Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 8B / a

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 $k$.