# Linear Algebra: Diagonalisability

Problem

Let A be the matrix

$\Bigg(\begin{matrix} 0&0&1\\ 1&0&0 \\0&1&0 \end{matrix} \Bigg)$

Giving brief justifications, determine whether A is diagonalizable over (a) the complex field; (b) the field $Z_2$ with two elements; (c) the field $Z_3$ with three elements; (d) the field $Z_7$ with 7 elements.

Progress

The characteristic polynomial $\chi_A(x)=x^3-1$, which in this case is equal to the minimum polynomial, $m_A(x)$. We make use of the fact that A is diagonalisable $\Leftrightarrow$ $m_A(x)$ can be expressed as the product of disjoint linear factors.

(a) In $\mathbb{C}$, $m_A(x)=x^3-1=(x-1)(x+\alpha)(x-\beta)$, where $\alpha \neq \beta$ and so A is diagonalisable.

Not sure how to apply the argument to the fields $\mathbb{Z}_n$ for $n=2,3,7$ however. Any help would be appreciated. Regards.

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Factorization in $\mathbb{C}$ is not right, but there are three distinct roots. There are also three distinct roots in $\mathbb{Z}_7$. For the other two, will have to work. It is not true that diagonalizable implies three distinct roots. Maybe look at whether there is a basis of eigenvectors? – André Nicolas Dec 23 '11 at 15:21
According to my calcualtion, $$(x-1)(x+i)(x-i)=x^3-x^2+x-1$$ – yohBS Dec 23 '11 at 15:22
@AndréNicolas: I've edited the factorisation in $\mathbb{C}$; I don't think you need to explicitly identify the factors, just to show their existence. We have that $x^3=1(mod7)$ for $x=1,2,4$, and so $m_A(x)=(x-1)(x-2)(x-4)$ in $\mathbb{Z}_7$ and we are done. I'll continue with working for $\mathbb{Z}_1,2$. Thanks! – Mathmo Dec 23 '11 at 15:45

Your minimal polynomial $x^3-1$ factors $x^3-1=(x-1)(x^2+x+1)$ over any field. The second factor $x^2+x+1$ certainly splits over $\mathbb C$ (roots $-\frac12\pm\frac{\sqrt3}2\mathbb i$), over $\mathbb Z/2\mathbb Z$ it is irreducible, over $\mathbb Z/3\mathbb Z$ it splits $(x-1)^2$, and over $\mathbb Z/7\mathbb Z$ it splits $(x-2)(x-4)$. So the matrix is diagonalizable over $\mathbb C$ and over $\mathbb Z/7\mathbb Z$ (three distinct roots in both cases), not over the other two fields. It does diagonalize over the extension field $\mathbb F_4$ over $\mathbb Z/2\mathbb Z$ where $x^2+x+1$ has two distinct roots, but not over any extension field of $\mathbb Z/3\mathbb Z$ since the minimal polynomial $x^3-1=(x-1)^3$ has a triple root there.
Many thanks! Probably should've looked into splitting fields more. How would you demonstrate that $x^3-1$ is irreducible over $\mathbb{Z}_2$? – Mathmo Dec 23 '11 at 16:31
@TJO: A quadratic (or cubic) polynomial is irreducible iff it has no roots. In $\mathbf Z_2$ there are only two candidates, and they aren't roots of $x^2+x+1$. However $x^3-1$ is reducible (over any field) as $1$ is a root. – Marc van Leeuwen Dec 23 '11 at 16:58
Over the complex numbers and over $\mathbb{Z}_7$, the characteristic polynomial has three distinct roots, so our matrix is diagonalizable.
Over $\mathbb{Z}_2$, the only eigenvalue is $1$. A short computation shows that the only eigenvector is $(1,1,1)^T$, so the matrix is not diagonalizable. Over $\mathbb{Z}_3$, again the only eigenvalue is $1$, and again the eigenspace is $1$-dimensional. Indeed over any field $K$, the eigenspace for any fixed eigenvalue is $1$-dimensional.