Past exam of the linear algebra course of the University of Oxford / 2023 / 1

The primary decomposition theorem for linear maps starts with a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$. Factor its minimal polynomial aswhere the $p_j$ are distinct monic polynomials that are irreducible polynomials and $a_j\ge 1$. Define $V_j=\ker p_j(T{)}^{a_j}$. ThenThus the direct sum separates the irreducible polynomial factors into uniquely determined invariant subspaces. The statement works even when the minimal polynomial does not split into linear polynomials over $F$.

All calculations here take place in the finite field $F={\mathbb{F}}_3$. For the given matrix $S$, its characteristic polynomial isThe values $q(0),q(1),q(2)$ are $2,2,1$, so this quadratic polynomial has no polynomial root in $F$ and is an irreducible polynomial. Direct calculation givesNeither $S$ nor $q(S)$ is zero. The Cayley-Hamilton theorem says that the minimal polynomial divides $xq(x)$; its proper monic polynomial divisors cannot annihilate $S$, so $m_S=xq$.

The two kernels give the required primary decomposition theorem for linear maps explicitly:The determinant of the matrix with these three basis vectors as columns is $1$ in $F$, proving $W=W_1\oplus W_2$. To check these are invariant subspaces, writing $u=(1,2,0{)}^T$ and $z=(0,0,1{)}^T$ gives $Su=u+2z$ and $Sz=2u$. Hence the matrix of the restriction to $W_2$ in the basis $(u,z)$ iswith minimal polynomial $q$; the nonzero vector space $W_1$ has zero restriction and minimal polynomial $x$.

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The dual map sends an output measurement back along $T$:The function composition of these linear maps is linear, and $T^{\prime }$ is linear in $\phi$. This is the algebraic dual map, so its definition involves no complex conjugation or inner product.

For a vector subspace $U\subseteq V$, its annihilator of a vector subspace isLet $\dim U=k$ and $\dim V=n$. Extend a basis $u_1,\dots ,u_k$ of $U$ to a basis $u_1,\dots ,u_n$ of $V$, and let ${\epsilon }^1,\dots ,{\epsilon }^n$ be the dual basis. Every linear functional has the unique expression $\phi ={\sum }_{j=1}^n\phi (u_j){\epsilon }^j$. It belongs to $U^0$ precisely when its first $k$ coefficients vanish. ConsequentlyThe listed dual basis elements are linearly independent, so counting them proves the formula, including $U=0$ and $U=V$.

Suppose $U$ is an invariant subspace for $T$. For $\phi \in U^0$ and $u\in U$, the dual map satisfiesbecause $Tu\in U$ and the annihilator of a vector subspace consists of the linear functionals vanishing there. ThusThis is the correspondence between invariant subspaces and annihilators.

We prove by mathematical induction on $n=\dim V$ that every linear map on an $n$-dimensional complex vector space admits both triangular forms. For $n=0$ or $1$ this is immediate. When $n>1$, the fundamental theorem of algebra supplies an eigenvector $v$; its span $U=⟨v⟩$ is an invariant subspace.

By parts (ii) and (iii), $U^0$ has dimension $n-1$ and is an invariant subspace of the dual map $T^{\prime }$. Apply the induction hypothesis to $T^{\prime }{|}_{U^0}$ to choose a basis ${\phi }_1,\dots ,{\phi }_{n-1}$ with an upper triangular matrix. Extend it by a linear functional ${\phi }_n$ satisfying ${\phi }_n(v)=1$. Such a functional exists by basis extension, and cannot belong to $U^0$. In this basis of $V^{\prime }$, the matrix of $T^{\prime }$ has the formwhere $B$ is an upper triangular matrix; here the last diagonal entry follows by evaluating $T^{\prime }{\phi }_n$ at $v$, where $Tv=\lambda v$.

There is a basis $e_1,\dots ,e_n$ of $V$ with ${\phi }_i(e_j)={\delta }_{ij}$, since the dual basis construction gives a finite-dimensional bidual evaluation map isomorphism. In particular $e_n=v$. The matrix transpose relationship for a dual map givesa lower triangular matrix. Reversing this basis turns a lower triangular matrix into an upper triangular matrix. If $R$ is the permutation matrix reversing the order, we have provedThis completes the induction by triangularization using dual spaces.