Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2006 / 1

Yes. The matrix is a cyclic shift operator on three coordinates. Since $e_1,A{e}_1,A^2{e}_1$ are the three standard basis vectors, no polynomial of degree below three annihilates $A$. Also $A^3=I$, so its minimal polynomial is $X^3-1$. Over the complex numbers this has the distinct roots $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. The minimal polynomial criterion for diagonalizability therefore proves diagonalizability.

Explicitly, $v_{\lambda }=(1,{\lambda }^2,\lambda {)}^{\mathsf{T}}$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$ when ${\lambda }^3=1$. These three eigenvectors form an eigenbasis. This is a special case of diagonalization of a cyclic shift over a field.

No. The same cyclic-vector argument gives minimal polynomial $X^3-1=(X-1)(X^2+X+1)$ over the finite field ${\mathbb{F}}_2$. The quadratic has no root in ${\mathbb{F}}_2$: its value is one at both zero and one. Thus the minimal polynomial does not split into linear factors over the specified field, violating the minimal polynomial criterion for diagonalizability. It becomes diagonalizable over a suitable larger finite field, but that is a different question.

No. In characteristic three, $X^3-1=(X-1{)}^3$. This is still the minimal polynomial, because the three cyclic basis vectors remain independent. Its repeated factor violates the minimal polynomial criterion for diagonalizability. Equivalently, the only possible eigenvalue is one; a diagonalizable matrix with that sole eigenvalue would be the identity matrix, whereas this matrix is not the identity.

Yes. In ${\mathbb{F}}_7$, the minimal polynomial splits asThe three factors are distinct, so the minimal polynomial criterion for diagonalizability applies. For example the eigenvectors $v_{\lambda }=(1,{\lambda }^2,\lambda {)}^{\mathsf{T}}$ for $\lambda =1,2,4$ give an eigenbasis over this finite field. The contrast with characteristic three is that the shift polynomial now splits without repeated roots.