Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 1 / 6A (Vectors and Matrices)

Using the convention of the question, the characteristic polynomial and characteristic equation areForone hasDirect multiplication givesentry by entry. Thus ${\chi }_A(A)=0$, verifying the Cayley-Hamilton theorem for $2\times 2$ matrices.

If $A_{ji}$ denotes the matrix obtained by deleting row $j$ and column $i$, then the adjugate matrix isFor nonsingular $A$,Therefore, when $A$ and $B$ are nonsingular,

The determinant $\det (M-tI)$ is a polynomial in $t$ with leading term $(-t{)}^n$, so it is not the zero polynomial and has only finitely many roots. Hence one may choose $\alpha >0$ smaller than every positive root, if any. Then

Each entry of $\mathrm{adj}(M-tI)$ is, up to sign, the determinant of an $(n-1)\times (n-1)$ minor whose entries are affine polynomials in $t$. The Leibniz formula for determinants therefore makes each adjugate entry a polynomial in $t$ of degree at most $n-1$.

For arbitrary $A,B$, choose positive sequences $s_j,t_j\to 0$ for which $A-s_{j}I$ and $B-t_{j}I$ are nonsingular. The nonsingular identity from part (b) applies to their product:Every entry is a polynomial, hence continuous, in the matrix entries. Letting $j\to \infty$ provesfor all square matrices.

DefineThe Cayley-Hamilton theorem forgivesIf $A$ is nonsingular, multiplication by $A^{-1}$ shows $P(A)=\det (A)A^{-1}=\mathrm{adj}(A)$.

For arbitrary $A$, apply the nonsingular result to $A-tI$ for a sequence of nonzero $t\to 0$ avoiding the finitely many singular values. Both the characteristic coefficients and the adjugate entries depend polynomially on the matrix entries, so the limit gives