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Direct multiplication and the identities ${\cos }^2\theta +{\sin }^2\theta =1$ and ${\cosh }^2\theta -{\sinh }^2\theta =1$ giveThus $R$ is an orthogonal matrix, while $S$ preserves the indefinite quadratic form $x_1^2-x_2^2$.

LetSince $J_{-}^2=-I$ and $J_{+}^2=I$, separating the even and odd terms of the matrix exponential givesand

Over $\mathbb{C}$, triangularize $B$ and denote its eigenvalues, with algebraic multiplicity, by ${\beta }_1,\dots ,{\beta }_n$. ThenThe numbers ${\omega }_l=e^{2\pi il/m}$ are all the root of unity solutions of $w^m=1$, soConsequentlyIf $\lambda$ is an eigenvalue of $B^m$, choose $z$ with $z^m=\lambda$. The displayed identity makes at least one factor ${\chi }_B({\omega }_{l}z)$ zero. Thus $\mu ={\omega }_{l}z$ is an eigenvalue of $B$ and

Eigenvectors belonging to distinct eigenvalues are linearly independent. Under the hypothesis they therefore form a basis $v_1,\dots ,v_n$. For every $j$,because ${\chi }_A({\lambda }_j)=0$. A linear map that vanishes on a basis is zero, henceThis proves the Cayley-Hamilton theorem in the distinct-eigenvalue case.

ChooseThe Leibniz formula for determinants shows that its leading term is $(-z{)}^n$, as required. A scalar $\lambda$ is an eigenvalue exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, soThe fundamental theorem of algebra gives a complex root of the degree-$n$ polynomial ${\chi }_A$, so every nonempty complex square matrix has at least one complex eigenvalue.

Let the vertex position vectors be $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$. SupposeSubtracting the second scalar product from the first givesso the remaining pair of opposite edges of a tetrahedron is also perpendicular.

PutDirect expansion gives, cyclically,with the other two differences equal to minus twice the scalar products of the other opposite-edge pairs. All three scalar products vanish, hence

WriteThe first equation is then automatic, while the second becomesThus the general solution isAs $\mathbf{u}$ varies, $\mathbf{x}$ and $\mathbf{y}$ are antipodal points of the sphere with center $\mathbf{a}$ and radius $\sqrt{\mathbf{a}\cdot \mathbf{a}-c}$, so they are opposite ends of a diameter.

The inequality $(1+1/n{)}^n\le e$ gives, for $|x|<1$,For $|x|>1$, the terms have magnitude at least $|x{|}^n$ and therefore do not tend to zero, so the series diverges by the term test for divergence. Hence

LetThe set $S$ contains zero. If $r\in S$ and $0\le s<r$, then $s\in S$ by comparison. Moreover, if the original power series diverges at $x_0$, no $r>|x_0|$ can lie in $S$, because absolute convergence at $r$ would imply absolute convergence at $x_0$. Thus $R=\sup S$ is finite and nonnegative.

If $|x|<R$, choose $r\in S$ with $|x|<r$; comparison gives absolute convergence at $x$. If $|x|>R$ and the series converged at $x$, then its terms $a_n{x}^n$ would be bounded. For any $y$ with $R<|y|<|x|$,so the geometric series comparison would put $|y|$ in $S$, contradicting the definition of $R$. This proves the radius of convergence property.

For ${\sum }_{n\ge 1}{x}^n/3^n$, the ratio is $x/3$, so the geometric series criterion gives

Condition (iii) is also sufficient. If $|a_n|$ did not tend to infinity, some $M>0$ would contain infinitely many terms with $|a_n|\le M$. Those terms form a bounded subsequence, and the Bolzano-Weierstrass theorem would give it a convergent subsequence, contrary to the hypothesis. Hence $|a_n|\to \infty$ and

Condition (ii) is sufficient. If $a_n\to \ell \ne 0$, then eventually $|a_n|\ge |\ell |/2$. Thereforeso

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

Condition (i) is insufficient because the allowed limit may be zero. For example, $a_n=1/n$ is nonzero and converges to zero, whereas $1/a_n=n$ does not converge in $\mathbb{R}$.

Let the columns be $C_1,\dots ,C_n$, and replace $C_k$ by $C_k+\lambda C_l$, where $k\ne l$. By multilinearity of the determinant,The second determinant vanishes because it has two equal columns. Hence

Interchanging two rows reverses the sign of an alternating multilinear map, soIn the Leibniz formula for determinants, the same result follows by composing every permutation with the corresponding transposition, which reverses its sign.

The scalar multiplication $\lambda A$ multiplies every one of the $n$ rows by $\lambda$. Applying multilinearity of the determinant once to each row gives

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives