$$${A}\in{M}_{{n}×{n}} \left(\mathbb{R}\right). \\ $$$$\Rightarrow\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\:{are}\:{eigenvectors}\:{corresponding} \\ $$$${to}\:{eigenvalue}\:{w}^{\ast} \neq{w}. \\ $$$$\Rightarrow{span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right)\leqslant{ker}\left({A}−{w}^{\ast} {E}\right) \\ $$$${let}\:{u}\in{ker}\left({A}−{w}^{\ast} {E}\right)/{span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right)\:{and} \\ $$$${u}\neq\mathrm{0} \\ $$$${then},\:{Au}={w}^{\ast} {u} \\ $$$$\Rightarrow{Au}^{\ast} ={wu}^{\ast} \Rightarrow{u}^{\ast} \in{span}\left(\left\{{v}_{\mathrm{1}} ,…,{v}_{{k}} \right\}\right) \\ $$$$\Rightarrow{u}\in{span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right) \\ $$$$\Rightarrow{u}\in\left[{span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right)\right]\cap\left[{ker}\left({A}−{w}^{\ast} {E}\right)/{span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right)\right]=\left\{\mathrm{0}\right\} \\ $$$$\Rightarrow{u}=\mathrm{0} \\ $$$$\Rightarrow\:{contradiction} \\ $$$$ \\ $$$$\Rightarrow{ker}\left({A}−{w}^{\ast} {E}\right)={span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right) \\ $$$$ \\ $$$$\Rightarrow{rg}\left({A}\right)={dim}\left({ker}\left({A}−{wE}\right)\right)+{dim}\left({ker}\left({A}−{w}^{\ast} {E}\right)\right)= \\ $$$$={dim}\left({span}\left(\left\{{v}_{\mathrm{1}} ,…,{v}_{{k}} \right\}\right)\right)+{dim}\left({span}\left(\left\{{v}_{\mathrm{1}} ^{\ast} ,…,{v}_{{k}} ^{\ast} \right\}\right)\right)= \\ $$$$={k}+{k}=\mathrm{2}{k} \\ $$$$\Rightarrow\mathrm{2}\mid{rg}\left({A}\right) \\ $$
Commented by aleks041103 last updated on 05/Dec/23
$${We}\:{could}'{ve}\:{solved}\:{it}\:{in}\:{an}\:{even}\:{easier}\:{manner} \\ $$$${let}\:{the}\:{minimal}\:{polynomial}\:{of}\:{A}\:{be}\:{m}_{{A}} \left({x}\right). \\ $$$$\left({by}\:{def}\:{m}_{{A}} \left({x}\right)\:{is}\:{the}\:{lowest}\:{order}\:{polynomial}\right. \\ $$$$\left.{that}\:{has}\:{the}\:{property}\:{m}_{{A}} \left({A}\right)=\mathrm{0}\right) \\ $$$${since}\:{f}\left({A}\right)={A}^{\mathrm{3}} +{A}^{\mathrm{2}} +{A}=\mathrm{0},\:{then}\:{m}_{{A}} \mid{f}. \\ $$$${the}\:{spectrum}\:{of}\:{f}\left({x}\right)=\mathrm{0}\:{is}\:{x}=\mathrm{0},{w},{w}^{\ast} . \\ $$$$\Rightarrow{f}\:{has}\:{a}\:{simple}\:{spectrum}. \\ $$$${since}\:{m}_{{A}} \mid{f}\Rightarrow{m}_{{A}} \:{has}\:{a}\:{simple}\:{spectrum}\:{too}. \\ $$$${There}\:{is}\:{a}\:{theorem},\:{which}\:{we}\:{essentially} \\ $$$${proved}\:{for}\:{this}\:{specific}\:{case},\:{that}\:{states} \\ $$$${that}\:{the}\:{matrix}\:{A}\:{is}\:{diagonalizable} \\ $$$${if}\:{and}\:{only}\:{if}\:{the}\:{minimal}\:{polynomial} \\ $$$${of}\:{A}\:{has}\:{simple}\:{spectrum}. \\ $$$${Thus},\:{since}\:{m}_{{A}} \:{has}\:{simple}\:{spectrum} \\ $$$${then}\:{A}\:{is}\:{diahlgonalisable}. \\ $$$${From}\:{here}\:{on}\:{forward}\:{it}\:{is}\:{the}\:{same}. \\ $$