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Iterating an affine map gives $x_k=M^{k-1}{x}_1+{\sum }_{j=0}^{k-2}{M}^{j}c$. A fixed point exists exactly when $c$ is in the image of a linear map $I-M$. If the spectral radius of $M$ is below one, its matrix powers tend to zero and every orbit converges to the unique fixed point $(I-M{)}^{-1}c$.

The cokernel records output information that a linear map cannot produce. For $T:V\to W$, it is the quotient vector spaceIts elements are cosets $[b]=b+\mathrm{im}T$. Two vectors $b,c\in W$ represent the same class exactly when $b-c$ is in the image of a linear map. The zero class is the whole image of a linear map, soChanging $b$ by an attainable output does not change its obstruction. This is useful because one can test compatibility before trying to find an input. The quotient vector space construction is described in Axler's §3E. ^[ref]

For the linear map $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ given by $T(x,y)=(x+2y,0)$, the image of a linear map is the horizontal line. All outputs $(b_1,b_2)$ with the same second coordinate have the same cokernel class, because their difference is horizontal. Thus this cokernel is isomorphic to $\mathbb{R}$ via $[(b_1,b_2)]\mapsto b_2$. In particular, $(3,0)$ has zero class and is attainable, while $(3,1)$ has nonzero class and is unattainable.

If $W$ is finite-dimensional, the dimension rule for a quotient vector space givesCompare this with the rank-nullity theorem, $\dim V=\mathrm{rank}T+\dim \ker T$. The kernel counts invisible inputs; the cokernel counts missing output directions. They need not have the same dimension: $B(x,y,z)=(x+2y,0)$ has a two-dimensional kernel and a one-dimensional cokernel. A linear map is surjective exactly when its cokernel is zero.

The cokernel is a quotient, rather than a chosen perpendicular vector subspace. With a finite-dimensional inner product space as output, one may represent each class by its unique component in the orthogonal complement of the image of a linear map. That representation depends on the chosen inner product. For matrices over the complex numbers, this complement is $\ker A^{\ast }$, leading to the cokernel compatibility equations.

The same quotient definition applies to a group homomorphism of abelian groups. For $T:\mathbb{Z}\to \mathbb{Z}$ given by $T(n)=2n$, the cokernel is $\mathbb{Z}/2\mathbb{Z}$. The equation $2n=b$ is solvable exactly when $b$ is an even integer; the two cosets record integer parity. A quotient captures the obstruction here just as it captures a missing output coordinate for a linear map.

The column rank of a matrix is the dimension of its image of a linear map, equivalently the dimension of the linear span of its columns. Multiplication on the left by an invertible matrix preserves all column linear relations, and elementary column operations preserve their linear span. In row echelon form the pivot columns form a basis of the column space. Its dimension equals the number of nonzero rows, and hence the row rank.

The column space of a matrix $A$ is the linear span of its columns. It is the image of a linear map represented by $A$.

If an $m\times n$ matrix has columns $c_1,\dots ,c_n$, then matrix multiplication gives $Ax=x_1{c}_1+\cdots +x_n{c}_n$. Thus its column space consists of exactly the attainable output vectors in ${\mathbb{F}}^m$, and $Ax=b$ has a solution exactly when $b$ belongs to this column space. Its dimension is the matrix rank; the kernel instead describes input vectors in ${\mathbb{F}}^n$ whose contributions cancel.^[ref]

The monomial $x^i{y}^j$ is an eigenvector of the differential linear operator $x^2{\partial }_x^2+y^2{\partial }_y^2$, with eigenvalue $i(i-1)+j(j-1)$. Over the real or complex numbers this vanishes exactly for $i,j\in \{0,1\}$. On polynomials of bounded total degree, the kernel is therefore spanned by the permitted members of $1,x,y,xy$, while the image of a linear map is spanned by all remaining monomials. The trace is the sum of those diagonal eigenvalues.

