Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 9G (Linear Algebra)

Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

Suppose first that $\alpha (U)\subseteq U$. If $y\in U^{\perp }$ and $u\in U$, thenThus ${\alpha }^{\ast }(U^{\perp })\subseteq U^{\perp }$.

Conversely, assume that latter inclusion. For $u\in U$ and $y\in U^{\perp }$,Hence $\alpha u\in (U^{\perp }{)}^{\perp }=U$. This proves the adjoint criterion for an invariant orthogonal complement:

Write $B(x,y)=⟨\alpha x,y⟩$. The hypothesis says $B(x,x)=0$. Expanding at $x+y$ giveswhereas expanding at $x+iy$ givesTherefore $B(x,y)=0$ for every $x,y$, soThis is the complex polarization argument for a vanishing quadratic form.

The conclusion is false over a real inner product space. On ${\mathbb{R}}^2$, the nonzero operatorsatisfies $⟨Jx,x⟩=0$ for every $x$.

For every $x$,If $\alpha$ is normal, these quantities are equal. Conversely, equality of the norms for all $x$ givesThe operator in parentheses is self-adjoint, so polarization makes it zero. ThereforeThe same equivalence holds over a real inner product space: for a self-adjoint operator $T$, the real polarization identityshows that a vanishing quadratic form forces $T=0$.

Proceed by induction on $\dim V$. Over $\mathbb{C}$, $\alpha$ has an eigenvector $v$, say $\alpha v=\lambda v$. The normal operator $\alpha -\lambda I$ satisfies the norm equality from part (c), soThus ${\alpha }^{\ast }v=\overline{\lambda }v$. Part (a) now shows that $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. The restriction of $\alpha$ to $v^{\perp }$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized vector $v$ proves the finite-dimensional spectral theorem for normal operators.

Hence