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A root of a polynomial $f$ is a scalar $a$ satisfying $f(a)=0$ in the chosen field or ring. For $p(X)=(X+1)(X-2{)}^2$ over the real numbers, the distinct roots of a polynomial are $-1$ and $2$. The second has multiplicity of a root two, so there are three roots when counted with multiplicity.

The domain matters. $X^2+1$ has no real polynomial root, but has the two complex number roots $i,-i$. The fundamental theorem of algebra guarantees a polynomial root in the complex numbers for every nonconstant complex polynomial, and complete factorization counts its degree of a polynomial many roots with multiplicity of a root. It does not promise that those roots are real or that they have a simple expression.

Consider the diagonal matrix $D=2I_2$ and a Jordan blockBoth have characteristic polynomial $(t-2{)}^2$, trace $4$, determinant $4$, and eigenvalue $2$ with algebraic multiplicity two. But $\ker (D-2I)$ is the whole plane, whereas $(J-2I)(a,b{)}^T=(b,0{)}^T$ vanishes only when $b=0$. Their eigenspaces have different dimensions, so eigenvectors under similarity rules out matrix similarity. There is an even shorter proof: $S^{-1}(2I)S=2I$ for every invertible matrix $S$.

The missing information is how the linear operator acts beyond its eigenvectors. Here $D-2I=0$, while $J-2I\ne 0$ but $(J-2I{)}^2=0$. Their minimal polynomials are respectively $t-2$ and $(t-2{)}^2$. Over a field where the characteristic polynomial splits, the full sizes of the Jordan blocks, encoded by the Jordan normal form, determine the matrix similarity class.

For a nonzero element of a Weyl algebra over a characteristic-zero field, choose a monomial $x^{\alpha }{D}^{\beta }$ of maximal total degree with coefficient $c_{\alpha \beta }\ne 0$. The displayed iterated commutators kill every lower-degree monomial. Another monomial of the same degree survives only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. Hence the result is a nonzero scalar. If the element has filtered degree at most $j$, expansion of these commutators gives a sum of terms $uav$ with both $u$ and $v$ of degree at most $j$. This proves faithful finite-step action of a Weyl algebra over any characteristic-zero field.

For a nonzero multivariate polynomial of total degree $d$ over a field, and a finite subset $S$ of that field, at most $d|S{|}^{n-1}$ points of $S^n$ are zeros. For one variable this is the root bound for a polynomial. If $d\ge |S|$, the bound is already trivial. Otherwise, in the inductive proof write the polynomial as degree $k$ in its last variable, with a nonzero leading coefficient of degree at most $d-k$ in the other variables. The leading coefficient vanishes on at most $(d-k)|S{|}^{n-2}$ fibers; outside those fibers there are at most $k$ roots of a polynomial per fiber. Counting the bad fibers with the trivial $|S|$ bound gives the asserted inequality.

Fix an $R$-bimodule $M$. A unital ground-ring-linear split extension with square-zero kernel $M$ can be identified with $R\oplus M$, with multiplication $(a,u)(b,v)=(ab,av+ub+f(a,b))$. Associativity is exactly $\delta f=0$. Changing a unital linear section by $g:R\to M$ changes $f$ by $\delta g$. Thus equivalence classes of these split extensions are $H{H}^2(R,M)$. Over a field every such vector-space extension splits; over a general ground ring a splitting hypothesis is essential for this ordinary cochain classification.

A ring is semiprime when it has no nonzero nilpotent ideals. Equivalently, a two-sided ideal with square zero must be zero. This is weaker than having no nonzero nilpotent elements: a matrix ring over a field is semiprime and can have nilpotent matrices.

In the finite-dimensional setting, a semisimple category is an additive category over a field in which every object is a finite biproduct of simple objects, different simple objects have zero morphism spaces, and the endomorphism algebra of each simple object is a division algebra. Such categories have finite matrix blocks and all their idempotents split. The semisimplicity of the reduced Temperley-Lieb category gives an example whose simple objects are the surviving Jones-Wenzl colors.

A filling $T$ of shape $\lambda$ and content $\alpha$ defines a module homomorphism ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: given a numbered tableau, sum over distinct assignments obtained by permuting $T$ within its rows, and put each numbered entry into output row $j$ when its assigned label is $j$. Restricting to $S^{\lambda }$ gives ${\hat{\Theta }}_T$. For Semistandard Young tableaux $T$, these restrictions are linearly independent over every field. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ if the characteristic of a field is not two, or if $\lambda$ is a two-regular partition of an integer. For two-singular shapes in characteristic two, they need not span. Column antisymmetrization and triangular leading tabloids establish independence; straightening the images of a cyclic polytabloid establishes spanning under the stated hypotheses.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.