For an idempotent linear map $P$, induction gives $P^j=P$ for all integers $j\ge 1$. Summing the matrix exponential series therefore proves the displayed identity. The direct sum of its kernel of a linear map and image of a linear map is fixed and scaled by $e^{\alpha }$, respectively. The result requires no Hermitian matrix hypothesis and is valid for real or complex $\alpha$.

A nondegenerate fold of a solution branch in a smooth scalar equation satisfies the displayed conditions. The implicit function theorem still gives $\lambda =\lambda (x)$ because $F_{\lambda }\ne 0$. Implicit differentiation gives ${\lambda }^{\prime }(x_{\ast })=0$ and ${\lambda }^{\prime \prime }(x_{\ast })=-F_{xx}/F_{\lambda }\ne 0$: the parameter has a genuine local extremum on the curve.

In the cubic polynomial, $F_x=0$ gives $x_{\ast }=\pm 1/\sqrt{3}$, and substituting into $F=0$ gives ${\lambda }_{\ast }=2x_{\ast }/3$. Furthermore, $F_{xx}=6x_{\ast }\ne 0$. Yet $\nabla F=(3x^2-1,1)$ never vanishes, so the full solution branch remains a regular level set everywhere. The obstruction belongs to the projection onto $\lambda$, not to the curve itself.

For systems, the cited continuation account uses a one-dimensional kernel of $F_u$ and $F_{\lambda }$ outside its image of a linear map to define a simple fold (its Definition 2.2). These conditions make $[F_u{F}_{\lambda }]$ have full rank and support a transverse correction. A nonzero second-order term along the kernel is additionally needed to describe a nondegenerate quadratic turning point; in the scalar example this is $F_{xx}\ne 0$.^[ref]

For a complex $m$-by-$n$ matrix $A$, the linear system $Ax=b$ has a solution precisely when $b$ is orthogonal to every vector in $\ker A^{†}$, where $A^{†}$ is the conjugate transpose. For a real matrix, use the transpose $A^T$ instead. Indeed,because the defining adjoint operator identity turns orthogonality to every $Ax$ into $A^{†}y=0$. Taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimensions. Thus a linear functional in the annihilator of a vector subspace gives the exact obstruction to solvability, rather than just a necessary test.

For $A=\mathrm{diag}(1,0)$, its image of a linear map is the first coordinate axis and $\ker A^{†}$ is the second. Consequently $Ax=(b_1,b_2{)}^T$ is solvable exactly when $b_2=0$.

For matrices representing linear maps on finite-dimensional vector spaces, $AX=B$ has a solution precisely when the image of a linear map represented by $B$ is contained in that represented by $A$. Necessity follows by applying the equation to vectors. For sufficiency choose an $A$-preimage of each column of $B$ and assemble those vectors as the columns of $X$. All solutions are $X_0+N$ with $AN=0$, equivalently every column of $N$ lies in the kernel of a linear map represented by $A$. When $B$ has at least one column, a solution is unique precisely when $A$ has trivial kernel.

If a linear functional $\ell$ vanishes on the image of a linear map $T$ but $\ell (b)\ne 0$, the equation $Tx=b$ has no solution. In Euclidean coordinates, a vector $n$ with $n^{T}A=0$ provides such an obstruction whenever $n^{T}b\ne 0$. In finite dimensions these annihilating functionals exactly characterize membership in the image: $b\in \mathrm{im}A$ precisely when it is orthogonal to every vector in $\ker A^T$.

For an idempotent linear map $P$ on a finite-dimensional real vector space, take any positive definite inner product on that space. The displayed new inner product is positive definite since its squared norm vanishes only if both $Pu$ and $(I-P)u$ vanish. It makes the kernel of a linear map and image of a linear map orthogonal. Since $P^2=P$, one has $⟨Pu,v{⟩}_P=⟨u,Pv{⟩}_P$; thus $P$ becomes an orthogonal projection.

A linear map between Lie algebra representations is a representation homomorphism when it commutes with the action of every element. Its kernel and image of a linear map are invariant subspaces. This is a map of modules, distinct from a Lie algebra homomorphism between the acting algebras themselves.

