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On all real-valued functions on $\mathbb{R}$, define $D_{r}f(x)=f(x+r)-f(x)$. The operators commute, and every real number except $-1$ is an eigenvalue. For eigenvalue $\lambda \ne -1$, values may be chosen independently on the cosets of $r\mathbb{Z}$ and extended by $f(x+r)=(1+\lambda )f(x)$, so every eigenspace is infinite-dimensional.

Let $C=[B,A]$ and suppose $AC=CA$. Then $C$ is nilpotent. For the derivation $D(X)=[B,X]$, one has $D(p(A))=p^{\prime }(A)C$. If $f(A)=0$, induction givesIndeed, if $u{C}^m=C^{m}u=0$, differentiating $u{C}^m=0$ and multiplying on the left by $C^m$ gives $C^{m}D(u)C^m=0$; for $u=f^{(k)}(A)$ this replaces $m$ by $2m+1$. Taking $k=\mathrm{deg}f$ makes $f^{(k)}$ a nonzero constant and proves nilpotence.

The minimal polynomial $m_A$ is the unique monic polynomial of least degree satisfying $m_A(A)=0$. It divides every polynomial that annihilates $A$.

For a square matrix $A$, its exponential is the convergent power seriesIt is the fundamental matrix for $y^{\prime }=Ay$ and satisfies $e^{(s+t)A}=e^{sA}{e}^{tA}$.

Eigenvalue interlacing bounds the ordered eigenvalues of a compressed or rank-modified operator between those of the original operator.

The characteristic polynomial isIts roots are precisely the eigenvalues of $A$.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

A densely defined operator $T$ on a Hilbert space is self-adjoint when it equals its adjoint operator, including equality of their operator domains:For a bounded linear operator defined on the whole Hilbert space, this is equivalent tofor all vectors $v,w$. For an unbounded self-adjoint operator, the domain condition is essential: the inner-product identity on $D(T)$ alone gives only a symmetric operator.

If $V$ is a subspace of a space with an inner product and $P_V$ is its orthogonal projection, the compression of an operator $A$ to $V$ isIf the columns of $Z$ are an orthonormal basis of $V$, its matrix in that basis is $Z^{\ast }AZ$.

For $d^{T}x>0$, setThen $d^{T}y=1$, and $Ax=b$ becomes $Ay=bt$, turning a linear-fractional program into a linear program with objective $c^{T}y$.

This quantitative refinement belongs to the theory of Baker's theorem; its dependence on the scaled coefficient bound is more specific than that broad theorem.

Choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_n$, putand choose $A_i\ge 10$ large enough to bound the naive polynomial height of the minimal polynomial of ${\alpha }_i$ and $\exp (|\log {\alpha }_i|)$. After relabelling the terms if useful, setThere is an effectively computable constant $C$, depending only on $n$ and the degree of the number field generated by the ${\alpha }_i$, such that

The division by $\log A_n$ is the useful refinement over the general lower bound for a linear form in logarithms when one algebraic number has a large height correlated with a coefficient.

Choose logarithms of nonzero algebraic numbers ${\alpha }_1,\dots ,{\alpha }_n$ and algebraic numbers ${\beta }_0,\dots ,{\beta }_n$, and putLet $A_j$ bound $10$, the naive polynomial height of the minimal polynomial of ${\alpha }_j$, and $\exp (|\log {\alpha }_j|)$. Let $B$ bound all the corresponding heights of the ${\beta }_j$ and all $\log A_j$. There is an effective constant $C$, depending only on $n$ and the degree of the number field generated by the data, such that

Let $K={\mathbb{F}}_{2^d}$, let $\alpha$ generate $K^{\times }$, and let $T:K\to {\mathbb{F}}_2$ be nonzero linear with nondegenerate bilinear form $(x,y)\mapsto T(xy)$. If the minimal polynomial of $\alpha$ is $P(X)=X^d+{\sum }_{j<d}{c}_j{X}^j$, then $x_n=T({\alpha }^n)$ satisfiesso it is produced by an LFSR of length at most $d$. Its period is exactly $2^d-1$: any period $r$ would imply $T(({\alpha }^r-1)y)=0$ for every $y\in K$, hence ${\alpha }^r=1$ by nondegeneracy.

If an output prefix contains $d$ consecutive zeros followed by a one, no register of degree at most $d$ can generate it: the all-zero state would have been reached and would remain zero forever.

A maximal-period feedback polynomial cannot have $1$ as a root. Over ${\mathbb{F}}_2$, this saysso an even number of the coefficients $a_i$ equal one and the full monic polynomial has an odd number of nonzero coefficients.

A degree-$d$ binary register has $2^d$ states. The zero state is fixed, so a nonzero periodic orbit has length at most $2^d-1$.

The Berlekamp-Massey algorithm finds the shortest linear recurrence satisfied by a finite sequence over a field. It processes symbols incrementally, updating the connection polynomial whenever the current recurrence has a nonzero discrepancy.

A feedback shift register updates a finite state by shifting its stored symbols and inserting the value of a feedback function of the previous state. The feedback function need not be linear.