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The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

A matrix is a diagonalizable matrix if there is an invertible matrix $P$ for which $P^{-1}MP$ is diagonal. Its columns then form a basis of eigenvectors, and the diagonal entries are their corresponding eigenvalues.

Here the given eigenvectors form an orthonormal basis. Put them into the columns of $Q$:Define $\Lambda =Q^{T}MQ$. Its entries areThis proves the claimed diagonalization of a matrix, rather than merely guessing it:The trace follows directly from this factorization:because each column of $Q$ has unit length. Trace is the sum of the eigenvalues, with multiplicity.

For the particular matrix, $(M^2{)}_{ij}={\sum }_k{M}_{ik}{M}_{kj}$. If $i=j$, every $k\ne i$ contributes one and $k=i$ contributes zero. If $i\ne j$, precisely $k=i$ and $k=j$ contribute zero; the other $n-2$ terms contribute one. ThusApplying this identity to a nonzero eigenvector givesEvery eigenvalue is therefore $n-1$ or $-1$.

If $r$ is the multiplicity of $n-1$, the trace formula for a diagonalizable matrix and the zero diagonal giveso $r=1$. Up to ordering, the required diagonal form isFor $n=1$ this means just $(0)$. For $n\ge 2$, the eigendirections are also transparent: the all-ones vector has eigenvalue $n-1$, and the $(n-1)$-dimensional subspace whose coordinate sum is zero has eigenvalue $-1$, since $(M\mathbf{x}{)}_i={\sum }_j{x}_j-x_i$. The spectrum consists of one collective mode and $n-1$ zero-sum modes.

For the reality of the eigenvalues, one may initially allow a nonzero complex eigenvector $\mathbf{v}$ with $M\mathbf{v}=\lambda \mathbf{v}$. Since $M$ is real symmetric, it is a Hermitian matrix, andis real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf{v}$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.

For two such eigenvectors, symmetry givesDistinct eigenvalues therefore give zero dot product. Together with the prescribed unit lengths,These $n$ vectors form an orthonormal basis of ${\mathbb{R}}^n$.

Expand a real unit vector as $\mathbf{x}={\sum }_a{c}_a{\mathbf{e}}_a$. Then ${\sum }_a{c}_a^2=1$, and its Rayleigh quotient isMoreover,Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge 2$. Equality occurs precisely along the top eigendirection:

To obtain a unit vector in $S$ using just ${\mathbf{e}}_1,{\mathbf{e}}_2$, let $u=({\mathbf{e}}_1{)}_1$ and $v=({\mathbf{e}}_2{)}_1$ denote their first coordinates. If $u^2+v^2>0$, takeThen ${\alpha }_{1}u+{\alpha }_{2}v=0$ and ${\alpha }_1^2+{\alpha }_2^2=1$. Hence ${\alpha }_1{\mathbf{e}}_1+{\alpha }_2{\mathbf{e}}_2$ has first coordinate zero and unit length. If $u=v=0$, simply take ${\alpha }_1=1,{\alpha }_2=0$. In either case the Rayleigh quotient of the constructed vector isThe continuous quadratic form attains its maximum on the compact space $S$, so

For $\mathbf{x}=(0,\mathbf{y}{)}^T$, deletion of the first row and column gives ${\mathbf{x}}^{T}M\mathbf{x}={\mathbf{y}}^{T}A\mathbf{y}$, while $|\mathbf{x}|=|\mathbf{y}|$. The spectral theorem for real symmetric matrices applied to $A$ identifies its largest eigenvalue with this maximum:Combining the upper bound for every unit vector with the constructed lower bound provesThis is largest-eigenvalue interlacing for a principal submatrix: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.

Every counted point has integer coordinates and lies in the square $[-n,n{]}^2$, while the origin is always counted. HenceNo precise circle-area estimate is needed; this elementary bound controls the growth of the power series coefficients.

The comparison test for series states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.

For $|z|=r<1$,For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the ratio test and the comparison test for series show absolute convergence. The case $r=0$ is immediate.

If $|z|\ge 1$, then $|a_n{z}^n|\ge 1$, so the terms do not tend to zero. The term test for divergence therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the radius of convergence isThe series converges exactly in the open unit disc. This is an example of polynomial coefficient growth with unit power-series radius: the lattice-count bound is polynomial rather than exponential.

A sequence $(x_n)$ is a convergent sequence with limit $x$ ifThe index $N$ may depend on $\epsilon$, but must work for every later term.

Fix $\epsilon >0$. Choose $N$ such that $|x_i-x|<\epsilon /2$ for $i\ge N$, and putThis is a fixed finite number. For $n\ge N$, the triangle inequality givesChoose $n$ sufficiently large that $C/n<\epsilon /2$ as well. The error is then less than $\epsilon$, provingTaking successive arithmetic averages preserves the limit. This is the Cesàro theorem for convergent sequences: any troublesome initial terms contribute only a fixed numerator divided by $n$.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.