Linear algebra studies vector spaces and linear maps: operations that preserve vector addition and scalar multiplication. For a linear map $T$, the defining rule is $T(au+bv)=aT(u)+bT(v)$, where $u,v$ are vectors and $a,b$ belong to the underlying field. The vectors might be arrows, lists of measurements, or polynomials. What matters is that their linear combinations make sense and that the same rule governs their transformation.

A concrete linear map is $T(x,y)=(x+y,x-y)$. Its two outputs record a sum and a difference. Since $T(1,0)=(1,1)$ and $T(0,1)=(1,-1)$, we immediately obtain $T(2,1)=2(1,1)+(1,-1)=(3,1)$. Knowing the outputs of the basis vectors determines the output of every vector; a matrix stores exactly that information in its columns.

The attraction of linear algebra is that one small collection of ideas answers several different questions. A system of linear equations asks which inputs produce a prescribed output. The image of a linear map describes the attainable outputs, while the kernel describes the invisible input changes. A basis gives every vector an address, and change of basis lets us choose addresses that make a linear map easier to understand. This shared language turns a collection of separate calculations into a picture of what is possible and how much freedom remains. Gilbert Strang's MIT course emphasizes these connections and their usefulness across other disciplines.^[ref]

A linear map $T:V\to W$ between finite-dimensional vector spaces admits a linear left inverse $S:W\to V$ if and only if its kernel of a linear map is $\{0\}$, equivalently if and only if it is injective.

Necessity follows from $Tv=0⟹v=STv=0$. For sufficiency, choose a basis $v_1,\dots ,v_n$ of $V$. Triviality of the kernel of a linear map implies that $T{v}_1,\dots ,T{v}_n$ are linearly independent. By basis extension, they extend to a basis $T{v}_1,\dots ,T{v}_n,w_{n+1},\dots ,w_m$ of $W$. Define $S(T{v}_i)=v_i$ and $S(w_j)=0$, and extend by linearity. Then $ST{v}_i=v_i$ on a basis, proving $ST=I_V$.

The freedom in $S$ lies in its values on a direct-sum complement of the image of a linear map. For example, for $T(x,y)=(x,y,0)$, every linear map $S(u,v,w)=(u+aw,v+bw)$ is a left inverse, with arbitrary scalars $a,b$. A right inverse instead requires $T$ to be a surjective linear map.

The nullity of a linear map $T:V\to W$ is $n(T)=\dim \ker T$, the dimension of its kernel. It counts independent input directions that can change without changing the output. For a finite-dimensional vector space $V$, the rank-nullity theorem givesFor an $m\times n$ matrix $A$ of matrix rank $r$, this is $n-r$, using the number of columns, not the number of rows. ^[ref]

The nullity answers a practical question about a linear system: once one input is known to produce the requested output, how much freedom remains? If $T(x_0)=b$, its fiber of a function over $b$ isIndeed, $T(x)=T(x_0)$ exactly when $T(x-x_0)=0$. Thus every nonempty fiber is an affine subspace of dimension $n(T)$, with the same direction space as the kernel. The kernel itself is the fiber over zero. Outputs outside the image of a linear map have empty fibers; the nullity does not guarantee existence, and we assign no solution-space dimension to an empty fiber here.

In the diagram, $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ has matrix $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, matrix rank one and nullity one. Its kernel has basis $(1,-1)$, and the full solution of $F(x,y)=(c,0)$ is $(x,y)=(c,0)+s(1,-1)$. One free scalar moves the solution along a straight line; the output $(c,1)$ is impossible.

The existing three-input example $T(x,y,z)=(x+y,z)$ instead has rank two and nullity one. Its image of a linear map is all of ${\mathbb{R}}^2$, andEvery output has a solution straight line. In contrast, $U(x,y,z)=(x+y,0)$ has rank one and nullity two, with kernel basis $(1,-1,0),(0,0,1)$. Its reachable output $(a,0)$ has the solution affine subspacea plane, while $(a,b)$ with $b\ne 0$ has no solution. These two linear maps have the same input and output dimensions but different freedom and reachability. Computing the kernel separates what an output determines from what an extra constraint would have to fix. For either example, a basis of the kernel gives one independent free scalar per basis vector, which is why nullity measures freedom rather than the number of solutions.

