Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 3 / b (Eigenvalue comparisons)

Take the nonzero vector $1=(1,\dots ,1{)}^T$. Its $i$th image entry under matrix multiplication is the sum of row $i$:Thus $A1=\lambda 1$, exhibiting $1$ as an eigenvector with eigenvalue $\boxed{\lambda }$.

Yes. The sums of the columns of $A$ are the sums of the rows of its transpose $A^T$. By the previous argument, $A^{T}1=\lambda 1$, so $\det (\lambda I_n-A^T)=0$. The determinant is unchanged by transpose, givingTherefore $\boxed{\lambda \text{ is an eigenvalue of }A}$. The vector $1$ need not itself be an eigenvector of $A$ in the column-sum case.

If $A$ is an invertible matrix, thenIf instead $B$ is an invertible matrix, thenThus in either case $AB$ and $BA$ are similar matrices. Explicitly, whenever $Y=P^{-1}XP$ and $Xu=\lambda u$ with $u\ne 0$, one has $Y(P^{-1}u)=\lambda P^{-1}u$ and $P^{-1}u\ne 0$. Applying the same argument with $P^{-1}$ in place of $P$ proves equality of the sets of eigenvalues, including zero.

The same matrix similarity preserves being a diagonalisable matrix. If $S^{-1}(AB)S=D$ is a diagonal matrix, thenTherefore $BA$ is a diagonalisable matrix whenever $AB$ is, and the converse follows by interchanging $A,B$. Thus they are either both diagonalisable matrices or neither is. In fact the previous part shows that one invertible factor suffices.