Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 14G (Linear Mathematics)

After choosing a basis of $V$, a linear endomorphism is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. ThereforeThe $n^2+1$ endomorphisms $I,\alpha ,\dots ,{\alpha }^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero polynomial $p$ satisfying $p(\alpha )=0$.

For nonzero $V$, the minimal polynomial $m_{\alpha }$ is the monic polynomial of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_{\alpha }$ gives $p=q{m}_{\alpha }+r$, with $\mathrm{deg}r<\mathrm{deg}{m}_{\alpha }$, and evaluation implies $r(\alpha )=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_{\alpha }$. If $V=\{0\}$, the convention $m_{\alpha }=1$ handles the degenerate case.

The arguments work over the underlying scalar field. In part (b), roots in that field correspond to eigenvalues there; to discuss other roots one extends scalars, for example by complexifying a real vector space.

If $\alpha v=\lambda v$ with $v\ne 0$, then $m_{\alpha }(\alpha )v=m_{\alpha }(\lambda )v$. The left side vanishes, so $m_{\alpha }(\lambda )=0$.

Conversely, if $m_{\alpha }(\lambda )=0$, write $m_{\alpha }(X)=(X-\lambda )q(X)$. Then $(\alpha -\lambda I)q(\alpha )=0$. If $\alpha -\lambda I$ were invertible, multiplying by its inverse would imply $q(\alpha )=0$, contradicting the minimal degree. Thus $\alpha -\lambda I$ is not invertible. In finite dimension this means it has a nonzero kernel, furnishing an eigenvector with eigenvalue $\lambda$. Hence the eigenvalues are exactly the roots of the minimal polynomial, with the scalar-field convention stated in part (a).

Invariance of $W$ means every power of $\alpha$ restricts to the corresponding power of $\beta$. Therefore $m_{\alpha }(\beta )$ is the restriction of $m_{\alpha }(\alpha )$, and it is zero. Divide $m_{\alpha }=q{m}_{\beta }+r$ with $\mathrm{deg}r<\mathrm{deg}{m}_{\beta }$. Evaluation at $\beta$ gives $r(\beta )=0$. By minimality of $m_{\beta }$, $r=0$, provingThe zero-subspace case has $m_{\beta }=1$ and satisfies the same conclusion.

Take $V={\mathbb{R}}^3$, $\alpha =\mathrm{diag}(0,1,0)$ and $W=\mathrm{span}(e_1,e_2)$. Then $W$ is a proper invariant subspace, and the restriction has matrix $\mathrm{diag}(0,1)$. In each case $X(X-1)$ annihilates the operator, while any annihilator must vanish at both zero and one. ConsequentlyThis same example works over any scalar field.