Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 2 / 10E (Linear Algebra)

The determinant isEquivalently it is the alternating multilinear function of the columns with value one on the identity matrix. To prove multiplicativity, write column $j$ of $AB$ as ${\sum }_k{b}_{kj}{A}_k$, where $A_k$ is column $k$ of $A$. Multilinear expansion gives a sum over choices $k_1,\dots ,k_n$; terms with repeated choices vanish by alternation. The remaining permutations contributeIf $C$ is invertible, $CB=C(BC)C^{-1}$, so multiplicativity makes $\det (CB{C}^{-1}-\lambda I)=\det (BC-\lambda I)$ and $p_{CB}=p_{BC}$. For arbitrary $A$, put $C=A+tI$. Its determinant is a nonzero polynomial in $t$, so it is invertible except for finitely many values. Thus for all other $t$,For each fixed $\lambda$ both sides are polynomials in $t$, and equality at infinitely many $t$ makes them identical. Setting $t=0$ proves the Characteristic polynomials of AB and BA identity $\boxed{p_{BA}=p_{AB}}$ without an invertibility assumption on $A$.

In the convention $p_A(\lambda )=\det (A-\lambda I)$, the coefficient of ${\lambda }^{n-1}$ is $(-1{)}^{n-1}\mathrm{tr}A$: in the determinant expansion such a term chooses one diagonal entry of $A$ and $n-1$ factors $-\lambda$; nonidentity permutations have at least two nonfixed positions and contribute lower degree. Therefore $\boxed{p_A=p_B⟹\mathrm{tr}A=\mathrm{tr}B}$.

This assertion is false in general. In dimension two take $A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ and $C=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$. Then $ABC=A$, whereas $ACB=0$. Their characteristic polynomials are $\lambda (\lambda -1)$ and ${\lambda }^2$, respectively, so $\boxed{p_{ABC}\ne p_{ACB}}$.

This assertion is true. Apply the proved Characteristic polynomials of AB and BA identity to the two matrices $A$ and $BC$:It holds whether or not any of the three matrices is invertible.