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The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every vector in the null space as $t(2,-9,-3,21{)}^{\top }$. Thus a basis for the kernel of the linear map isDirect multiplication gives $A(2,-9,-3,21{)}^{\top }=0$. The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.

Gaussian elimination gives the row operations $R_2\leftarrow R_2-3R_1$, $R_3\leftarrow R_3-R_1$, $R_4\leftarrow R_4-2R_1$, producing the lower three rows $(0,1,4,1)$, $(0,1,4,1)$, $(0,-3,2,-1)$. Subtract the new row two from row three and add three times row two to row four; the remaining nonzero row is $(0,0,14,2)$. Move it above the zero row and finish elimination to obtainThe pivot columns are one, two and three. Row operations preserve the linear dependence relations among columns, so the corresponding columns of the original matrix form a basis for the column space, which is the image of the linear map:In particular, the matrix rank is three. Columns of the reduced matrix itself would generally give a different subspace, so the original columns must be used.

Define $f:W_1\cap W_2\to U$ by $f(u)=(u,-u)$. It is well-defined by the definition of $U$, andThus $f$ is a linear map. If $f(u)=f(v)$, comparing first components gives $u=v$, proving injectivity. Every element of $U$ has the form $(u,-u)=f(u)$, proving surjectivity. Hence $f$ is a linear isomorphism, with inverse $(u,-u)\mapsto u$, and $\dim U=\dim (W_1\cap W_2)$.

Every output $u_1+u_2$ lies in the sum of subspaces $W_1+W_2$, since it is a linear combination of members of the two vector subspaces. For $u=(u_1,u_2)$, $v=(v_1,v_2)$ and scalars $a,b$,Therefore $g$ is a linear map. Every vector in $W_1+W_2$ is such an output: group the terms of any linear combination from $W_1\cup W_2$ according to their originating vector subspace. Thus the addition map of subspaces is also surjective.

Take $V={\mathbb{R}}^2$ and the linear map $h(x,y)=(x,0)$. ThenThese two vector subspaces have zero intersection and together span ${\mathbb{R}}^2$, so $V=\mathcal{R}(h)\oplus \mathcal{N}(h)$ as a direct sum of vector spaces.

Take $V={\mathbb{R}}^2$ and the linear map $h(x,y)=(y,0)$. ThenTheir intersection is nonzero and their sum is only one-dimensional. Hence $V\ne \mathcal{R}(h)\oplus \mathcal{N}(h)$. This nilpotent linear map has $h^2=0$, illustrating why taking sufficiently high powers changes the situation.

Apply $h$ to both sides of $\mathcal{R}(h^m)=\mathcal{R}(h^{m+1})$. Because $h(\mathcal{R}(h^j))=\mathcal{R}(h^{j+1})$, we obtain $\mathcal{R}(h^{m+1})=\mathcal{R}(h^{m+2})$. Repeating this argument proves equality at every later step. More formally, an induction on $k$ givesThe base case is $k=0$; the induction step follows from applying the linear map $h$ and then the assumed first equality. Thus one equality permanently stabilizes the descending sequence of images of a linear map.

Regard $A$ as a linear map ${\mathbb{R}}^4\to {\mathbb{R}}^4$. By the rank-nullity theorem and matrix rank $3$,All three are vector subspaces of ${\mathbb{R}}^4$, with the row space identified with four-component row vectors in the usual way. The null space lies in the input vector space of concentrations; the column space lies in the output vector space of concentration-change rates. Both have four coordinates, although they play different physical roles.

In this particular matrix, the row sums and column sums both vanish. Put $1=(1,1,1,1{)}^T$. Then $A1=0$ and $1^{T}A=0$, and the two three-dimensional vector subspaces coincide:This equality follows here from both sum identities and the dimensions of a vector space; it does not make $A$ a symmetric matrix and is not true for every square matrix.

The Moore--Penrose inverse is characterized by

Here $A^{†}=A^{+}$ denotes the Moore-Penrose inverse, while the star denotes the conjugate transpose. The first two identities express consistency when applying and reversing the linear map; the last two make the resulting orthogonal projection matrices Hermitian. They must hold together. For example, $A=(12)$ and $B=(10{)}^T$ satisfy $ABA=A$ and $BAB=B$, butis not Hermitian. This $B$ is a right inverse, yet $Bb=(3,0{)}^T$ is not the minimum-norm least-squares solution for $b=3$. The last two Penrose equations impose the orthogonality that selects the pseudoinverse.

