Choose the $z$-axis along the line of intersection of the two planes. Their unit normals lie in the $xy$-plane; after choosing the $x$-axis, takewhere $\theta$ is the oriented angle between the planes. The two reflection matrices restrict toon the $xy$-plane and both fix the $z$-axis. Their product isThus the composition of two plane reflections is a rotation about the intersection line through angle $2\theta$.

Writing $n$ as a column vector, the reflection isso its reflection matrix is

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.

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

Apply the root test to the terms $a_n{z}^{n^2}$:This tends to zero when $|z|<1$ and to infinity when $|z|>1$. On $|z|=1$, the series converges absolutely because $\sum |a_n|$ converges: the point $1$ lies strictly inside the original radius two. Thus the lacunary power series has radiusand in fact converges on its entire boundary circle.

The series can be written asThe original power series converges when $|z{|}^k<2$ and diverges when $|z{|}^k>2$. Hence its radius as a power series in $z$ is

Let $L={\mathrm{limsup}}_{n\to \infty }|a_n{|}^{1/n}$. The Cauchy-Hadamard theorem, proved by applying the root test, gives a radius of convergencewith the usual conventions. For $|z|<R$, choose $q<1$ eventually bounding $|a_n{z}^n{|}^{1/n}$, which gives absolute convergence by comparison with a geometric series. For $|z|>R$, infinitely many terms have $n$th root greater than one, so the terms fail to tend to zero and the series diverges.

Here $R=2$ and the assumed limit gives $|a_n{|}^{1/n}\to 1/2$. ThereforeThus