Diagonalizability over $mathbb{C}$ and $mathbb{R}$ respectively












3












$begingroup$


I am new to Linear Algebra, and would love some feedback regarding the following question, which I found a bit confusing:



$$A = begin{Bmatrix}0&1&0&0\0&0&1&0\0&0&0&1\1&0&0&0end{Bmatrix}$$



Which of the following two statements is true and/or false?





  1. $A$ is diagonalizable over $mathbb{C}$


  2. $A$ is diagonalizable over $mathbb{R}$


I calculated the characteristic polynomial to be $λ^4+1=0$, and I doublechecked the calculations. This can of course be rewritten as $(λ-1)(λ+1)(λ^2+1)$.



This leads me to the conclusion that the eigenvalues over $mathbb{C}$ are $1, -1, i$, and $-i$, while the eigenvalues over $mathbb{R}$ are $1$ and $-1$.



I then calculated the eigenvectors, which came out to be:



for $λ=1: (1,1,1,1)$



for $λ=-1: (-1,1,-1,1)$



for $λ=i: (i,-1,-i,1)$



for $λ=-i: (-i,-1,i,1)$



Now, clearly the geometric multiplicity is 1 in each of these cases (while only the first two cases are applicable "over $mathbb{R}$"). I assume the algebraic multiplicity is also one in each case.



Therefore, $A$ is diagonalizable both over $mathbb{C}$ and over $mathbb{R}$.



Thank you!






complex-numbers real-numbers diagonalization







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$endgroup$








  • 2




    $begingroup$
    $lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 8:26
















3












$begingroup$


I am new to Linear Algebra, and would love some feedback regarding the following question, which I found a bit confusing:



$$A = begin{Bmatrix}0&1&0&0\0&0&1&0\0&0&0&1\1&0&0&0end{Bmatrix}$$



Which of the following two statements is true and/or false?





  1. $A$ is diagonalizable over $mathbb{C}$


  2. $A$ is diagonalizable over $mathbb{R}$


I calculated the characteristic polynomial to be $λ^4+1=0$, and I doublechecked the calculations. This can of course be rewritten as $(λ-1)(λ+1)(λ^2+1)$.



This leads me to the conclusion that the eigenvalues over $mathbb{C}$ are $1, -1, i$, and $-i$, while the eigenvalues over $mathbb{R}$ are $1$ and $-1$.



I then calculated the eigenvectors, which came out to be:



for $λ=1: (1,1,1,1)$



for $λ=-1: (-1,1,-1,1)$



for $λ=i: (i,-1,-i,1)$



for $λ=-i: (-i,-1,i,1)$



Now, clearly the geometric multiplicity is 1 in each of these cases (while only the first two cases are applicable "over $mathbb{R}$"). I assume the algebraic multiplicity is also one in each case.



Therefore, $A$ is diagonalizable both over $mathbb{C}$ and over $mathbb{R}$.



Thank you!






complex-numbers real-numbers diagonalization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 8:26














3












3








3


1



$begingroup$


I am new to Linear Algebra, and would love some feedback regarding the following question, which I found a bit confusing:



$$A = begin{Bmatrix}0&1&0&0\0&0&1&0\0&0&0&1\1&0&0&0end{Bmatrix}$$



Which of the following two statements is true and/or false?





  1. $A$ is diagonalizable over $mathbb{C}$


  2. $A$ is diagonalizable over $mathbb{R}$


I calculated the characteristic polynomial to be $λ^4+1=0$, and I doublechecked the calculations. This can of course be rewritten as $(λ-1)(λ+1)(λ^2+1)$.



This leads me to the conclusion that the eigenvalues over $mathbb{C}$ are $1, -1, i$, and $-i$, while the eigenvalues over $mathbb{R}$ are $1$ and $-1$.



I then calculated the eigenvectors, which came out to be:



for $λ=1: (1,1,1,1)$



for $λ=-1: (-1,1,-1,1)$



for $λ=i: (i,-1,-i,1)$



for $λ=-i: (-i,-1,i,1)$



Now, clearly the geometric multiplicity is 1 in each of these cases (while only the first two cases are applicable "over $mathbb{R}$"). I assume the algebraic multiplicity is also one in each case.



Therefore, $A$ is diagonalizable both over $mathbb{C}$ and over $mathbb{R}$.



Thank you!






complex-numbers real-numbers diagonalization







share|cite|improve this question











$endgroup$




I am new to Linear Algebra, and would love some feedback regarding the following question, which I found a bit confusing:



$$A = begin{Bmatrix}0&1&0&0\0&0&1&0\0&0&0&1\1&0&0&0end{Bmatrix}$$



Which of the following two statements is true and/or false?





  1. $A$ is diagonalizable over $mathbb{C}$


  2. $A$ is diagonalizable over $mathbb{R}$


I calculated the characteristic polynomial to be $λ^4+1=0$, and I doublechecked the calculations. This can of course be rewritten as $(λ-1)(λ+1)(λ^2+1)$.



