Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 7B / iv / Solution

For $t\ne 0$, factor the matrix asThe determinant of a scalar multiple of an $n\times n$ matrix gains the scalar's $n$th power. ThusThe right side extends to a polynomial at $t=0$: multiplying ${\chi }_A(t^{-1})$ by $t^n$ removes all negative powers. Its value there is $(\det A{)}^{-1}$, since the leading coefficient of ${\chi }_A(s)=\det (A-sI)$ is $(-1{)}^n$. This equals ${\chi }_{A^{-1}}(0)$, so the identity holds at zero with that interpretation.

For the relation between eigenvalues and the characteristic polynomial,The first equivalence is the definition of an eigenvector; the others use the kernel of a linear map and the criterion for an invertible matrix to have nonzero determinant. Since ${\chi }_A$ has degree $n$, the fundamental theorem of algebra gives $n$ roots over $\mathbb{C}$ counted with multiplicity. The multiplicity of a root is its algebraic multiplicity, and the eigenspace is exactly $\ker (A-\lambda I)$; its dimension is the geometric multiplicity, which need not equal the algebraic multiplicity.

Finally, factor $p(s)=s^4-\mu$ over $\mathbb{C}$ as $p(s)={\prod }_{j=1}^4(s-{\lambda }_j)$, including repeated roots when $\mu =0$. Substituting $A$ into this polynomial is valid because the resulting factors commute. If $\mu$ is an eigenvalue of $A^4$, thenAt least one factor must be zero, so its ${\lambda }_j$ is an eigenvalue of $A$ and satisfies ${\lambda }_j^4=\mu$. Thus the reverse direction of this instance of the spectral mapping theorem isThis argument also covers $\mu =0$, when the conclusion is that $A$ is singular and has eigenvalue zero.