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

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.