This leads me to the conclusion that the eigenvalues over $mathbb{C}$ are $1, -1, i$, and $-i$, while the eigenvalues over $mathbb{R}$ are $1$ and $-1$.



I then calculated the eigenvectors, which came out to be:



for $λ=1: (1,1,1,1)$



for $λ=-1: (-1,1,-1,1)$



for $λ=i: (i,-1,-i,1)$



for $λ=-i: (-i,-1,i,1)$



Now, clearly the geometric multiplicity is 1 in each of these cases (while only the first two cases are applicable "over $mathbb{R}$"). I assume the algebraic multiplicity is also one in each case.



Therefore, $A$ is diagonalizable both over $mathbb{C}$ and over $mathbb{R}$.



Thank you!






complex-numbers real-numbers diagonalization




complex-numbers real-numbers diagonalization






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edited Jan 18 at 8:40









idriskameni

749321




749321










asked Jan 18 at 8:04









daltadalta

1578




1578








  • 2




    $begingroup$
    $lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 8:26














  • 2




    $begingroup$
    $lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 8:26








2




2




$begingroup$
$lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:26




$begingroup$
$lambda^4 - 1 = (lambda^2 - 1)(lambda^2 + 1) = (lambda - 1)(lambda + 1)(lambda^2 + 1)$; $lambda^4 + 1$ is irreducible over $Bbb R$. Check again! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:26










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$begingroup$

All your work seems correct to me, but $A$ is not diagonalisable over $mathbb{R}$, since your eigenvectors aren't over $mathbb{R}$, but over $mathbb{C}$. The diagonalisation would be $PDP^{-1}=A$ where:
$$
P=begin{pmatrix}
1 & -1 & i & -i \
1 & 1 & -1 & -1 \
1 & -1 & -i & i \
1 & 1 & 1 & 1
end{pmatrix},
D=operatorname{diag}(1, -1, i, -i)
$$
And since $P$ is a complex matrix, $A$ is diagonalisable over $mathbb{C}$, and not $mathbb{R}$.






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    $begingroup$

    All your work seems correct to me, but $A$ is not diagonalisable over $mathbb{R}$, since your eigenvectors aren't over $mathbb{R}$, but over $mathbb{C}$. The diagonalisation would be $PDP^{-1}=A$ where:
    $$
    P=begin{pmatrix}
    1 & -1 & i & -i \
    1 & 1 & -1 & -1 \
    1 & -1 & -i & i \
    1 & 1 & 1 & 1
    end{pmatrix},
    D=operatorname{diag}(1, -1, i, -i)
    $$
    And since $P$ is a complex matrix, $A$ is diagonalisable over $mathbb{C}$, and not $mathbb{R}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      All your work seems correct to me, but $A$ is not diagonalisable over $mathbb{R}$, since your eigenvectors aren't over $mathbb{R}$, but over $mathbb{C}$. The diagonalisation would be $PDP^{-1}=A$ where:
      $$
      P=begin{pmatrix}
      1 & -1 & i & -i \
      1 & 1 & -1 & -1 \
      1 & -1 & -i & i \
      1 & 1 & 1 & 1
      end{pmatrix},
      D=operatorname{diag}(1, -1, i, -i)
      $$
      And since $P$ is a complex matrix, $A$ is diagonalisable over $mathbb{C}$, and not $mathbb{R}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        All your work seems correct to me, but $A$ is not diagonalisable over $mathbb{R}$, since your eigenvectors aren't over $mathbb{R}$, but over $mathbb{C}$. The diagonalisation would be $PDP^{-1}=A$ where:
        $$
        P=begin{pmatrix}
        1 & -1 & i & -i \
        1 & 1 & -1 & -1 \
        1 & -1 & -i & i \
        1 & 1 & 1 & 1
        end{pmatrix},
        D=operatorname{diag}(1, -1, i, -i)
        $$
        And since $P$ is a complex matrix, $A$ is diagonalisable over $mathbb{C}$, and not $mathbb{R}$.






        share|cite|improve this answer









        $endgroup$



        All your work seems correct to me, but $A$ is not diagonalisable over $mathbb{R}$, since your eigenvectors aren't over $mathbb{R}$, but over $mathbb{C}$. The diagonalisation would be $PDP^{-1}=A$ where:
        $$
        P=begin{pmatrix}
        1 & -1 & i & -i \
        1 & 1 & -1 & -1 \
        1 & -1 & -i & i \
        1 & 1 & 1 & 1
        end{pmatrix},
        D=operatorname{diag}(1, -1, i, -i)
        $$
        And since $P$ is a complex matrix, $A$ is diagonalisable over $mathbb{C}$, and not $mathbb{R}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 8:31









        Yuval GatYuval Gat

        9741213




        9741213






























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