Real n-by-n Matrices…












5














Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question






















  • Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    – A.Γ.
    Jan 1 at 20:44
















5














Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question






















  • Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    – A.Γ.
    Jan 1 at 20:44














5












5








5







Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces







share|cite|improve this question













Let $M_n(mathbb{R})$ denote the vector space of real $ntimes n$ matrics, and let $A in M_n(mathbb{R})$.



Part (a) of this question says: Suppose $B in M_n(mathbb{R})$ such that $AB = I_n$ (the $n times n$ identity matrix. If $C in M_n(mathbb{R})$ such that $CA = 0$, then prove $C = 0$.



I have already proven part (a).



Part (b) asks: Assume there exists a least positive integer $m$ such that $t_0I + t_1A + dots + t_mA^m = 0$ for some $t_0, dots, t_m in mathbb{R}$ with $t_m neq 0$. Also, suppose that $AB = I_n$ for some $B in M_n(mathbb{R})$.Prove that $t_0 neq 0$. (Hint: Use the result from part (a)).



My idea is to use induction on $m$. That is, suppose



$$t_0I = 0.$$



But this implies that $t_0 = 0$ since $I$ is the identity. But we could see this as



begin{align*}
t_0I &= 0
\
t_0AB &= 0
\
t_0CAB &= 0
\
t_0C &= 0.
end{align*}



But I don't think this is the right approach. So I have a few questions: (i) Is this the correct approach? (ii) How do I use the result from part (a) properly? (iii) What's a good resource for these types of questions? The book that I am using is "Linear Algebra Done Right by Sheldon Axler".



I am studying for my linear algebra comp in a few weeks so any help is appreciated!






linear-algebra matrices vector-spaces




linear-algebra matrices vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 20:25









Taylor McMillan

433




433












  • Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    – A.Γ.
    Jan 1 at 20:44


















  • Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
    – A.Γ.
    Jan 1 at 20:44
















Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
– A.Γ.
Jan 1 at 20:44




Once you have got $t_0=0$ then anything you multiply it with gonna be zero. Since we assume $t_mne 0$ then $mge 1$.
– A.Γ.
Jan 1 at 20:44










3 Answers
3






active

oldest

votes


















7














Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
$$
t_0 I + t_1 A + cdots + t_m A^m = 0
$$



If $t_0=0$ you have,
$$
t_1 A + cdots + t_m A^m = 0
$$



Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
$$
t_0'I + t_1'A + cdots t_n ' A^n = 0
$$
where $n<m$.



Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






share|cite|improve this answer





















  • Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    – Taylor McMillan
    Jan 1 at 21:05










  • That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    Jan 1 at 21:07



















4














Hint: assume $t_0=0$ then
$$
t_1A+ldots+t_mA^m=0.
$$
Multiply by $B$ and get the contradiction with the minimality of $m$.






share|cite|improve this answer





























    1














    Here is another hint/solution:




    • assume it to equal $0$,

    • multiply your equation by $A$, and factor by $A$,

    • using part (a), get a contradiction.






    share|cite|improve this answer










    New contributor




    Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


















    • Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
      – timtfj
      Jan 1 at 22:18











    Your Answer





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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    7














    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer





















    • Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      – Taylor McMillan
      Jan 1 at 21:05










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      Jan 1 at 21:07
















    7














    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer





















    • Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      – Taylor McMillan
      Jan 1 at 21:05










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      Jan 1 at 21:07














    7












    7








    7






    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.






    share|cite|improve this answer












    Suppose exists $B$ such that $AB=I$ and that $m$ is the smallest integer such that, exist coefficients $t_i$ such that,
    $$
    t_0 I + t_1 A + cdots + t_m A^m = 0
    $$



    If $t_0=0$ you have,
    $$
    t_1 A + cdots + t_m A^m = 0
    $$



    Now, see if you can use the fact that $AB = I$ to find new coefficients $t_i'$ satisfying
    $$
    t_0'I + t_1'A + cdots t_n ' A^n = 0
    $$
    where $n<m$.



    Since by assumption $m$ is the smallest integer such that this type of combination exists, then you have a contradiction.



    I liked that book as a first course in linear algebra. You could also see if past exam problems are available to study from. To me, studying for exams is somewhat dependent on the format of exam itself so its hard to give general advice.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 1 at 20:46









    tch

    584210




    584210












    • Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      – Taylor McMillan
      Jan 1 at 21:05










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      Jan 1 at 21:07


















    • Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
      – Taylor McMillan
      Jan 1 at 21:05










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      Jan 1 at 21:07
















    Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    – Taylor McMillan
    Jan 1 at 21:05




    Thank you for this. My goto for using contradiction is typically when the problem screams "this is obviously true". Didn't really think that this result was intuitive so I didn't take this route. When in doubt try it out, I guess.
    – Taylor McMillan
    Jan 1 at 21:05












    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    Jan 1 at 21:07




    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    Jan 1 at 21:07











    4














    Hint: assume $t_0=0$ then
    $$
    t_1A+ldots+t_mA^m=0.
    $$
    Multiply by $B$ and get the contradiction with the minimality of $m$.






    share|cite|improve this answer


























      4














      Hint: assume $t_0=0$ then
      $$
      t_1A+ldots+t_mA^m=0.
      $$
      Multiply by $B$ and get the contradiction with the minimality of $m$.






      share|cite|improve this answer
























        4












        4








        4






        Hint: assume $t_0=0$ then
        $$
        t_1A+ldots+t_mA^m=0.
        $$
        Multiply by $B$ and get the contradiction with the minimality of $m$.






        share|cite|improve this answer












        Hint: assume $t_0=0$ then
        $$
        t_1A+ldots+t_mA^m=0.
        $$
        Multiply by $B$ and get the contradiction with the minimality of $m$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 1 at 20:46









        A.Γ.

        22.5k32656




        22.5k32656























            1














            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer










            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















            • Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              – timtfj
              Jan 1 at 22:18
















            1














            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer










            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.


















            • Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              – timtfj
              Jan 1 at 22:18














            1












            1








            1






            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.






            share|cite|improve this answer










            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            Here is another hint/solution:




            • assume it to equal $0$,

            • multiply your equation by $A$, and factor by $A$,

            • using part (a), get a contradiction.







            share|cite|improve this answer










            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 1 at 22:14









            amWhy

            192k28224439




            192k28224439






            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered Jan 1 at 21:25









            Cauchy is my master

            112




            112




            New contributor




            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





            New contributor





            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            Cauchy is my master is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.












            • Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              – timtfj
              Jan 1 at 22:18


















            • Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
              – timtfj
              Jan 1 at 22:18
















            Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
            – timtfj
            Jan 1 at 22:18




            Bullet lists need a blank line before them in order to appear—I've added one. Welcome to the site!
            – timtfj
            Jan 1 at 22:18


















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            8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
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            Алькесар

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            Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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