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Write $y=(y_1,y_2,y_3{)}^T$. From the coordinate formula for the cross product,If $y=c_1{e}_1$, this becomesThus $c_1\ne 0$ gives rank three and a zero-dimensional kernel, while $c_1=0$ gives rank one and kernel dimension two.

If $y\cdot e_1=0$, then $y_1=0$. When $y\ne 0$, the equations $Tx=0$ force $x_1=0$ and $y_3{x}_2-y_2{x}_3=0$, soWhen $y=0$, again $\mathrm{rank}T=1$ and $\dim \ker T=2$. These dimensions also agree with the rank-nullity theorem.

In odd dimension,so $\det A=0$. Since $A$ is real, its kernel contains a nonzero real vector $a$.

The plane $a^{\perp }$ is invariant under $A$, because $a\cdot Ab=-(Aa)\cdot b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta |b|$ with $\theta >0$. Then $A^{2}b$ is parallel to $b$, andsoThe matrix exponential fixes the kernel vector:Separating even and odd powers in its series and using $A^{2j}b=(-1{)}^j{\theta }^{2j}b$ givesThus $e^A$ acts as a rotation through angle $\theta$ on $a^{\perp }$ and fixes its axis $\mathbb{R}a$.

A real skew-symmetric matrix satisfies $A^{†}=A^T=-A$, so it is a skew-Hermitian matrix. If $Ax=\lambda x$, thenTaking the complex conjugate and using $A^{†}=-A$ shows that this scalar is the negative of its conjugate. Henceso every eigenvalue is purely imaginary or zero.

If $Ax=\lambda x$ and $Ay=\mu y$, thenSince ${\lambda }^{\ast }=-\lambda$, this becomes $(\mu -\lambda )x^{†}y=0$. Distinct eigenvalues therefore have orthogonal eigenvectors:

If $A^T=-A$, then the scalar $x^{T}Ax$ equals its own transpose and henceso it vanishes.

Conversely, suppose $x^{T}Ax=0$ for every real $x$. Taking $x=e_i$ gives $A_{ii}=0$. Taking $x=e_i+e_j$ then gives $A_{ij}+A_{ji}=0$. Thus every entry satisfies $A^T=-A$, and

The cross product $x\times y$ is the vector perpendicular to $x$ and $y$, of magnitude $|x||y|\sin \alpha$ and orientation fixed by the right-hand rule. In suffix notation,where ${\epsilon }_{ijk}$ is the Levi-Civita symbol. ThereforeUsinggivesHence the vector triple product identity is

Similarly,Thus, for $0<\theta <\pi$,At $\theta =0$, $1-z=0$, whose complex argument is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin (\theta /2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.

By Euler's formula, $z=e^{i\theta }$. Factoring out the half-angle givesSince $0\le \theta <\pi$, the cosine is positive. ThereforeOn the Argand diagram, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta /2$; resolving either side along the diagonal gives its length $2\cos (\theta /2)$.

We first prove by induction thatIt holds for $n=1$, and if it holds at $n$, thenConsequentlyPart (c) and the squeeze theorem now give

For $u_n=2^{-n}{n}^k$,Thus for all sufficiently large $n$ the ratio is, for example, at most $3/4$. Comparison with a decaying geometric sequence, equivalently the ratio test, yields

If $s\le 0$, the terms $n^{-s}$ do not tend to zero, so the series diverges by the term test for divergence. If $s>0$, the terms form a nonnegative decreasing sequence, and the Cauchy condensation test gives the condensed seriesThis geometric series converges exactly when $1-s<0$. Hence the p-series criterion is

For each $k\ge 0$, monotonicity gives the dyadic-block boundsSumming the upper bounds proves convergence of $\sum x_n$ whenever $\sum 2^k{x}_{2^k}$ converges. Summing the lower bounds gives, up to the first term,so convergence of $\sum x_n$ forces convergence of the condensed series. This is the Cauchy condensation test:

The upper bound follows immediately from convexity, sinceFor the lower bound, writeApplying convexity and rearranging givesChoose a fixed $r_0>0$ with $[y_0-r_0,y_0+r_0]\subset (a,b)$. Applying these bounds with $r=r_0$ and $r=-r_0$ shows that the difference $f(y_0+h)-f(y_0)$ is trapped between quantities of order $|h|/r_0$. Both tend to zero, so every convex function on an open interval is continuous.

Now suppose $c$ is a local minimum but some $x$ satisfies $f(x)<f(c)$. For sufficiently small $\lambda >0$, the point $z=(1-\lambda )c+\lambda x$ lies in the neighbourhood on which $c$ is minimal, whereas convexity givesa contradiction. ThusDifferentiability at the minimizer is unnecessary: $f(x)=|x-c|$ is convex, has its global minimum at $c$, and is not differentiable there.