Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 3 (Determinants and eigenvalues)

For the linear map $T_A:x\mapsto Ax$ on the finite-dimensional vector space ${\mathbb{R}}^n$, the rank-nullity theorem implies that injectivity and surjectivity are equivalent. Consequently $T_A$ is invertible if and only if $\ker T_A=\{0\}$.

Using the permitted equivalence between invertibility and a nonzero determinant, we obtain

We use the monic polynomial convention for the characteristic polynomial of an $n\times n$ matrix:It has degree $n$, since the only contribution to the coefficient of $t^n$ in the determinant is the product of the $n$ diagonal terms $t$. The alternative convention $\det (A-t{I}_n)=(-1{)}^n{\chi }_A(t)$ has the same polynomial roots.

A scalar $\lambda$ is an eigenvalue of $A$ if and only if there is a nonzero vector $v$ with $Av=\lambda v$. By the preceding kernel criterion,The last condition is ${\chi }_A(\lambda )=0$, exactly the definition of a polynomial root. For a real matrix, this proves the assertion for real eigenvalues; the same proof over the complex numbers gives the usual complex version.

Write $B=[b_1,\dots ,b_n]$. Since $AB=[A{b}_1,\dots ,A{b}_n]$ and $b\mapsto Ab$ is a linear map, the multilinearity of the determinant gives, for any column position $j$,If $b_i=b_j$ for $i\ne j$, the corresponding columns $A{b}_i,A{b}_j$ also agree, so $D(B)=0$. Thus $D$ is an alternating multilinear map of the columns of $B$, even when $A$ is not invertible.

We have $D(I_n)=\det A$. Define a function of the columns of $B$ byIt is an alternating multilinear map, because both summands are, and $F(I_n)=\det A+1-\det A=1$. Hence $F$ is a normalized determinantal function. By the asserted uniqueness of the determinant, $F(B)=\det B$ for every $B$. Rearranging yieldsThis argument also covers $\det A=0$ without dividing by it.

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.