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A vector space over the real numbers is a set $V$ with an addition making it an abelian group and a scalar multiplication by $\mathbb{R}$ satisfying, for scalars $a,b$ and vectors $u,v$, the identities $a(u+v)=au+av$, $(a+b)v=av+bv$, $a(bv)=(ab)v$ and $1v=v$. A vector subspace is a nonempty subset closed under addition and scalar multiplication; a proper vector subspace is one unequal to $V$. A spanning set is a set whose finite linear combinations give every vector of $V$. A basis is a linearly independent spanning set. The dimension is the number of vectors in a basis, or its cardinality when the basis is infinite.

The sum of vector subspaces and their intersection of vector subspaces areBoth are vector subspaces: addition and scalar multiplication preserve the displayed forms or simultaneous membership. The intersection of vector subspaces is never empty because $0\in U\cap W$.

For the two given hyperplanes, adding and subtracting their defining equations gives $x_1=x_2$ and $x_3=x_4$. HenceThese two nonzero vectors are orthogonal, and consequently form an orthogonal basis of the intersection of vector subspaces. Put $u=(1,-1,-1,1)$ and $w=(1,-1,1,-1)$. Direct substitution gives $u\in U$, $w\in W$. Their dot products with $b_1,b_2$ vanish, and $u\cdot w=0$. Each defining equation of $U$ or $W$ is one nonzero linear constraint in four variables, so each vector subspace has dimension three. Thus each space has the indicated orthogonal basis:The last four vectors are nonzero and mutually orthogonal, so they are a basis of ${\mathbb{R}}^4$. They lie in $U+W$, proving $U+W=V$. The dimension identity is therefore verified numerically:

Here is a nested-interval proof of the Bolzano-Weierstrass theorem. Put every term of the bounded real sequence in a closed interval $I_0=[-M,M]$, with $M>0$. Bisect this interval and choose a closed half containing infinitely many terms. Repeating gives nested closed intervals $I_j=[l_j,r_j]$, each containing infinitely many sequence terms, with length $2M/2^j$.

By the least-upper-bound property, $x_{\ast }={\sup }_j{l}_j$ exists. For each $j$, every lower endpoint lies at most $r_j$, while $l_j\le x_{\ast }$; hence $x_{\ast }\in I_j$. Choose increasing indices $n_j$ with $a_{n_j}\in I_j$, possible because each interval contains infinitely many terms. ThenThus every bounded real sequence has a convergent subsequence.

For a sequence with no convergent subsequence, take $a_n=n$: any subsequence has $n_j\ge j$, so its terms tend to positive infinity rather than to a finite limit of a sequence. For an unbounded sequence with a convergent subsequence, take $a_{2n}=0$ and $a_{2n-1}=n$. Its odd terms are unbounded, while its even subsequence converges to zero.

The limit of a sequence of real numbers is $a$ if, for every $\epsilon >0$, there is an integer $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$. It tends to positive infinity if, for every real $M$, there is an integer $N$ such that $n\ge N$ implies $a_n>M$.

If $a_n\to +\infty$, choose the latter threshold with $M=1/\epsilon$. Then $a_n>1/\epsilon >0$ eventually, so $|1/a_n|<\epsilon$. Thus $1/a_n\to 0$. The converse is false: $a_n=-n$ has reciprocal tending to zero and tends to negative infinity. The precise sign-free statement is given by reciprocal limits and escape in absolute value: $1/a_n\to 0$ is equivalent to $|a_n|\to \infty$.

If $a_n\to a\ne 0$, eventually $|a_n-a|<|a|/2$, and the triangle inequality gives $|a_n|>|a|/2$. ConsequentlyGiven $\epsilon >0$, take $N$ large enough that $|a_n-a|<\min (|a|/2,\epsilon |a{|}^2/2)$ for all $n\ge N$. This proves $1/a_n\to 1/a$ directly from the definition of the limit of a sequence.

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

For a function with continuous derivatives through order $m+1$ on the interval between $0$ and $x$, the Taylor formula with integral remainder isThis is Taylor's theorem about zero; translation gives the formula about any other point. To prove it, the case $m=0$ is the fundamental theorem of calculus. For $m\ge 1$, integration by parts givesSubstitution into the formula at order $m-1$ proves the formula at order $m$ by induction. The integral estimate, valid also for negative $x$, isFor the smooth function satisfying $f^{(3)}=f$, each fixed derivative is continuous on $[-R,R]$ and hence is bounded there by the extreme value theorem. Thus a bound $M_j$ exists for every $j$. Differentiating the differential equation repeatedly gives $f^{(j+3)}=f^{(j)}$. Only the three functions $f,f^{\prime },f^{\prime \prime }$ are needed: take $M$ to be the maximum of their absolute-value bounds on $[-R,R]$. Then $|f^{(j)}(x)|\le M$ simultaneously for every $j\ge 0$ and $|x|\le R$.

