Real n-by-n Matrices…












2














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.
    1 hour ago
















2














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.
    1 hour ago














2












2








2







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 1 hour ago









Taylor McMillan

203




203












  • 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.
    1 hour ago


















  • 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.
    1 hour ago
















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.
1 hour ago




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.
1 hour ago










3 Answers
3






active

oldest

votes


















3














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
    51 mins ago










  • That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    49 mins ago



















2














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 to = 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.


















    • I like this too, thanks!
      – Taylor McMillan
      19 mins ago











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    3














    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
      51 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      49 mins ago
















    3














    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
      51 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      49 mins ago














    3












    3








    3






    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 1 hour ago









    tch

    429210




    429210












    • 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
      51 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      49 mins ago


















    • 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
      51 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      49 mins ago
















    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
    51 mins ago




    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
    51 mins ago












    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    49 mins ago




    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    49 mins ago











    2














    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


























      2














      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
























        2












        2








        2






        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 1 hour ago









        A.Γ.

        22k22455




        22k22455























            1














            Here is another Hint/solution.
            - assume to = 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.


















            • I like this too, thanks!
              – Taylor McMillan
              19 mins ago
















            1














            Here is another Hint/solution.
            - assume to = 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.


















            • I like this too, thanks!
              – Taylor McMillan
              19 mins ago














            1












            1








            1






            Here is another Hint/solution.
            - assume to = 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 to = 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






            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 32 mins ago









            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.












            • I like this too, thanks!
              – Taylor McMillan
              19 mins ago


















            • I like this too, thanks!
              – Taylor McMillan
              19 mins ago
















            I like this too, thanks!
            – Taylor McMillan
            19 mins ago




            I like this too, thanks!
            – Taylor McMillan
            19 mins ago


















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            Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
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