Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 8B (Vectors and Matrices)

Construct the matrix of a linear map by expressing the images of the basis vectors in that same basis:Thus the $j$th column is $[L{e}_j{]}_B$. For $v=v_1{e}_1+v_2{e}_2$, linearity givesThe column $[Lv{]}_B$ is a coordinate vector; reconstructing the actual vector requires $Lv=(A[v{]}_B{)}_1{e}_1+(A[v{]}_B{)}_2{e}_2$. This distinguishes the linear map from its basis-dependent matrix.

By the given matrix similarity, $A^{\prime }=P^{-1}AP$ for an invertible $P$. HenceThe characteristic polynomials are identical. Therefore the characteristic equations and eigenvalues, including algebraic multiplicities, agree. The conclusion is independent of the choice between $\det (tI-A)$ and $\det (A-tI)$ conventions.

Rescaling the first basis vector gives the change-of-basis matrix $P=\mathrm{diag}(k,1)$, invertible because $k\ne 0$. Direct multiplication givesThus the matrices are related by matrix similarity. In terms of basis images, the coefficient along the rescaled first basis vector is divided by $k$, while the second component of its image is multiplied by $k$.

Let the repeated eigenvalue be $a$, and choose an eigenvector $v_1\ne 0$. Complete it to a basis $(v_1,v_2)$ of ${\mathbb{C}}^2$. In this basis the matrix is an upper triangular matrix, with first column $(a,0{)}^T$, say $T=\left(\begin{array}{cc}a& c\\ 0& d\end{array}\right)$. Its characteristic polynomial is $(t-a)(t-d)$. By part (a)(ii), this must equal $(t-a{)}^2$, so $d=a$.

If $c=0$, the matrix is already $aI$. If $c\ne 0$, rescale the first basis vector by $k=c$ and apply part (a)(iii); the upper-right entry becomes one. ThusThe second possibility is a size-two Jordan block. It has only one independent eigenvector, whereas $aI$ has a two-dimensional eigenspace, so the two types cannot be related by matrix similarity. This repeated-eigenvalue classification in dimension two proves the required result without assuming the general Jordan normal form theorem.

Use the orthonormal right-handed basisDirect multiplication gives $B(r)u=u$, $B(r)v=rv+w/\sqrt{2}$ and $B(r)w=-v/\sqrt{2}+rw$. Thus the matrix in this basis isThe last two columns are orthogonal and both have squared norm $r^2+1/2$. Therefore $B(r)$ is an orthogonal matrix precisely when $r^2+1/2=1$, giving the unique positive valueIts determinant is $r_0^2+1/2=1$, so it is a proper rotation matrix, not a reflection. The transverse block has cosine and sine both $1/\sqrt{2}$. ConsequentlyWith the oriented axis $u$, the angle is positive by the right-hand rule, since $u\times v=w$. The axis also follows from the unit-eigenvalue eigenspace. As a check, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.