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

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

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!

  • 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

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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