Past exam of the linear algebra course of the University of Oxford / 2017 / 1

For a linear transformation $T$ of a finite-dimensional vector space over a field $F$, its minimal polynomial $m_T$ is the unique monic polynomial of least degree satisfying $m_T(T)=0$. There is at least one nonzero polynomial with this property because the powers $I,T,T^2,\dots$ lie in the finite-dimensional vector space of linear maps $V\to V$. Dividing any other such polynomial by $m_T$ leaves a remainder of smaller degree that also vanishes at $T$, so that remainder must be zero. This also proves uniqueness.

If $C$ represents $T$ in any basis, the paper's characteristic polynomial isIt is independent of the basis, since changing the basis replaces $C-xI$ by a similar matrix, preserving its determinant. Its degree is $\dim V$ and its leading coefficient is $(-1{)}^{\dim V}$; $m_T$ is always monic.

The Cayley-Hamilton theorem statesLet $d=\dim V$. Extend the field to an algebraically closed field and choose a basis in which the matrix of $T$ is an upper triangular matrix, as permitted. Write its diagonal entries as ${\lambda }_1,\dots ,{\lambda }_d$ and put $V_i=\mathrm{span}(e_1,\dots ,e_i)$. Each $V_i$ is an invariant subspace, andApply $T-{\lambda }_{d}I$ first, then $T-{\lambda }_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are commuting operators, soThe determinant of an upper triangular matrix gives ${\chi }_T(x)=(-1{)}^d{\prod }_i(x-{\lambda }_i)$. Thus ${\chi }_T(T)=0$ after the field extension. Its entries were in the original field, whose embedding is injective, so they were already zero there. The zero-dimensional case is immediate as well.

The displayed matrix is the transpose convention for a companion matrix. Expanding $\det (xI-A)$ along its last row givesIndeed, the last entry contributes $(x+{\alpha }_{n-1})x^{n-1}$ and the entry in column $j<n$ contributes ${\alpha }_{j-1}{x}^{j-1}$; the cofactor signs cancel those from the superdiagonal entries $-1$. ConsequentlyThe Cayley-Hamilton theorem gives $f(A)=0$, so $m_T$ divides $f$. To exclude a smaller minimal polynomial, let $e_1,\dots ,e_n$ be the standard basis. The first $n-1$ rows of $A$ showIf ${\sum }_{j=0}^{n-1}{c}_j{A}^j=0$, multiplying on the left by $e_1^{\mathsf{T}}$ gives ${\sum }_j{c}_j{e}_{j+1}^{\mathsf{T}}=0$, so every coefficient is zero. No nonzero polynomial of degree below $n$ vanishes at $A$. Since $f$ and $m_T$ are monic polynomials,

Write $B=D+K$, where $D$ has $A$ in each diagonal block and zero elsewhere, while $K$ has $I$ immediately above the diagonal and zero elsewhere. These block matrices satisfy $DK=KD$: either product has $A$ immediately above the diagonal. Also, $K^i$ has $I$ on its $i$th superdiagonal and is zero elsewhere.

The binomial theorem for commuting matrices therefore yieldsOnly $r=i$ can contribute to block $(k,k+i)$, provided that block lies in the matrix. HenceThus ${\beta }_{i,j}=\left(\genfrac{}{}{0ex}{}{j}{i}\right)$, interpreted as an element of the field. The formula remains valid in positive field characteristic: some of these coefficients can then vanish. This possibility matters in the next part.

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ coprime. Equivalently, $f$ is a separable polynomial. This includes every irreducible polynomial over a field of characteristic zero, and every irreducible polynomial over a perfect field.

Under this separability hypothesis, Bézout's identity for polynomials gives $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ is an invertible matrix. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a nilpotent matrix with $f(B{)}^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe minimal polynomial divides this characteristic polynomial. Since $f$ is an irreducible polynomial, it must be a power of $f$; the exact nilpotency index just found forces

The arbitrary-field qualification is necessary. In the rational function field $F={\mathbb{F}}_2(t)$, take $f(x)=x^2-t$. This is an irreducible polynomial: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its formal derivative is zero. With $N=2$,The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative polynomial $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general field.

There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ are coprime. The binomial theorem in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the Hasse derivatives, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.