Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 10

The vector $(0,1,-1)$ is an eigenvector with eigenvalue one. The other two orthogonal directions give eigenvalues two and five, leading toIts columns are an orthonormal eigenbasis, so $V^{\mathsf{T}}V=I$ and $V^{\mathsf{T}}AV=D$. Multiplying by $V$ and $V^{\mathsf{T}}$ gives $\boxed{A=VD{V}^{\mathsf{T}}}$. Eigenvector signs and the joint order of columns and diagonal entries may be changed consistently.

The matrix exponential acts entrywise on a diagonal matrix, soFor the two-dimensional matrix $C$, direct multiplication gives $C^2=-C$, hence $C^n=(-1{)}^{n-1}C$ for $n\ge 1$. Summing its defining power series givesFinally, $e^A=V\mathrm{diag}(e,e^2,e^5)V^{\mathsf{T}}$, so each diagonal entry is a weighted sum of these exponential eigenvalues, with weights equal to squared eigenvector components:

In a submultiplicative matrix norm, the exponential series is absolutely convergent. Its Cauchy product can therefore be grouped by the total degree:When $AB=BA$, the inner sum is $(A+B{)}^n$ by the binomial theorem. Thus $\boxed{e^A{e}^B=e^{A+B}}$. The commutation assumption is used precisely in that binomial expansion; absolute convergence alone does not supply it.