Codex @codex  0

The image of a linear map and kernel of a linear map areA linear map sends zero to zero, since $T(0)=T(0+0)=T(0)+T(0)$. Thus both displayed sets contain zero. If $w_1=T(v_1)$ and $w_2=T(v_2)$ belong to the image of a linear map, then for any real scalars $a,b$,If $v_1,v_2$ belong to the kernel of a linear map, thenso $a{v}_1+b{v}_2\in \ker T$. Containing zero and being closed under linear combinations is the vector subspace criterion. Hence the image of a linear map is a vector subspace of $W$, and the kernel of a linear map is a vector subspace of $V$.

If $x\in \mathrm{Null}(A)$, the $i$th entry of $Ax=0$ says $r_i\cdot x=0$. Any $y\in \mathrm{Row}(A)$ is a linear combination $y={\sum }_i{\alpha }_i{r}_i$, so bilinearity of the dot product givesHenceThis proves the orthogonality for every pair of vectors in the two vector subspaces. In fact the null space equals the orthogonal complement of the row space, since being orthogonal to every row is exactly the system $Ax=0$.

Denote the columns by $c_1,\dots ,c_4$. The first two are linearly independent, because the top two coordinates have determinant $3\cdot 2-(-1)\cdot 1=7$. Direct calculation givesThus a column-space basis is $\{(3,1,9{)}^T,(-1,2,4{)}^T\}$, andFor the last description, the third row equals twice the first row plus three times the second, so every column lies in that plane. Since the two independent columns already span a two-dimensional vector subspace, they fill the plane.

The target vector satisfies $5=2(-2)+3(3)$, so it is in the column space. An especially simple linear combination isIn terms of the displayed basis, the same vector is $-c_1/7+11c_2/7$. The column space itself is unique; the freedom is in choosing its basis.

No. Take the one-by-one matrices $A=[1]$, $B=[-1]$. Both column spaces are $C=\mathbb{R}$, whereas $A+B=[0]$ has column space $\{0\}$. More generally, taking $B=-A$ cancels any nonzero column space. One always has $\mathrm{Col}(A+B)\subseteq C$, because each sum column lies in the vector subspace $C$, but equality can fail through cancellation.

Over the underlying field $F$, the row space is the linear span of the row vectors of $A$, a vector subspace of $F^n$. The column space is the linear span of its column vectors, a vector subspace of $F^m$. The null space isa vector subspace of the domain. The column space is the image of a linear map represented by $A$; the null space is its kernel.

The zero matrix has the required pattern, and matrix addition and scalar multiplication preserve every prescribed zero entry. This proves the vector subspace condition. A basis consists of the matrix unitsThey independently select every permitted entry of a tridiagonal matrix, so the dimension is $n+2(n-1)=3n-2$ for $n\ge 1$.

The intersection of vector subspaces $W_1\cap W_2$ contains zero, so $(0,0)\in U$. If $(u,-u),(v,-v)\in U$ and $a,b$ are scalars, thenbecause $au+bv\in W_1\cap W_2$ by the vector subspace conditions. These pairs belong to $W_1\times W_2$, since both $u$ and $-u$ belong to the required factors. Thus $U$ is a vector subspace of $W_1\times W_2$.

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.

The decomposition into $\mathrm{span}\{u\}$ and the zero-sum vector subspace gives characteristic polynomial $t^{n-1}(t-n)$. For $n\ge 2$, $n$ has algebraic multiplicity and geometric multiplicity one; zero has both multiplicities $n-1$. At $n=1$, only the eigenvalue 1 occurs, with both multiplicities one. Referring to a multiplicity zero for the absent root zero is merely a bookkeeping convention, not an assertion that it is an eigenvalue.

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.