For finite-dimensional vector spaces $V,W$ over a field $F$, a bilinear pairing $B:V\times W\to F$ is perfect when the induced linear maps $V\to W^{\ast }$ and $W\to V^{\ast }$ are isomorphisms. Equivalently, both dimensions agree and the matrix of the pairing in any two bases is invertible. The evaluation pairing between a finite-dimensional vector space and its dual space is perfect. This notion allows two different spaces; a nondegenerate bilinear form is the case $V=W$.

A planar rotation about the origin is the linear map represented by the rotation matrix $R_{\theta }=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. It preserves inner products, lengths, oriented angles and area, with positive angles measured anticlockwise. Its inverse is $R_{-\theta }$ and its determinant is one.

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

Evaluation of bounded-degree multivariate polynomials on a finite set defines a linear map $E_P:f\mapsto (f(p){)}_{p\in P}$. Its rank is at most $|P|$ and its kernel is the space of polynomials vanishing on $P$. The rank-nullity theorem gives $\dim \ker E_P\ge \left(\genfrac{}{}{0ex}{}{d+n}{n}\right)-|P|$ for degree at most $d$ in $n$ variables. Dependent point constraints only enlarge the kernel.

If a bounded linear map $P$ reproduces every element of an approximating polynomial space $V$, then $f-Pf=(f-v)-P(f-v)$ for every $v\in V$. The triangle inequality and infimum over $v$ prove the displayed error bound. The same argument works for any linear approximating space, and applies in particular to de la Vallée Poussin sums.

A linear map between ordered vector spaces is positive if it carries the positive cone of the domain into the positive cone of the codomain. For function spaces this means $f\ge 0⟹Tf\ge 0$. A Feller semigroup consists of such positive maps on continuous functions.

For a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$, factor its minimal polynomial as $m_T={\prod }_i{p}_i^{a_i}$ with distinct monic polynomials $p_i$ that are irreducible polynomials. ThenEach $V_i$ is an invariant subspace, since $T$ commutes with every polynomial in $T$. These are canonical vector subspaces, determined by $T$, without requiring any eigenvalues in $F$.

For a proof, put $m_i=p_i^{a_i}$ and $M_i=m_T/m_i$. The coprime polynomials $m_i,M_i$ admit a Bézout's identity for polynomials $b_i{M}_i+c_i{m}_i=1$. Set $E_i=b_i(T)M_i(T)$. Its image is contained in $V_i$ because $m_i(T)E_i=0$. The polynomial ${\sum }_i{b}_i{M}_i-1$ is divisible by every $m_i$, hence by their product, so ${\sum }_i{E}_i=I$. On $V_i$, $E_i=I$; on $V_j$ with $j\ne i$, $E_i=0$ because $m_j$ divides $M_i$. These facts prove the direct sum and identify $E_i$ as its projection operators. Finally, if the minimal polynomial of any restriction had a smaller exponent, $m_T/p_i$ would annihilate every summand and hence all of $V$, a contradiction.

For a bounded linear operator into a Banach space, compactness is equivalent to the displayed condition, where ${\pi }_Y$ is the quotient map by a finite-dimensional subspace of the target. A finite net for the image of the unit ball gives such a subspace by taking its span. Conversely, if the image lies within $\epsilon$ of a finite-dimensional subspace, choose approximants there; they remain bounded because the original image is bounded. A finite net for these approximants gives a slightly larger finite net for the image. This proves total boundedness and hence compactness. This condition does not require choosing the approximants by a linear map, and therefore does not assume an approximation property for the Banach space.

For composable linear maps, $\mathrm{im}(AB)=A(\mathrm{im}B)$ lies in $\mathrm{im}A$ and has dimension at most that of $\mathrm{im}B$. This proves the bound on the rank of a matrix product. When $A$ is $m\times p$, $B$ is $p\times n$, and $m,n\ge p$, the bound $p$ is attained by $A=\left(\genfrac{}{}{0ex}{}{I_p}{0}\right)$ and $B=(I_{p}0)$.

Every matrix of rank $r$ is equivalent to a block matrix with $I_r$ in its upper-left corner and zeros elsewhere. This follows by choosing bases adapted to the kernel and image of the represented linear map.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.