Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 5

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

An invertible matrix is a square matrix for which there is a matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I$. Write its assumed diagonalization as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. ThenIf the determinant is nonzero, every ${\lambda }_j$ is nonzero, so ${\Lambda }^{-1}=\mathrm{diag}({\lambda }_j^{-1})$ exists and $A^{-1}=P{\Lambda }^{-1}{P}^{-1}$. If the determinant is zero, at least one eigenvalue is zero; its nonzero eigenvector is mapped to zero, which an invertible map cannot do. HenceThe equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

For the positive integer $k$ in the nilpotence condition, diagonalize $B=P\Lambda P^{-1}$. ThenEvery diagonal entry must vanish, so each ${\lambda }_j=0$. Consequently $\Lambda =0$ and $\boxed{B=0}$. This proves that a diagonalizable nilpotent matrix is zero.

Without diagonalizability a counterexample isIts only eigenvalue is zero but its kernel has dimension one, so it cannot have a basis of eigenvectors. It is the simplest nontrivial nilpotent Jordan block.