Find all 2 by 2 complex matrix with the following condition












1












$begingroup$



Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
$A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
$A$ with $A^2 = -I$. Deduce that for every even $n$ there are
infinitely many $n times n$ real matrices with $A^2 = -I$.




Try:



We have



$$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



$$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



Therefore, we have



$$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






linear-algebra







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$endgroup$

















    1












    $begingroup$



    Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
    $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
    $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
    infinitely many $n times n$ real matrices with $A^2 = -I$.




    Try:



    We have



    $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



    So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



    If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



    if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



    $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



    now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



    Therefore, we have



    $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



    are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



    Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






    linear-algebra







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
      $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
      $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
      infinitely many $n times n$ real matrices with $A^2 = -I$.




      Try:



      We have



      $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



      So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



      If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



      if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



      $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



      now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



      Therefore, we have



      $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



      are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



      Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






      linear-algebra







      share|cite|improve this question











      $endgroup$





      Find all $A in mathrm{Mat}_{2 times 2} ( mathbb{C})$ such that
      $A^2 = -I$ and prove that there is no 2 by 2 real diagonal matrices
      $A$ with $A^2 = -I$. Deduce that for every even $n$ there are
      infinitely many $n times n$ real matrices with $A^2 = -I$.




      Try:



      We have



      $$ A = begin{bmatrix} a & b \ c & d end{bmatrix} implies A^2 = begin{bmatrix}a^2+bc & b(a+d) \ c(a+d) & cb+d^2 end{bmatrix} $$



      So $A^2 = - I$ iff $-1 = a^2+bc = cb + d^2$ and $0=b(a+d)=c(a+d)$



      If $b=0$, then we have $a^2=-1 $ , $d^2=-1$.



      if $a^2=-1$, then $a= pm i$ and similarly $b= pm i$. As long as $a+d neq 0$, we see that $c=0$ otherwise $c in mathbb{C}$ is arbitrary and so we have



      $$ begin{bmatrix} i & 0 \ 0 & i end{bmatrix}, begin{bmatrix} -i & 0 \ 0 & -i end{bmatrix},begin{bmatrix} -i & 0 \ c & i end{bmatrix},begin{bmatrix} i & 0 \ c & -i end{bmatrix} $$



      now if $b neq 0$, then $a+d=0$ and so $a=-d$ and thus $c = -(1+a^2)/b$



      Therefore, we have



      $$ begin{bmatrix} a & b \ -(1+a^2)/b & -a end{bmatrix} $$



      are all posible matrices. Now, if the $a,b,c,d$ were real, we see that $b neq 0$ and so we get matrices like the one above only.



      Now, is this correct so far? I'm stuck for the $ntimes n$ case. Any help?






      linear-algebra




      linear-algebra






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      edited Jan 4 at 6:01









      Lee

      320111




      320111










      asked Jan 4 at 5:52









      JamesJames

      2,638325




      2,638325






















          1 Answer
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          $begingroup$

          Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



          For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



          Bonus question: What happens for odd $n$ with real matrices?






          share|cite|improve this answer











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            $begingroup$

            Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



            For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



            Bonus question: What happens for odd $n$ with real matrices?






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



              For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



              Bonus question: What happens for odd $n$ with real matrices?






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



                For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



                Bonus question: What happens for odd $n$ with real matrices?






                share|cite|improve this answer











                $endgroup$



                Looking very good so far. However, for the $a=pm i, d=-a$ case, you missed the possibility $c=0$ with $b$ being anything.



                For even $n$, take the real $2times2$ matrix you've found, make $frac n2$ copies of it all along the diagonal of an $ntimes n$ matrix (whether you make them all equal, or just make sure they are all of that form is up to you), and otherwise make all entries $0$. This easily gives you infinitely many different $ntimes n$ matrices.



                Bonus question: What happens for odd $n$ with real matrices?







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 6:16

























                answered Jan 4 at 6:04









                ArthurArthur

                122k7122210




                122k7122210






























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