If $A$ is a finite module over a commutative central subring $R$ and $M$ is a simple left $A$-module, then $M=Am$ is finite over $R$. A central element not annihilating $M$ acts bijectively, since its kernel and image are $A$-submodules. Put $B=R/{\mathrm{Ann}}_R(M)$. For $0\ne r\in B$, write generators as $m_i=r{\sum }_j{b}_{ij}{m}_j$. The adjugate matrix identity shows $\det (I-r(b_{ij}))$ annihilates $M$, hence is zero in $B$. Since this determinant has form $1-rq$, $r$ is invertible. Thus $B$ is a field. If $R$ is a finite-type algebra over an algebraically closed field $k$, the residue field is $k$, and every such simple module is finite-dimensional over $k$.

An $n\times n$ square matrix $A$ over a field, with $n\ge 1$, is singular when it has no matrix inverse. Geometrically, its linear map loses at least one input direction: some nonzero vector $v$ satisfies $Av=0$, so $A(x+v)=Ax$. Over the real numbers, the following family makes that loss visible:

For this diagonal matrix, $\det A(t)=t$. Whenever $t\ne 0$, the input can be recovered as $(x,y)=(b_1,b_2/t)$ and the kernel is $\{0\}$. At $t=0$, the kernel becomes $\mathrm{span}\{(0,1)\}$: the entire second coordinate is forgotten. Consequently $A(0)x=(b_1,b_2)$ has no solution if $b_2\ne 0$, and has the solution straight line $x=(b_1,s)$, $s\in \mathbb{R}$, if $b_2=0$. A singular matrix therefore need not make a particular linear system inconsistent; it prevents unique recovery when that linear system is consistent.

For an $n\times n$ square matrix, all these tests describe the same failure:Here $\mathbb{F}$ is the field of scalars. Equivalently, the columns are linearly dependent, the linear map is neither injective nor surjective, and some outputs cannot be reached. See the invertible matrix equivalences and Strang's MIT 18.06SC lecture summary, Multiplication and inverse matrices, p. 2. Over an infinite field, every consistent linear system for a singular matrix has infinitely many solutions; over a finite field with $q$ elements and nullity $d$, it has exactly $q^d$ solutions. This follows by adding each kernel vector to one solution.

The square hypothesis is essential. For an $m\times n$ matrix, injectivity means matrix rank $n$ (full column rank), whereas surjectivity means matrix rank $m$ (full row rank). A rectangular matrix can satisfy one condition without the other; its kernel and image of a linear map give the useful tests without calling it a singular matrix.

Exact singularity also differs from sensitivity. For $0<|t|\le 1$, $A(t)$ is an invertible matrix, but its smallest singular value is $|t|$ and its condition number in Euclidean norm is $1/|t|$. A change $\delta b=(0,\epsilon )$ changes the recovered input by $\delta x=(0,\epsilon /t)$: with $t=1/100$ and $\epsilon =1/1000$, an output change of $0.001$ causes an input change of $0.1$. These are exact toy values, not measurements. The condition number quantifies this sensitivity before a direction is lost; see Steve Marschner, Condition numbers, §2. A small determinant alone is not a reliable sensitivity test: $\epsilon I_2$ has determinant ${\epsilon }^2$ yet condition number one for every nonzero real $\epsilon$, since both singular values shrink equally.

Over a field $k$, the trace-zero $2\times 2$ matrices form the special linear Lie algebra ${\mathfrak{sl}}_2(k)$. The matrix trace of a commutator is zero, so this is a Lie subalgebra of the general linear Lie algebra. A basis isIn fact $\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)=aH+bE+cF$, uniquely. The individual products areSubtracting in the stated order givesFor characteristic different from two, this is the standard sl2 triple; see Etingof, Section 11.4. The products remain valid in characteristic two, but there $H=I$ is central and $2E=2F=0$, so the familiar characteristic zero structure changes.

The following table gives all nine basis brackets. Each entry is $[\text{row},\text{column}]$; the diagonal vanishes by alternation, and exchanging row and column reverses the sign.

Bilinearity now computes any commutator, for exampleThe three structure constants of a Lie algebra relations encode all of this multiplication. They also show why the sl2 triple is useful in Lie algebra representations: the actions of $E$ and $F$ move between eigenvalues of the action of $H$ in steps of two.

For a singular projective Weierstrass cubic, its smooth locus of a variety contains the point at infinity and carries the chord-and-tangent group law. The singular point is excluded. Over an algebraic closure, a node gives the multiplicative algebraic group and a cusp gives the additive group. A node whose branches are not defined over the base field gives a nonsplit one-dimensional algebraic torus. This group is distinct from an elliptic curve, whose projective model is smooth.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.