Real n-by-n Matrices…












1














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.
    37 mins ago
















1














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.
    37 mins ago














1












1








1







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






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asked 56 mins ago









Taylor McMillan

153




153












  • 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.
    37 mins 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.
    37 mins 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.
37 mins 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.
37 mins ago










2 Answers
2






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
    16 mins ago










  • That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    14 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





















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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

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    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
      16 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      14 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
      16 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      14 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 35 mins 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
      16 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      14 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
      16 mins ago










    • That is what I am doing. My university has several past exams posted I am doing all of those problems.
      – Taylor McMillan
      14 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
    16 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
    16 mins ago












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




    That is what I am doing. My university has several past exams posted I am doing all of those problems.
    – Taylor McMillan
    14 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 35 mins ago









        A.Γ.

        22k22455




        22k22455






























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