Past exam of the linear algebra course of the University of Oxford / 2023 / 3

Throughout Question 3, the inner product is linear in its first argument and conjugate-linear in its second. Over the real numbers, conjugation has no effect. The symbol $T^{\ast }$ now denotes the adjoint operator for this inner product.

An adjoint operator of $T:V\to V$ is a linear map $T^{\ast }:V\to V$ satisfyingThis identity transfers $T$ from the first input of the inner product to the second.

If $R$ and $S$ both satisfy the defining identity of an adjoint operator, thenfor every $u,w\in V$. Choosing $u=(R-S)w$ gives $\Vert (R-S)w{\Vert }^2=0$. The positive definiteness of the inner product forces $(R-S)w=0$ for every $w$, proving

For the normal operator $T$, the adjoint operator identity and $T^{\ast }T=T{T}^{\ast }$ giveFor the last equality, $⟨T{T}^{\ast }v,v⟩=⟨T^{\ast }v,T^{\ast }v⟩$ and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots provesIn particular the two kernels coincide.

Let $w=Tv$. If $T^{2}v=0$, then $Tw=0$, and part (i) gives $\Vert T^{\ast }w\Vert =\Vert Tw\Vert =0$. Thus $T^{\ast }w=0$ andThe positive definiteness of the inner product gives $w=0$. Since the reverse kernel inclusion always holds, we obtainThis is kernel stabilization for a normal operator.

For a polynomial $f(x)={\sum }_{j=0}^d{a}_j{x}^j$, the defining identity and uniqueness of the adjoint operator giveIndeed, the adjoint operator of a sum is the sum of the adjoint operators, the adjoint of $aT$ is $\bar{a}{T}^{\ast }$, and the adjoint of a product $AB$ is $B^{\ast }{A}^{\ast }$. Since the normal operator $T$ commutes with $T^{\ast }$, each power of $T$ commutes with each power of $T^{\ast }$. Multiplying the two finite sums therefore givesHence polynomials of a normal operator are normal, including constant polynomials.

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,This contradicts minimality: $m_T/f_j$ is a monic polynomial of smaller degree annihilating $T$. Therefore the minimal polynomial of a normal operator is a square-free polynomial over either of the given fields:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.

Take $N=T-\lambda I$. It is a polynomial in the normal operator $T$, so part (b)(iii) makes $N$ a normal operator. Also $N^{\ast }=T^{\ast }-\bar{\lambda }I$. If $Tv=\lambda v$, part (b)(i), applied to $N$, givesThe norm vanishes only at the zero vector, soThis is the adjoint eigenvector identity for a normal matrix expressed independently of coordinates.

Let $u\in \ker (T-\lambda I)$ and $w\in \ker (T-\mu I)$, with $\lambda \ne \mu$. By part (i) and the adjoint operator identity,The last step uses conjugate-linearity in the second argument of the inner product. Therefore $(\lambda -\mu )⟨u,w⟩=0$, proving the orthogonality of the two eigenspaces:

Over the complex numbers, the fundamental theorem of algebra makes every irreducible polynomial factor $f_j(x)$ equal to $x-{\lambda }_j$. Part (b)(iv) giveswith distinct eigenvalues. The primary decomposition theorem for linear maps consequently expresses $V$ as the direct sum of the actual eigenspaces $E_{{\lambda }_j}=\ker (T-{\lambda }_{j}I)$, with no higher generalized eigenspaces needed. Part (ii) makes these summands mutually orthogonal.

Apply the Gram-Schmidt process to a basis within each eigenspace. Its linear combinations stay within that eigenspace, so all resulting vectors are still eigenvectors. Their union is orthonormal by part (ii) and spans $V$ by the primary decomposition theorem for linear maps. Thusfor an orthonormal basis $\mathcal{B}$ of eigenvectors. This proves the finite-dimensional complex spectral theorem for normal operators by the requested decomposition argument.