Matrix demonstration $A^k$












1












$begingroup$


Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $



Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far? What approach do you think will work here?
    $endgroup$
    – user3482749
    Jan 13 at 0:37






  • 2




    $begingroup$
    Have you tried proof by induction?
    $endgroup$
    – user7530
    Jan 13 at 0:37










  • $begingroup$
    yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
    $endgroup$
    – PTSONIC
    Jan 13 at 0:40






  • 1




    $begingroup$
    Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
    $endgroup$
    – Git Gud
    Jan 13 at 0:46










  • $begingroup$
    yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
    $endgroup$
    – PTSONIC
    Jan 13 at 1:01
















1












$begingroup$


Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $



Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried so far? What approach do you think will work here?
    $endgroup$
    – user3482749
    Jan 13 at 0:37






  • 2




    $begingroup$
    Have you tried proof by induction?
    $endgroup$
    – user7530
    Jan 13 at 0:37










  • $begingroup$
    yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
    $endgroup$
    – PTSONIC
    Jan 13 at 0:40






  • 1




    $begingroup$
    Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
    $endgroup$
    – Git Gud
    Jan 13 at 0:46










  • $begingroup$
    yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
    $endgroup$
    – PTSONIC
    Jan 13 at 1:01














1












1








1


1



$begingroup$


Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $



Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$






linear-algebra







share|cite|improve this question











$endgroup$




Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $



Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 0:36









Robert Lewis

49k23168




49k23168










asked Jan 13 at 0:36









PTSONICPTSONIC

83




83












  • $begingroup$
    What have you tried so far? What approach do you think will work here?
    $endgroup$
    – user3482749
    Jan 13 at 0:37






  • 2




    $begingroup$
    Have you tried proof by induction?
    $endgroup$
    – user7530
    Jan 13 at 0:37










  • $begingroup$
    yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
    $endgroup$
    – PTSONIC
    Jan 13 at 0:40






  • 1




    $begingroup$
    Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
    $endgroup$
    – Git Gud
    Jan 13 at 0:46










  • $begingroup$
    yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
    $endgroup$
    – PTSONIC
    Jan 13 at 1:01


















  • $begingroup$
    What have you tried so far? What approach do you think will work here?
    $endgroup$
    – user3482749
    Jan 13 at 0:37






  • 2




    $begingroup$
    Have you tried proof by induction?
    $endgroup$
    – user7530
    Jan 13 at 0:37










  • $begingroup$
    yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
    $endgroup$
    – PTSONIC
    Jan 13 at 0:40






  • 1




    $begingroup$
    Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
    $endgroup$
    – Git Gud
    Jan 13 at 0:46










  • $begingroup$
    yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
    $endgroup$
    – PTSONIC
    Jan 13 at 1:01
















$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37




$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37




2




2




$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37




$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37












$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40




$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40




1




1




$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46




$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46












$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01




$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01










2 Answers
2






active

oldest

votes


















2












$begingroup$

Try a simple induction on $k$; it is clear that for $k = 1$,



$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$



then assuming that for some $k$



$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$



we find



$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$



which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$



    How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Very nice indeed, endorsed, +1!!!
      $endgroup$
      – Robert Lewis
      Jan 13 at 1:47












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Try a simple induction on $k$; it is clear that for $k = 1$,



    $A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$



    then assuming that for some $k$



    $A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$



    we find



    $A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
    $= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$



    which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Try a simple induction on $k$; it is clear that for $k = 1$,



      $A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$



      then assuming that for some $k$



      $A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$



      we find



      $A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
      $= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$



      which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Try a simple induction on $k$; it is clear that for $k = 1$,



        $A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$



        then assuming that for some $k$



        $A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$



        we find



        $A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
        $= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$



        which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.






        share|cite|improve this answer









        $endgroup$



        Try a simple induction on $k$; it is clear that for $k = 1$,



        $A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$



        then assuming that for some $k$



        $A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$



        we find



        $A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
        $= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$



        which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 0:56









        Robert LewisRobert Lewis

        49k23168




        49k23168























            2












            $begingroup$

            Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$



            How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very nice indeed, endorsed, +1!!!
              $endgroup$
              – Robert Lewis
              Jan 13 at 1:47
















            2












            $begingroup$

            Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$



            How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Very nice indeed, endorsed, +1!!!
              $endgroup$
              – Robert Lewis
              Jan 13 at 1:47














            2












            2








            2





            $begingroup$

            Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$



            How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.






            share|cite|improve this answer











            $endgroup$



            Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$



            How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 16 at 1:10

























            answered Jan 13 at 1:26









            amdamd

            31.8k21053




            31.8k21053












            • $begingroup$
              Very nice indeed, endorsed, +1!!!
              $endgroup$
              – Robert Lewis
              Jan 13 at 1:47


















            • $begingroup$
              Very nice indeed, endorsed, +1!!!
              $endgroup$
              – Robert Lewis
              Jan 13 at 1:47
















            $begingroup$
            Very nice indeed, endorsed, +1!!!
            $endgroup$
            – Robert Lewis
            Jan 13 at 1:47




            $begingroup$
            Very nice indeed, endorsed, +1!!!
            $endgroup$
            – Robert Lewis
            Jan 13 at 1:47


















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