Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 2

Use the inner product convention that the first argument is linear and the second is conjugate-linear. This is the convention corresponding to the conjugated eigenvalue in part (c).

An orthonormal basis $(q_1,\dots ,q_n)$ is a basis with $⟨q_i,q_j⟩={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta. Each vector has norm one and distinct vectors are orthogonal.

The Gram-Schmidt process starts from a basis $(e_1,\dots ,e_n)$ and definesThe coefficients are in this order because the inner product is linear in its first argument. For $i<k$, taking the inner product with $q_i$ gives $⟨u_k,q_i⟩=0$. The residual $u_k$ cannot be zero: otherwise $e_k$ would belong to the linear span of $e_1,\dots ,e_{k-1}$, contradicting the initial basis. Normalization makes $\Vert q_k\Vert =1$, while induction preservesThus the process produces an orthonormal basis of all of $V$.

In an orthonormal basis, every vector $w$ has the coordinate expansion $w={\sum }_i⟨w,e_i⟩e_i$. If $⟨Su,v⟩=⟨u,Rv⟩$ for all $u,v$, apply it to $u=e_i$, $v=e_j$ and take complex conjugates:ConsequentlyConversely, this formula givesLinearity in the first argument and conjugate-linearity in the second extend the equality from basis vectors to all $u,v$. This proves the equivalence.

The displayed formula specifies exactly one image for each basis vector, so it defines a unique linear transformation $S^{\ast }$ and proves the existence and uniqueness of the adjoint on an inner product space.

Conjugate symmetry givesBy uniqueness of the adjoint operator, $\boxed{(S^{\ast }{)}^{\ast }=S}$. Alsoso uniqueness gives $\boxed{(ST{)}^{\ast }=T^{\ast }{S}^{\ast }}$. The reversal of order follows directly from moving the two factors across the inner product one at a time.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.