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

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.