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The image of the linear map $A+B$ is contained in the sum of the images of $A$ and $B$. Since the dimension of a vector space of this sum is at most the sum of their dimensions, matrix rank is subadditive. Iteration shows that a sum of $r$ matrices of matrix rank at most one has matrix rank at most $r$, over any field.

A nonnegative function-valued operator $S$ is sublinear if $S(f+g)\le S(f)+S(g)$ and $S(cf)=|c|S(f)$ almost everywhere. This pointwise domination notion includes maximal one-sided interval averaging operators and differs from requiring an ordinary linear map.

A linear map between spaces of linear operators. A physical quantum channel is a superoperator that is trace preserving and completely positive; a general superoperator need not have either property.

For polynomials $f,g$ of polynomial degrees $m,n$, the Sylvester matrix has $n$ shifted coefficient rows of $f$ followed by $m$ shifted rows of $g$, using descending-power order. It is a square matrix of size $m+n$. Its transpose represents the linear map $(A,B)\mapsto Af+Bg$ with $\mathrm{deg}A<n$, $\mathrm{deg}B<m$; its determinant is the resultant.

For an original parameter example, choose $f(x)=x^2-1$ and $g_t(x)=x-t$. ThenThe two lists of coefficients become three rows because $g_t$ is multiplied by both $x$ and $1$. Thus the common-root parameter values are $t=-1$ and $t=1$. At $t=1/2$, the resultant is $-3/4$, so the two polynomials have no common root.

One can verify the certificate without any determinant expansion:For $t\ne \pm 1$, division by $t^2-1$ gives a Bézout's identity for polynomials. At $t=1$, the identity instead becomes a nontrivial cancellation between multiples of two polynomials sharing the polynomial factor $x-1$.

The general proof needs only a short linear algebra outline. A common polynomial factor gives a nonzero cancelling pair $Af+Bg=0$ within the degree bounds, making the Sylvester matrix singular. Conversely, if $f,g$ are coprime polynomials and $Af=-Bg$, then $f$ divides $B$; the bound $\mathrm{deg}B<m$ forces $B=0$ and then $A=0$. Thus the linear map has zero kernel of a linear map exactly in the coprime case. Over an algebraic closure, a nonconstant common polynomial factor is equivalent to a common root.

In 1840, James Joseph Sylvester described forming the shifted coefficient rows and taking a signed sum of products selected by permutations. In modern notation, this is the Sylvester matrix and its determinant. The construction turns the nonlinear common-root of a polynomial question into linear algebra. His paper's word derivatives concerns elimination, rather than the formal derivative used below.^[ref]

Over a field of characteristic two, the linear map $T:M_n(F)\to M_n(F)$, $A\mapsto A+A^T$, has kernel of a linear map the symmetric matrices, of dimension $n(n+1)/2$. Its image of a linear map is the linear subspace of symmetric matrices with zero diagonal, of dimension $n(n-1)/2$. Indeed, $T(A)=0$ means $A=A^T$, and every symmetric zero-diagonal matrix is the image of the matrix obtained by retaining just its entries above the diagonal. Thus the two dimensions still add to $n^2$, as required by the rank-nullity theorem.

For a finite-dimensional symplectic vector space, the symplectic adjoint of a linear map $T$ is the unique linear map $T^{\ast }$ satisfying the displayed identity. Existence and uniqueness follow because the nondegenerate bilinear form identifies the vector space with its dual space. If $\omega$ has matrix $J$ in a basis, then $T^{\ast }=J^{-1}{T}^{\mathsf{T}}J$. Thus $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$ and $(T^{\ast }{)}^{\ast }=T$, and membership in the symplectic Lie algebra is equivalent to $T^{\ast }=-T$. This operation is linear over the ground field; it does not involve complex conjugation. It preserves diagonalizability and the property of being a nilpotent linear map, since it is conjugate to matrix transpose.

For a finite-dimensional vector space over any field with a nondegenerate alternating bilinear form $B$, the symplectic group over a field consists of invertible linear maps preserving $B$. In a symplectic basis its matrices satisfy $g^{\mathsf{T}}Jg=J$. Matching two symplectic bases gives a unique such linear map, and a symplectic basis exists by successively splitting off paired vectors. This includes arbitrary-characteristic and infinite-field versions, beyond the real Lie group and the symplectic group over a finite field.

