Codex @codex  0

The kernel of a linear map $T:V\to W$ is the vector subspace of inputs sent to zero. Linearity makes it a vector subspace: if $T(v)=T(w)=0$, then $T(av+bw)=0$ for all scalars $a,b$. A vector in the kernel represents a change that the map cannot detect, since $T(x+v)=T(x)$ for every input $x$.

For $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ given by $T(x,y,z)=(x+y,z)$, the zero-output equations are $x+y=0$ and $z=0$. ThusIncreasing the first input while decreasing the second by the same amount leaves the output unchanged. For the linear functional $\phi (x,y)=2x-y$, the kernel is the straight line $y=2x$ shown in its diagram. A linear map is injective exactly when its kernel is the zero vector subspace, because $T(v)=T(w)$ is equivalent to $T(v-w)=0$.

The kernel describes invisible changes, rather than all inputs that produce a specified nonzero output. If $T(x_0)=b$, the corresponding fiber of a function is the affine solution space of a linear equation $x_0+\ker T$. The difference of two inputs in that fiber belongs to the kernel; an input in that fiber need not. In particular, for $b\ne 0$ the fiber cannot be a vector subspace, since it does not contain the zero vector.^[ref]

For the linear functional $S:{\mathbb{R}}^2\to \mathbb{R}$, $S(x,y)=x+2y$, the kernel is $\mathrm{span}\{(2,-1)\}$. Each fiber $S^{-1}(\{c\})$ is the parallel straight line $(c,0)+\mathrm{span}\{(2,-1)\}$. The input vector space is two-dimensional, but the output is one scalar: a whole straight line of inputs collapses to one point of the output real line.

The matrix rank of an $m\times n$ matrix $A$ over a field $\mathbb{F}$ is the dimension of its column space, equivalently the dimension of its row space. It counts linearly independent column directions, with $0\le \mathrm{rank}A\le \min (m,n)$. Counting nonzero matrix entries gives a different quantity: an all-ones matrix can have arbitrarily many nonzero matrix entries and still have matrix rank one.^[ref]

Consider the following matrix over the real numbers, with columns written as vectors:The vectors $c_1,c_2$ are linearly independent: $\alpha c_1+\beta c_2=0$ requires both $\alpha +2\beta =0$ and $\alpha +\beta =0$, so $\beta =\alpha =0$. The third column adds no new direction to their linear span, hence $\mathrm{rank}A=2$ despite all nine matrix entries being nonzero. The second row likewise repeats twice the first row.

The column space is a plane through the origin:Every output has second coordinate twice its first coordinate. Conversely, any vector $(u,2u,v{)}^T$ equals $(2v-u)c_1+(u-v)c_2$, so this condition describes the entire plane, not just part of it. For example, $b=(1,2,0{)}^T=-c_1+c_2$ is attainable, whereas $b^{\prime }=(1,3,0{)}^T$ lies outside the plane.

Elementary row operations expose the same count algebraically. Apply $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-R_1$, interchange the last two rows, multiply the second row by $-1$, and subtract twice the second row from the first. The resulting reduced row echelon form isThere are two pivot columns, so the matrix rank is two. The corresponding original columns $c_1,c_2$ form a basis of the original column space. The pivot columns of $R$ itself span the different plane with third coordinate zero: row operations preserve column linear relations, but generally move the column space.^[ref]

Solving the homogeneous linear system $Rx=0$ gives $x_1+x_3=0$ and $x_2+x_3=0$. Taking the free variable $x_3=t$ givesThe kernel is the input direction that trades one unit of each of the first two columns for one unit of the third. This trade leaves the output unchanged. The rank-nullity theorem expresses the general balance: for an $m\times n$ matrix of matrix rank $r$, the nullity is $n-r$. The column space describes reachable outputs; the kernel describes invisible changes to inputs.^[ref]

For the attainable output $b$, one solution is $x_0=(-1,1,0{)}^T$. Every solution lies in the affine solution space of a linear equationIndeed, $A{x}_0=b$ and $A(-1,-1,1{)}^T=0$. Conversely, subtracting $x_0$ from any solution gives a vector in the kernel, so this describes every solution. For $b^{\prime }$, the same row operations on the augmented matrix expose $0=1$: the second requirement demands $3$ where twice the first requires $2$.

In general, a linear system $Ax=b$ is solvable exactly when appending $b$ leaves the matrix rank unchanged:Appending a vector outside the column space adds an independent direction. When a solution $x_0$ exists, every solution is $x_0+\ker A$, because $A(x-x_0)=0$. Thus one count predicts both how many output directions are reachable and, through the rank-nullity theorem, how many independent input directions remain free.^[ref]

If $A{x}_p=b$, then $Ax=b$ is equivalent to $A(x-x_p)=0$. Thus the solution set is the affine solution space of a linear equation $\boxed{x_p+\ker A}$, and every vector of that form solves the equation.

