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 \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

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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