Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 3 / 9F (Linear Algebra)

The minimal polynomial $m_{\alpha }$ is the unique monic polynomial of least positive degree such that $m_{\alpha }(\alpha )=0$. The Cayley-Hamilton theorem says that $\alpha$ satisfies its characteristic polynomial, so such polynomials exist. Dividing any two monic candidates of least degree shows uniqueness.

If $m_{\alpha }(x)=x^m$, then ${\alpha }^m=0$ but ${\alpha }^{m-1}\ne 0$. The least $r$ for which $({\alpha }^3{)}^r=0$ is the least integer satisfying $3r\ge m$. Hence

If $\alpha v=\lambda v$ for nonzero $v$, thenso ${\lambda }^3$ is an eigenvalue. Let $\omega =e^{2\pi i/3}$. Since $\lambda \ne 0$, the three factors ofare pairwise coprime. The kernel decomposition for coprime polynomials therefore giveswhere an absent eigenspace is interpreted as zero.

SetFor every $i$,Consequently $q({\alpha }^3)$ contains $m_{\alpha }(\alpha )$ as a polynomial factor and is zero. By the divisibility property of the minimal polynomial,

If the nonzero ${\lambda }_i$ are real and distinct, their cubes are also distinct. Suppose the multiplicity of $x-{\lambda }_i^3$ in $m_{{\alpha }^3}$ were $d_i<c_i$. Sinceminimality makes $m_{\alpha }(x)$ divide $m_{{\alpha }^3}(x^3)$. At $x={\lambda }_i$, the factor $x^3-{\lambda }_i^3$ has a simple zero because its derivative is $3{\lambda }_i^2\ne 0$. Thus $m_{{\alpha }^3}(x^3)$ has multiplicity only $d_i$ there, contradicting the multiplicity $c_i$ in $m_{\alpha }$. Every exponent is therefore $c_i$, and