Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 3

The dual basis is the collection of linear functionals characterized byExplicitly, $e_i^{\prime }({\sum }_j{\alpha }_j{e}_j)={\alpha }_i$. Every linear functional $\ell \in V^{\prime }$ satisfies $\ell ={\sum }_i\ell (e_i)e_i^{\prime }$, proving that the displayed functionals span the dual space. Evaluating a zero linear combination at each $e_j$ makes all its coefficients zero, so they are also linearly independent. Thus they form a basis of $V^{\prime }$.

Define the bidual evaluation map by evaluation:For fixed $v$, this is a linear functional of $\ell$, so it belongs to the double dual. It is also linear as a function of $v$. If $v\ne 0$, some coordinate of $v$ in a basis is nonzero, and the corresponding dual basis functional detects it. Hence $J(v)\ne 0$, proving injectivity.

The spaces $V,V^{\prime },V^{\prime \prime }$ all have the same finite dimension. Therefore an injective linear map $J:V\to V^{\prime \prime }$ is surjective, proving the required linear isomorphism.

This linear isomorphism is natural because evaluation requires no basis. More explicitly, for any linear map $A:V\to W$, its dual map satisfiesso $A^{\prime \prime }{J}_V=J_{W}A$.

Writing $u={\sum }_i{u}_i{e}_i$ and $v={\sum }_i{v}_i{e}_i$, evaluation against the dual basis givesThis has codomain $\mathbb{C}$, not the printed $\mathbb{R}$. For a nonzero vector space, take $u=i{e}_1$ and $v=e_1$ to obtain $\varphi (u,v)=i$. Only the diagonal values $\varphi (v,v)$ must be real. The displayed formula defines the intended complex inner product.

It is linear in the first argument and conjugate-linear in the second:It has conjugate symmetry, since $\varphi (v,u)=\overline{\varphi (u,v)}$. Finally,with equality exactly when every coordinate vanishes, equivalently $v=0$. This proves positive definiteness. Since $\varphi (e_i,e_j)={\delta }_{ij}$, the original basis is orthonormal. This is the inner product determined by a basis.

Use the convention of part (a): the inner product is linear in its first argument. If $Tv=\lambda v$ with $v\ne 0$, the adjoint operator identity and $T^{\ast }=-T$ givePositive definiteness makes $⟨v,v⟩>0$, soThus every eigenvalue of a skew-adjoint operator is purely imaginary, including zero.

Take $v\in U^{\perp }$ and $u\in U$. Because $U$ is an invariant subspace, $Tu\in U$. The adjoint operator relation yieldsThis holds for every $u\in U$, so $Tv\in U^{\perp }$. Therefore the orthogonal complement is also an invariant subspace.

Proceed by induction on the dimension. For positive dimension, the fundamental theorem of algebra gives a root of the characteristic polynomial, hence a nonzero eigenvector $v$. Normalize it to have norm one. Its span is an invariant subspace, and part (ii) makes its orthogonal complement invariant too:The restriction of $T$ to the complement remains skew-adjoint, since the same inner product identity holds there. The induction hypothesis supplies an orthonormal eigenbasis for that smaller space. Adjoining $v$ gives an orthonormal eigenbasis for $V$.

The zero-dimensional base case is immediate. Thus $T$ is diagonalizable, indeed unitarily diagonalizable. This proves the finite-dimensional case of the spectral theorem for a skew-adjoint operator without assuming that distinct eigenvalues have multiplicity one.

Regard the real skew-symmetric matrix as acting on ${\mathbb{C}}^n$. Its Hermitian adjoint is $A^{\ast }=A^T=-A$. Parts (i) and (iii) show that its eigenvalues are purely imaginary and its complex minimal polynomial has no repeated roots.

Since $A$ has real entries, nonreal eigenvalues occur in conjugate pairs. A nonzero pair has the form $i\alpha ,-i\alpha$, with $\alpha >0$, and contributes the real quadratic polynomialZero, if present, contributes one factor $x$. Hence the real minimal polynomial isThe product annihilates $A$ after complexification of a real vector space, so it also annihilates the original real linear map; no factor can be removed because each displayed complex root is an eigenvalue. This proves both the asserted form and its minimality. The case $A=0$ has $\epsilon =1$ and an empty product.