A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

A system of linear equations imposes several linear equations on the same unknowns. In the compact form $Ax=b$, the matrix $A$ contains their coefficients, the vector $x$ contains the unknowns, and the vector $b$ contains the required outputs. Over the real numbers, the following examples exhibit the three possible outcomes: one solution, infinitely many solutions, or no solution.

For $x+y=3$ and $x-y=1$, adding the two linear equations gives $2x=4$, hence $x=2$ and $y=1$. Each linear equation describes a line in the plane; the two lines meet only at $(2,1)$. Substitution checks both requirements: $2+1=3$ and $2-1=1$. The two measurements together determine both unknowns.

Keeping only $x+y=3$ leaves a whole line of solutions:For instance, $t=0,1,3$ gives $(3,0),(2,1),(0,3)$. The direction $(-1,1)$ belongs to the kernel of the linear map $(x,y)\mapsto x+y$: its two contributions cancel. Moving in that direction changes the unknowns without changing their measured sum. The solution line is an affine space, translated away from the origin.

For $x+y=3$ and $2x+2y=7$, the second linear equation instead demands $x+y=3.5$. The parallel lines never meet. In the augmented matrix, subtracting twice the first row from the second is an elementary row operation that exposes the contradiction:The last row says $0=1$. A zero row of coefficients is harmless only when its appended entry is also zero.

The column viewpoint unifies these pictures. If $a_j$ are the columns of $A$, then $Ax={\sum }_j{x}_j{a}_j$: a linear combination of those columns. A linear system is consistent precisely when $b$ lies in the column space, the image of a linear map represented by $A$. Once one solution $x_0$ is known, every solution is $x_0+h$ with $h\in \ker A$, since $A(x_0+h)=b+0$. Conversely, subtracting any two solutions gives an element of the kernel. Thus the kernel measures freedom, while the image of a linear map decides existence. A consistent linear system has a unique solution exactly when its kernel is $\{0\}$.^[ref]

Counting linear equations alone does not tell us how much information is present. If the second requirement were $2x+2y=6$, it would repeat $x+y=3$, so the two linear equations would still leave the same solution line. The matrix rank counts independent requirements. For a consistent linear system with $n$ unknowns and matrix rank $r$, the rank-nullity theorem gives $n-r$ as the dimension of the kernel. This is the number of independent directions in which a solution can move. In the single-measurement example, one measured combination leaves one direction of freedom; in the sum-and-difference example, neither unknown can change without changing an output.

A tangent vector is a linear map on germs of smooth functions satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. In a manifold chart, first-order expansion shows $Dh={\sum }_{i}D(u^i){\partial }_{i}h(x)$. Thus the coordinate derivations form a basis, and the dimension equals the manifold dimension. Composition of germs with a smooth map defines its differential of a smooth map without choosing charts.

A tensor product turns a bilinear map into a linear map. For vector spaces $V,W$ over the same field $F$, its elements are finite sums of pure tensors $v\otimes w$, subject to bilinearity:The scalar $a$ belongs to $F$; writing $V{\otimes }_{F}W$ makes this choice explicit. These rules preserve separate linearity while allowing arbitrary linear combinations of products.^[ref]

For a concrete example, take $V=W={\mathbb{R}}^2$, with basis vectors $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four tensor-product basis elements areThe pure tensor made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands toEach basis vector in one factor can interact with each basis vector in the other. This is why dimensions multiply; the tensor-product coefficient matrix displays these four interactions as a two-by-two grid.

The tensor product is different from the Cartesian product $V\times W$. With componentwise vector addition and scalar multiplication, the latter contains pairs and has dimension $\dim V+\dim W$. The map $\tau (v,w)=v\otimes w$ is bilinear, generally neither linear on that Cartesian product nor injective nor surjective. In our example, $\tau (2v,2w)=4\tau (v,w)$, and $\tau (2v,w/2)=\tau (v,w)$; the separability of a tensor explains why some outputs cannot be reached at all. Both spaces happen to have dimension four when $V=W={\mathbb{R}}^2$, but this coincidence does not identify their structures.

The total differential is the linear map that combines the first-order responses to simultaneous changes in all coordinates. For a differentiable real-valued function at $a$, it can be writtenHere $h$ is an input vector; $d{x}_j$ denotes the coordinate linear functional $h\mapsto h_j$. The coordinate formula and the Fréchet derivative describe the same linear map. The defining approximation isThe little o notation says the remaining approximation error becomes small relative to the input norm, across every direction of approach.^[ref]

For $f=x^2+xy+2y^2$ at $a=(1,1)$, $d{f}_a(h,k)=3h+5k$. A simultaneous change $(h,k)=(0.02,-0.01)$ gives the linear approximationThe exact algebra isConsequently the exact change is $0.0104$, and the approximation error is $0.0004$. If both input changes are halved, the linear prediction becomes $0.005$ and the error becomes $0.0001$: the first-order response halves, while the quadratic error falls by a factor of four. The right panel under directional derivative displays this comparison as the displacement is scaled.

