Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 3 / a (Determinantal functions)

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.