At zero the derivatives consequently repeat the values $1,0,0$. Taylor's theorem therefore givesThe last limit of a sequence follows because the ratio of consecutive majorants is $R/(m+2)$, eventually less than $1/2$. Since $R$ is arbitrary, Taylor expansion from a periodic derivative equation provesThis proves convergence to the function using a remainder bound; it does not assume beforehand that a smooth function equals its Taylor series.

The mean value theorem states that, if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, there is $c\in (a,b)$ such thatTo deduce it from Rolle's theorem, subtract the secant line: $k(t)=f(t)-f(a)-\frac{f(b)-f(a)}{b-a}(t-a)$. This function has the same continuity and differentiability properties and satisfies $k(a)=k(b)=0$. Rolle's theorem gives an interior point with $k^{\prime }(c)=0$, which is precisely the required equality.

If $h^{\prime }=0$ on $\mathbb{R}$, apply the mean value theorem on any $[x,y]$ with $x<y$. It gives $h(y)-h(x)=(y-x)h^{\prime }(c)=0$. Hence $h$ is constant.

For the differential equation, the product rule givesThe preceding result makes this product constant. Thus all solutions are $f(x)=C{e}^{ax}$, and differentiation verifies every such solution.

For the integral assertion, continuity ensures that $f$ is Riemann integrable on every finite closed interval. Use the usual orientation for an integral with negative upper endpoint. For $h\ne 0$, additivity of the integral givesThe absolute value of this expression is at most the supremum of $|f(t)-f(x)|$ on the segment joining $x$ and $x+h$. Continuity at $x$ makes this bound tend to zero for either sign of $h$. Consequently $F^{\prime }(x)=f(x)$, proving this form of the fundamental theorem of calculus.

Finally, a differentiable solution of the integral equation is continuous, so the fundamental theorem of calculus yields $g^{\prime }=g$; setting $x=0$ gives $g(0)=A$. The differential-equation calculation above now forcesConversely, $A+{\int }_0^{x}A{e}^{t}dt=A+A(e^x-1)=A{e}^x$, so this function satisfies the original integral equation. It is unique because every differentiable solution has already been forced to this form and this initial condition.

The Einstein's principle of relativity says that the laws of physics take the same form in every inertial frame: no inertial frame is physically privileged, and uniform absolute motion cannot be detected by an internal experiment.

Assume $|u|<c$. Direct expansion of the transformed coordinates givesSince $y^{\prime }=y$ and $z^{\prime }=z$, the full Minkowski interval is invariant. Points of the outgoing light shell therefore satisfyAt each fixed $t^{\prime }>0$ this is a sphere of radius $c{t}^{\prime }$. The future branch remains future-directed: for a shell point with $t>0$, $|x|\le ct$ gives $t^{\prime }=\gamma (t-ux/c^2)>0$.

A Lorentz transformation is a linear coordinate transformation preserving the Minkowski metric, equivalently ${\Lambda }^T\eta \Lambda =\eta$. The calculation establishes this property for all events, not only points on the light cone. The displayed map has determinant one and preserves time orientation, so it is the proper orthochronous boost along the $x$ direction.

Multiplying numerator and denominator by the same nonzero constant does not change a Möbius transformation. Since $b\ne 0$, divide by $b$ and relabel the coefficients so that $b=1$. On $|z|=1$, preservation of the circle givesEquating the constant and first Fourier coefficients yieldsIf $c=0$, the second equality forces $a=0$, contradicting the nonzero determinant. Hence $d=\bar{a}/\bar{c}$. With $C=|c{|}^2>0$, the first equality becomesThe root $C=|a{|}^2$ would imply $ad=|a{|}^2/\bar{c}=c$, making the numerator and denominator proportional. It is excluded by nondegeneracy. Thus $|c|=1$ and $d=c\bar{a}$. Writing $1/c=e^{i\psi }$ givesHere $a$ is the normalized coefficient after division by the original $b$. The inequality $|a|\ne 1$ is again the determinant condition. For $|a|<1$ the map preserves the disc; for $|a|>1$ it interchanges its inside and outside. Circle preservation alone does not justify restricting to automorphisms of the unit disk.