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


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Question Number 208292 by universe last updated on 10/Jun/24

$let T be a n \times n matrix with integral$ $entries and Q = T + \frac{1}{2} I where I denote$ $the n \times n identity matrix then prove$ $that matrix Q is invertible$
	Answered by Berbere last updated on 10/Jun/24
	$Let T∈M_n (Z) ⇒ Q ∈GL_n (R)⇒Det(Q)≠0 ⇒T+(1/2)I is injective ⇒Ker(T+(I/2))=0 ⇒∀x∈R^n −0_R^n ⇒T(x)+(x/2)≠0_R^n ⇒−(1/2)∈ C−{Spect(T)} so it sufficient To show That −(1/2) is not a root of χ_T caracteristic Polynomial of T since T∈M_n (Z) χ_T =X^n +Σ_(k=0) ^(n−1) a_k X^k ; ∀k∈[0,n−1] a_k ∈Z Supose That (1/p);p prime is root of χ⇒1+Σ_(k=0) ^(n−1) p^(n−k) a_k =0 p∣1 absurd p>2⇒(1/p) cant bee root of χ⇒−(1/2) ∉spect (T) ⇒T+(1/2).I_n is injecrive since T∈M_n (Z)⇒is surjective by dim(ker)+dim(Im(T))=n ⇒T+(I_n /2) is bijective⇒Q is inversibl in M_n (Q) not M_n (Z) Q rational number Q is smallest Filed that contain Z exemple Q= ((((3/2) 1)),((1 (3/2))) ) ;Q^− =(4/5) ((((3/2) −1)),((−1 (3/2))) )= ((((6/5) −(4/5))),((−(4/5) (6/5))) )$
	$L e t T \in M_{n} (Z)$ $\Rightarrow Q \in {G L}_{n} (R) \Rightarrow D e t (Q) \neq 0$ $\Rightarrow T + \frac{1}{2} I i s i n j e c t i v e \Rightarrow K e r (T + \frac{I}{2}) = 0$ $\Rightarrow \forall x \in R^{n} - 0_{R^{n}} \Rightarrow T (x) + \frac{x}{2} \neq 0_{R^{n}} \Rightarrow - \frac{1}{2} \in C - {S p e c t (T)}$ $s o i t s u f f i c i e n t T o s h o w T h a t$ $- \frac{1}{2} i s n o t a r o o t o f χ_{T} c a r a c t e r i s t i c P o l y n o m i a l o f T$ $s i n c e T \in M_{n} (Z)$ $χ_{T} = X^{n} + \underset{k = 0}{\sum^{n - 1}} a_{k} X^{k}; \forall k \in [0, n - 1] a_{k} \in Z$ $S u p o s e T h a t \frac{1}{p}; p p r i m e i s r o o t o f χ \Rightarrow 1 + \underset{k = 0}{\sum^{n - 1}} p^{n - k} a_{k} = 0$ $p ∣ 1 a b s u r d p > 2 \Rightarrow \frac{1}{p} c a n t b e e r o o t o f χ \Rightarrow - \frac{1}{2} \notin s p e c t (T)$ $\Rightarrow T + \frac{1}{2} . I_{n} i s i n j e c r i v e s i n c e T \in M_{n} (Z) \Rightarrow i s s u r j e c t i v e$ $b y d i m (k e r) + d i m (I m (T)) = n$ $\Rightarrow T + \frac{I_{n}}{2} i s b i j e c t i v e \Rightarrow Q i s i n v e r s i b l i n M_{n} (Q) n o t M_{n} (Z)$ $Q r a t i o n a l n u m b e r Q i s s m a l l e s t F i l e d t h a t c o n t a i n Z$ $e x e m p l e Q = (\begin{matrix} \frac{3}{2} 1 \\ 1 \frac{3}{2} \end{matrix}); Q^{-} = \frac{4}{5} (\begin{matrix} \frac{3}{2} - 1 \\ - 1 \frac{3}{2} \end{matrix}) = (\begin{matrix} \frac{6}{5} - \frac{4}{5} \\ - \frac{4}{5} \frac{6}{5} \end{matrix})$
	Commented by Philton last updated on 13/Jun/24

	$- P h i l i p O n y e a k a : A p e r f e c t s o l u t i o n, s i r .$
	Commented by Berbere last updated on 16/Jun/24

	$T h a n k Y o u$
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