Question Number 208292 by universe last updated on 10/Jun/24
$$\:\mathrm{let}\:\mathrm{T}\:\mathrm{be}\:\mathrm{a}\:{n}×{n}\:\mathrm{matrix}\:\mathrm{with}\:\mathrm{integral}\: \\ $$$$\:\mathrm{entries}\:\mathrm{and}\:\:\mathrm{Q}\:=\:\mathrm{T}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\:\:\:\mathrm{where}\:\mathrm{I}\:\mathrm{denote} \\ $$$$\:\:\mathrm{the}\:\mathrm{n}×\mathrm{n}\:\mathrm{identity}\:\mathrm{matrix}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\mathrm{that}\:\mathrm{matrix}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{invertible} \\ $$
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))) )](https://www.tinkutara.com/question/Q208304.png)
$$\:{Let}\:{T}\in{M}_{{n}} \left(\mathbb{Z}\right) \\ $$$$\Rightarrow\:{Q}\:\in{GL}_{{n}} \left(\mathbb{R}\right)\Rightarrow{Det}\left({Q}\right)\neq\mathrm{0} \\ $$$$\Rightarrow{T}+\frac{\mathrm{1}}{\mathrm{2}}{I}\:{is}\:{injective}\:\:\Rightarrow{Ker}\left({T}+\frac{{I}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\forall{x}\in\mathbb{R}^{{n}} −\mathrm{0}_{{R}^{{n}} } \Rightarrow{T}\left({x}\right)+\frac{{x}}{\mathrm{2}}\neq\mathrm{0}_{\mathbb{R}^{{n}} } \Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\in\:{C}−\left\{{Spect}\left({T}\right)\right\} \\ $$$${so}\:{it}\:{sufficient}\:{To}\:{show}\:{That} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\:{is}\:{not}\:{a}\:{root}\:{of}\:\chi_{{T}} \:{caracteristic}\:{Polynomial}\:{of}\:{T} \\ $$$${since}\:{T}\in{M}_{{n}} \left({Z}\right) \\ $$$$\chi_{{T}} ={X}^{{n}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{k}} {X}^{{k}} ;\:\forall{k}\in\left[\mathrm{0},{n}−\mathrm{1}\right]\:{a}_{{k}} \in\mathbb{Z} \\ $$$${Supose}\:{That}\:\frac{\mathrm{1}}{{p}};{p}\:{prime}\:{is}\:{root}\:{of}\:\chi\Rightarrow\mathrm{1}+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{p}^{{n}−{k}} {a}_{{k}} =\mathrm{0} \\ $$$${p}\mid\mathrm{1}\:{absurd}\:{p}>\mathrm{2}\Rightarrow\frac{\mathrm{1}}{{p}}\:{cant}\:{bee}\:{root}\:{of}\:\chi\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\:\notin{spect}\:\left({T}\right) \\ $$$$\Rightarrow{T}+\frac{\mathrm{1}}{\mathrm{2}}.{I}_{{n}} \:{is}\:{injecrive}\:{since}\:{T}\in{M}_{{n}} \left({Z}\right)\Rightarrow{is}\:{surjective} \\ $$$${by}\:{dim}\left({ker}\right)+{dim}\left({Im}\left({T}\right)\right)={n} \\ $$$$\Rightarrow{T}+\frac{{I}_{{n}} }{\mathrm{2}}\:{is}\:{bijective}\Rightarrow{Q}\:{is}\:{inversibl}\:{in}\:{M}_{{n}} \left({Q}\right)\:{not}\:{M}_{{n}} \left({Z}\right) \\ $$$${Q}\:{rational}\:{number}\:{Q}\:{is}\:{smallest}\:{Filed}\:{that}\:{contain}\:\mathbb{Z} \\ $$$${exemple}\:{Q}=\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\end{pmatrix}\:;{Q}^{−} =\frac{\mathrm{4}}{\mathrm{5}}\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}}\end{pmatrix}=\begin{pmatrix}{\frac{\mathrm{6}}{\mathrm{5}}\:\:\:−\frac{\mathrm{4}}{\mathrm{5}}}\\{−\frac{\mathrm{4}}{\mathrm{5}}\:\:\:\:\frac{\mathrm{6}}{\mathrm{5}}}\end{pmatrix} \\ $$
Commented by Philton last updated on 13/Jun/24
$$\:\:\:\:\:−{Philip}\:{Onyeaka}\::\:{A}\:{perfect}\:{solution},{sir}. \\ $$
Commented by Berbere last updated on 16/Jun/24
$${Thank}\:{You} \\ $$