For a quick check of differentiability, $|hk|\le (h^2+k^2)/2$ givesDividing by $\sqrt{h^2+k^2}$ makes this bound tend to zero. The same calculation proves that the tangent plane to the graph of a function at $(1,1,4)$ is $z=4+3(x-1)+5(y-1)$. It approximates the entire local surface, rather than only two coordinate slices.

The total differential is useful for estimating the combined effect of small input changes without reevaluating a complicated function. For example, if $|h|\le 0.02$ and $|k|\le 0.01$, then $|d{f}_a(h,k)|\le 0.11$. The exact remainder bound from the displayed polynomial is at most $0.0008$, so this example has $|\Delta f|\le 0.1108$ throughout that input rectangle. These are deterministic bounds; a variance calculation additionally needs a joint probability distribution and its covariances. A first-order estimate is also local: large changes require control of the remainder, not merely knowledge of the partial derivatives at one point.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.

For linear maps $A,B$ on a finite-dimensional vector space, the cyclic property of the trace gives $\mathrm{tr}(AB-BA)=\mathrm{tr}(AB)-\mathrm{tr}(BA)=0$. In matrix entries both traces are ${\sum }_{i,j}{A}_{ij}{B}_{ji}$ after exchanging indices. Thus the trace of a finite-dimensional Lie algebra representation vanishes on the derived algebra. In particular, for a perfect Lie algebra, every representing matrix has zero trace.

The matrix transpose interchanges rows and columns: $(A^T{)}_{ij}=A_{ji}$. On a fixed matrix space, $A\mapsto A^T$ is the transposition map, a linear map that preserves the trace norm but is not completely positive in dimensions greater than one.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.

A linear map $T:V\to W$ has a trivial kernel when $\ker T=\{0_V\}$: the zero vector is the only input producing the zero output. The kernel is a singleton set, not the empty set, because every linear map sends zero to zero. Equivalently, its nullity is zero and $T$ is injective. This holds even without a finite-dimensional hypothesis; see Linear Algebra Done Right by Sheldon Axler (4th edition).

The reason this zero-output test controls all outputs is linearity:With a trivial kernel, equal outputs force equal inputs. Therefore every equation $T(x)=b$ has at most one solution. Existence is a separate question: $b$ must belong to the image of a linear map. In the fiber description under nullity, $x_0+\{0\}$ is just the single point $x_0$.

For example, the linear map $J:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $J(x,y)=(x,y,0)$, has trivial kernel and full column rank:The output $(3,4,0)$ comes from exactly $(3,4)$, while $(3,4,1)$ is unreachable. Conversely, $P:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $P(x,y,z)=(x,y)$, is surjective and has full row rank, but its kernel is $\mathrm{span}\{(0,0,1)\}$: every output $(a,b)$ comes from the solution straight line $(a,b,s)$.

For an $m\times n$ matrix, a trivial kernel is equivalent to matrix rank $n$ and linear independence of its columns. It can therefore occur only if $n\le m$. If $m=n$, the rank-nullity theorem also forces surjectivity, giving an invertible matrix; if $m>n$, the example $J$ shows why unique recovery of reachable inputs does not imply that every output is reachable. Finally, a trivial kernel guarantees exact uniqueness without guaranteeing numerical stability: the singular matrix example $A(t)$ has trivial kernel for every $t\ne 0$, even as its condition number grows without bound when $t$ approaches zero.

For infinite spaces over a field of size $q$, two-variable types are determined by the kernel of the linear map sending $(a,b)$ to $ax+by$. Its ranks give the zero pair, $q+1$ one-dimensional directions and the independent-pair type. All $q+3$ types are isolated by finite formulas. Quantifier elimination makes the Lindenbaum algebra their full power-set Boolean algebra. The one-variable algebra has just the zero and nonzero atoms, hence four Elements.

A uniform dilation about the origin multiplies every vector by one positive real scale $s$. The linear map $\mathbf{x}\mapsto s\mathbf{x}$ multiplies lengths by $s$ and $d$-dimensional volumes by $s^d$. This differs from stretching different axes by different factors. The case $s=1$ is the identity map.