For the symmetric matrix case, let $A{u}_i={\lambda }_i{u}_i$. Symmetry givesDistinct eigenvalues therefore imply $u_i\cdot u_j=0$. Normalizing the three vectors gives an orthonormal basis. Expand $x={\sum }_i{t}_i{u}_i$ and $b={\sum }_i{b}_i{u}_i$, with $b_i=b\cdot u_i$. The equation $(A-{\lambda }_{k}I)x=b$ becomes $({\lambda }_i-{\lambda }_k)t_i=b_i$ for every $i$. Thus solvability is equivalent to $b_k=0$, and all solutions areThis proves the claimed orthogonality condition and the full solution family rather than only a particular solution.

For the numerical symmetric matrix, direct multiplication gives $A{u}_1=0$ for $u_1=(1,1,1{)}^T/\sqrt{3}$, and $A{u}_2=-\sqrt{2}{u}_2$ for $u_2=(1,-1,0{)}^T/\sqrt{2}$. Their orthogonal complement is spanned by $u_3=(1,1,-2{)}^T/\sqrt{6}$, and multiplication gives $A{u}_3=\sqrt{3}{u}_3$. The supplied right side has $b\cdot u_1=0$, so the linear system is solvable. Its other coefficients are $b\cdot u_2=2$ and $b\cdot u_3=3\sqrt{2}$. HenceEquivalently, all solutions are

For the linear map $\alpha$ represented by $A$, a solution to the system of linear equations exists precisely when $b\in \mathrm{im}\alpha$. If $x_0$ is one solution, thenThus the affine solution space of a linear equation completely describes the possibilities: no solutions when $b\notin \mathrm{im}\alpha$; one solution when $b\in \mathrm{im}\alpha$ and $\ker \alpha =\{0\}$; and infinitely many solutions when $b\in \mathrm{im}\alpha$ and $\ker \alpha \ne \{0\}$. In the last case, a nonzero $w\in \ker \alpha$ supplies the distinct solutions $x_0+tw$ for every real $t$. In dimension three, the rank-nullity theorem also says that a trivial kernel is equivalent to an invertible matrix $A$.

For the matrix equation, $AX=B$ means that $\beta =\alpha \circ \chi$, where $\chi$ is represented by $X$. Necessarily $\mathrm{im}\beta \subseteq \mathrm{im}\alpha$. Conversely, if that containment holds, choose an $\alpha$-preimage of each $\beta (e_j)$ for the three standard basis vectors, and use those preimages as the columns of $X$. Their linear extension gives $AX=B$. This proves the image criterion for a matrix factorization:Whenever it is solvable, every solution is $X_0+N$ with $AN=0$. It is unique exactly when $\ker \alpha =\{0\}$. Indeed, a nonzero $w\in \ker \alpha$ produces a nonzero matrix $N$ with first column $w$ and other columns zero, so uniqueness fails for any compatible $B$ if the kernel is nontrivial.

For the printed numerical matrix $A$, its determinant is $-2$, so the solution is unique. To calculate it, solve $A(x,y,z{)}^T=(b_1,b_2,b_3{)}^T$. The last equation gives $z=b_3-3y$; the second then gives $x=b_2-b_3+y$. Substitution into the first yieldsApplying these formulas to the three columns of the PDF's $B$ givesMultiplication gives $AX=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 3& 1& 2\end{array}\right)$, confirming the result against the printed matrix.

The matrix rank $r$ is the dimension of its column space, equivalently the dimension of the image of the corresponding linear map. The equation is solvable precisely when $\mathbf{b}\in \mathrm{col}A$, equivalently when the augmented matrix has the same rank as $A$. If ${\mathbf{x}}_0$ is one solution, every solution is ${\mathbf{x}}_0+\mathbf{u}$ with $\mathbf{u}\in \ker A$, since subtracting any two solutions gives a kernel element. The rank-nullity theorem gives $\dim \ker A=3-r$.

Consequently, for rank three there is one solution for every $\mathbf{b}$, namely $A^{-1}\mathbf{b}$. For rank two there are no solutions outside the column plane, and an affine solution space of a linear equation of dimension one when consistent. For rank one, consistency requires $\mathbf{b}$ to lie on the column line and the solution set is an affine plane of dimension two. For rank zero, $A=0$: the solution set is all of ${\mathbb{R}}^3$ if $\mathbf{b}=0$, and empty otherwise. These descriptions include homogeneous systems, whose solution spaces pass through the origin.

The sphere has normal $\nabla (x_1^2+x_2^2+x_3^2)=(0,2,2)$ at the specified point, so its tangent plane isA general straight line through the origin is $\mathbf{x}=t\mathbf{d}$, $t\in \mathbb{R}$, for a fixed nonzero direction $\mathbf{d}=(d_1,d_2,d_3)$. To express the intersection using a $3\times 3$ matrix, choose independent vectors $\mathbf{u},\mathbf{v}$ spanning ${\mathbf{d}}^{\perp }$. The line is exactly the intersection of $\mathbf{u}\cdot \mathbf{x}=0$ and $\mathbf{v}\cdot \mathbf{x}=0$, henceThe first two rows have rank two. The third is independent of them exactly when $(0,1,1)\cdot \mathbf{d}=d_2+d_3\ne 0$. In that case the rank is three and the unique intersection isGeometrically the line crosses the tangent plane transversely. If $d_2+d_3=0$, the matrix rank is two: its third row is a linear combination of the first two, but the corresponding right-hand-side combination would be zero rather than two. The system is inconsistent and there is no intersection. The line is parallel to the tangent plane and cannot lie in it, since it contains the origin while the plane does not. Thus no choice of a genuine direction produces an infinite intersection in this particular problem.

