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let-T-be-a-n-n-matrix-with-integral-entries-and-Q-T-1-2-I-where-I-denote-the-n-n-identity-matrix-then-prove-that-matrix-Q-is-invertible-




Question Number 208292 by universe last updated on 10/Jun/24
 let T be a n×n matrix with integral    entries and  Q = T + (1/2)I   where I denote    the n×n identity matrix then prove    that matrix Q is invertible
letTbean×nmatrixwithintegralentriesandQ=T+12IwhereIdenotethen×nidentitymatrixthenprovethatmatrixQisinvertible
Answered by Berbere last updated on 10/Jun/24
 Let T∈M_n (Z)  ⇒ Q ∈GL_n (R)⇒Det(Q)≠0  ⇒T+(1/2)I is injective  ⇒Ker(T+(I/2))=0  ⇒∀x∈R^n −0_R^n  ⇒T(x)+(x/2)≠0_R^n  ⇒−(1/2)∈ C−{Spect(T)}  so it sufficient To show That  −(1/2) is not a root of χ_T  caracteristic Polynomial of T  since T∈M_n (Z)  χ_T =X^n +Σ_(k=0) ^(n−1) a_k X^k ; ∀k∈[0,n−1] a_k ∈Z  Supose That (1/p);p prime is root of χ⇒1+Σ_(k=0) ^(n−1) p^(n−k) a_k =0  p∣1 absurd p>2⇒(1/p) cant bee root of χ⇒−(1/2) ∉spect (T)  ⇒T+(1/2).I_n  is injecrive since T∈M_n (Z)⇒is surjective  by dim(ker)+dim(Im(T))=n  ⇒T+(I_n /2) is bijective⇒Q is inversibl in M_n (Q) not M_n (Z)  Q rational number Q is smallest Filed that contain Z  exemple Q= ((((3/2)   1)),((1     (3/2))) ) ;Q^− =(4/5) ((((3/2)     −1)),((−1     (3/2))) )= ((((6/5)   −(4/5))),((−(4/5)    (6/5))) )
LetTMn(Z)QGLn(R)Det(Q)0T+12IisinjectiveKer(T+I2)=0xRn0RnT(x)+x20Rn12C{Spect(T)}soitsufficientToshowThat12isnotarootofχTcaracteristicPolynomialofTsinceTMn(Z)χT=Xn+n1k=0akXk;k[0,n1]akZSuposeThat1p;pprimeisrootofχ1+n1k=0pnkak=0p1absurdp>21pcantbeerootofχ12spect(T)T+12.InisinjecrivesinceTMn(Z)issurjectivebydim(ker)+dim(Im(T))=nT+In2isbijectiveQisinversiblinMn(Q)notMn(Z)QrationalnumberQissmallestFiledthatcontainZexempleQ=(321132);Q=45(321132)=(65454565)
Commented by Philton last updated on 13/Jun/24
     −Philip Onyeaka : A perfect solution,sir.
PhilipOnyeaka:Aperfectsolution,sir.
Commented by Berbere last updated on 16/Jun/24
Thank You
ThankYou

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