Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 2 / 10C (Linear Algebra)

For a complex square matrix, define its determinant byThe cofactor is $C_{ij}=(-1{)}^{i+j}\det A^{(ij)}$, where $A^{(ij)}$ deletes row $i$ and column $j$. The adjugate is the transposed cofactor matrix, $\mathrm{adj}(A{)}_{ij}=C_{ji}$.

The $(i,j)$ entry of $\mathrm{adj}(A)A$ is ${\sum }_k{C}_{ki}{a}_{kj}$. For $j=i$, expansion along column $i$ gives $\det A$. For $j\ne i$, this same sum is the determinant of the matrix obtained by replacing column $i$ by column $j$: deleting the replaced column leaves all its cofactors unchanged. The new matrix has two equal columns, so its determinant is zero. ThusThis proof does not assume invertibility and therefore also covers singular matrices.

The cyclic matrix acts by $(v_1,v_2,v_3,v_4)\mapsto (v_2,v_3,v_4,v_1)$. For each fourth root of unity $\lambda$, the vector $v_{\lambda }=(1,\lambda ,{\lambda }^2,{\lambda }^3{)}^T$ satisfies $A{v}_{\lambda }=\lambda v_{\lambda }$, because ${\lambda }^4=1$. This supplies the four distinct eigenvalues and nonzero eigenvectors:Eigenvectors for distinct eigenvalues are linearly independent. Thus, with these columns in the indicated order,They form a basis over $\mathbb{C}$, so there are no other eigenvalues. Equivalently the characteristic polynomial is ${\lambda }^4-1$.