For a real matrix $A$ and a fixed vector $n$, every vector in the image of a linear map defined by $A$ is orthogonal to $n$ exactly when $A^{T}n=0$. Indeed, $n^{T}Ar=(A^{T}n{)}^{T}r$. If this vanishes for every $r$, choose $r=A^{T}n$ to obtain $|A^{T}n{|}^2=0$. Equivalently each column of $A$ is orthogonal to $n$. Requiring the statement only for nonzero $r$ changes nothing, since the zero input already has zero dot product. This is the orthogonal complement relation between the image of a linear map and the kernel of a linear map defined by its transpose.

A matrix represents a linear projection when applying it twice has the same effect as applying it once. For an orthogonal projection with the standard Hermitian inner product on ${\mathbb{C}}^m$, the additional condition is that the conjugate transpose equals the original matrix:Thus $P$ is both idempotent and a Hermitian matrix. Over the real numbers this says $P^2=P=P^T$. Its image of a linear map is the retained linear subspace, while its kernel of a linear map is the orthogonal complement: vectors in the image are unchanged, and vectors in the kernel become zero.

For a matrix $A$ with full column rank, projecting onto its column space givesThe invertibility of $A^{†}A$ follows from linear independence of the columns. If those columns are an orthonormal set, $A^{†}A=I$ and $P=A{A}^{†}$. These formulas and the conditions $P^2=P=P^T$ in the real case are developed in MIT 18.06SC, Projections onto Subspaces, pp. 1–3.

For the line spanned by $e=(1,2{)}^T$,In particular, $Pe=e$ and $P(2,-1{)}^T=0$. The eigenvalues are $1$ on the retained direction and $0$ on the removed direction. This exposes the meaning of an orthogonal projection as retaining selected components, rather than stretching them.

Symmetry refers to the specified inner product. In the real weighted setting, the formula becomes $P_M=A(A^{T}MA{)}^{-1}{A}^{T}M$ and satisfies $P_M^2=P_M$ and $P_M^{T}M=M{P}_M$. For the preceding weighted line example,It is not symmetric under the ordinary transpose, but it is an orthogonal projection for the weighted Euclidean inner product. The matrix criterion $P=P^{†}$ should therefore be used with the standard Hermitian inner product, not silently transferred to a different geometry.

Suppose $E^2=E$. Every vector has the decompositionIf $w=Ez$ also belongs to $\ker E$, then $w=E^{2}z=Ew=0$. HenceOn its image of a linear map, $E$ acts as the identity; on its kernel of a linear map, it acts as zero. Concatenate a basis of the image with a basis of the kernel. The resulting matrix is $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}E$.

Conversely, this block matrix squares to itself, so any linear map having that matrix in a basis satisfies $E^2=E$. Thus projection operators are precisely the linear maps with this block form. This is an algebraic projection; no inner product or orthogonality is required.

The identity $T^2=T$ gives the decompositionThe first summand lies in the kernel of a linear map $\ker T$, and the second in the image of a linear map $\mathrm{im}T$. If $w$ belongs to their intersection, write $w=Tu$; then $Tw=T^{2}u=Tu=w$, while $Tw=0$. Thus $w=0$, proving the direct sumOn these two invariant subspaces, $T$ is respectively zero and the identity map. Joining a basis of each produces a basis of eigenvectors, with matrixHence every linear projection is diagonalizable, over the original field, including in characteristic two.

Because the linear functionals ${\ell }_0,{\ell }_1$ agree with endpoint point evaluations on continuous functions,The endpoint values also make $e_0,e_1$ linearly independent. By its definition, the image of a linear map $T$ lies in $E=⟨e_0,e_1⟩$. Conversely $T{e}_j=e_j$, so $T$ acts as the identity operator on $E$. ThusFor every $f$, $Tf\in E$, whence $T^{2}f=Tf$. It is a linear projection, and an annihilating polynomial is