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

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!

• $\lambda^4 - 1 = (\lambda^2 - 1)(\lambda^2 + 1) = (\lambda - 1)(\lambda + 1)(\lambda^2 + 1)$; $\lambda^4 + 1$ is irreducible over $\Bbb R$. Check again! Cheers! – 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 $$\mathbb{R}$$.