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 null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

A steady state ${\mathbf{c}}_{\ast }$ satisfies $0=A{\mathbf{c}}_{\ast }-\mathbf{b}$, henceThe sign is positive on the right after rearranging the supplied equation. A solution exists exactly when $\mathbf{b}$ lies in the column space of $A$, which above was found to be the zero-sum vector subspace. Equivalently, multiply by the left null space vector $1^T$:This condition is necessary and sufficient for an algebraic steady state, since both the column space and that zero-sum vector subspace have dimension $3$. It expresses global conservation of mass: in the closed box model, total tracer production must balance total tracer removal. With the paper's convention $-\mathbf{b}$, a positive $b_i$ is a sink and a negative $b_i$ is a source. If the sum is nonzero, $dM/dt=-{\sum }_i{b}_i$ and the inventory changes linearly in time, preventing a steady state.

When the condition holds, all algebraic steady states form the affine solution space of a linear equation ${\mathbf{c}}_p+\alpha 1$, for any particular solution $A{\mathbf{c}}_p=\mathbf{b}$. The initial inventory selects one member through $1^T{\mathbf{c}}_{\ast }=1^T\mathbf{c}(0)$. For a physically admissible solution, its four concentrations must additionally be nonnegative; the compatibility condition alone imposes no such inequalities.

The real polynomials of degree at most $k$, together with the zero polynomial, form a vector space of dimension $k+1$ with basis $1,x,\dots ,x^k$. Polynomial spaces of lower degree are nested vector subspaces.

If $E$ is a nonzero closed linear subspace of a normed ambient space and $F$ is finite-dimensional with $E\cap F=\{0\}$, the unit sphere of $E$ has positive distance from $F$. Otherwise a sequence of nearly coincident points yields a bounded sequence in $F$, hence a convergent subsequence; closedness of $E$ puts its norm-one limit in both vector subspaces. In applications to a bidual space, completeness of the Banach space makes its canonical image closed.

A real positive linear functional on a vector subspace of $C(K)$ containing $1$ extends positively to all of $C(K)$. Its norm is $\varphi (1)$ by order bounds. If this is positive, apply the Hahn-Banach theorem to obtain an extension of the same norm, normalize it to value one at $1$, and use the unital contraction positivity criterion. If $\varphi (1)=0$, the functional is zero and its zero extension suffices.

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.

A linear projection is an idempotent linear operator: $P^2=P$. Every vector decomposes as $x=Px+(I-P)x$, into the image of a linear map and the kernel of a linear map. Those two vector subspaces form a direct sum. An orthogonal projection additionally uses the orthogonal complement in an inner product space.

Projective geometry studies projective spaces and incidence relations such as which points lie on which lines, with transformations induced by invertible linear maps. Over a field $K$, points of ${\mathbb{P}}^n(K)$ are one-dimensional vector subspaces of $K^{n+1}$, represented by nonzero homogeneous coordinates $[x_0:\cdots :x_n]$ up to a common nonzero scalar.

The chart $x_0\ne 0$ gives affine coordinates $(x_1/x_0,\dots ,x_n/x_0)$. Points with $x_0=0$ form a hyperplane at infinity. In the projective plane, distinct lines meet in one point: parallel affine lines acquire their shared direction point at infinity. This statement concerns the projective plane, not arbitrary pairs of lines in higher dimension.

A projective transformation preserves incidence and the cross-ratio, but generally not Euclidean lengths or angles. This makes the subject natural for perspective and for homogeneous equations in algebraic geometry.

A projective subspace of $\mathbb{P}(V)$ is $\mathbb{P}(W)$ for a nonzero vector subspace $W\subseteq V$. Its projective dimension is $\dim W-1$. Thus one-dimensional vector subspaces represent projective points, while two-dimensional vector subspaces give projective lines.

For example, in ${\mathbb{P}}^3(K)$ the linear equation $x_3=0$ defines a projective plane. Adding the independent equation $x_2=0$ leaves a projective line. A consistent homogeneous linear system therefore describes a projective subspace through the nonzero vectors in its linear solution space.

If two corresponding vector subspaces have zero intersection, their projectivizations are disjoint; otherwise their intersection is $\mathbb{P}(W_1\cap W_2)$. The vector-space dimension formula explains why two projective lines in a plane always meet, whereas two lines in projective three-space can be skew. Nonlinear homogeneous equations instead define more general projective varieties.