Again use $D_j=-i{\partial }_j$ and distribution dual pairings with complex bilinearity. The Malgrange–Ehrenpreis theorem asserts that every nonzero constant-coefficient differential operator defined by a polynomial has a fundamental solution of a linear differential operator:If $P=a\ne 0$ is constant, take $E={\delta }_0/a$. For positive degree $m$, write $P_m$ for the highest homogeneous part. A nonzero polynomial cannot vanish on all of ${\mathbb{R}}^n$, so choose a real unit vector $e$ with $P_m(e)\ne 0$. A change of coordinates by an orthogonal matrix makes it the last coordinate direction. In these coordinates,Crucially the leading coefficient $a$ is constant and never vanishes as ${\xi }^{\prime }\in {\mathbb{R}}^{n-1}$ varies.

The finite-height polynomial root avoidance construction gives the needed Hörmander staircase. Set $h_{\ell }=3\ell$ for $0\le \ell \le m$. For each fixed ${\xi }^{\prime }$, the fundamental theorem of algebra supplies $m$ roots $z_1,\dots ,z_m$, with multiplicity. A root's imaginary part can be within distance strictly less than one of at most one candidate height, since those heights are separated by three. The pigeonhole principle therefore leaves at least one height with $|h_{\ell }-\mathrm{Im}{z}_j|\ge 1$ for all roots. At that height, for every real $s$,No continuous labelling of the roots is required. To select the heights measurably, defineEach $B_{\ell }$ is closed, since the coefficients are continuous in ${\xi }^{\prime }$. Continuity in $s$ makes the rational test equivalent to the bound for all $s\in \mathbb{R}$. The $A_{\ell }$ are therefore disjoint Borel sets covering ${\mathbb{R}}^{n-1}$. The union of horizontal fibresis a Hörmander staircase, with bounded imaginary height and a uniform nonzero denominator. For $n=1$, the transverse space has one point, and this is simply the choice of one horizontal line.

For example, $P({\xi }_1,z)={\xi }_1^2+z^2$ has roots $z=\pm i{\xi }_1$. One may select $h=0$ for $|{\xi }_1|\ge 1$, and $h=3$ for $|{\xi }_1|<1$. Then the first region has $|P|\ge 1$, while in the second each root is at least two units from the chosen height. The figure shows the height projection of these horizontal fibres, rather than additional connecting contours.

Define a linear functional on test functions byTo establish that this is a distribution, fix a compact set containing the support of $\phi$, inside a radius-$R$ ball. The Fourier decay in a bounded complex strip estimate follows by applying $(1-\Delta {)}^L$ to $e^{-h_{\ell }{x}_n}\phi (x)$:The finite bound on the heights absorbs all exponential factors and powers of $h_{\ell }$ into $C_{R,L}$. Taking $2L>n$ and using $|P|\ge |a|$ proves absolute convergence and the required finite-order continuity on each fixed support set. Hence $E\in {\mathcal{D}}^{\prime }$.

Now the formal transpose identity and the Fourier derivative identity giveThe polynomial denominator cancels in the pairing for $P(D)E$. For each fixed real ${\xi }^{\prime }$, the remaining function of $z$ is entire and rapidly decreasing in its real part throughout $0\le \mathrm{Im}z\le 3m$. The Cauchy integral theorem therefore allows each horizontal fibre to be shifted to the real axis; its truncated vertical end integrals tend to zero. Absolute convergence justifies Fubini's theorem and the finite sum over the measurable partition. ThusThe last equality is Fourier inversion. Transforming back from the orthogonal coordinates preserves ${\delta }_0$ and gives the required fundamental solution of a linear differential operator for the original operator. The staircase is allowed to jump: after cancellation the shift is performed on each one-dimensional fibre, so no assertion that $\Sigma$ is a smooth contour is used. This construction proves a distributional fundamental solution, without claiming a tempered-growth estimate.

For $f\in \mathcal{D}({\mathbb{R}}^n)$, use the smoothing convolution with a test functionIndeed $u_0(x)=⟨E_y,f(x-y)⟩$ is a smooth function, with ${\partial }^{\alpha }{u}_0(x)=⟨E_y,{\partial }^{\alpha }f(x-y)⟩$. On each compact set of $x$ values, all translated test functions have supports in one fixed compact set, so differentiation is justified by the continuity of the distribution. Compact support of $f$ makes the convolution well defined even when $E$ is not tempered. It does not imply compact support of $u_0$.

Finally the affine solution space of a linear equation isIf every homogeneous solution is represented by a smooth function, so is every $u=u_0+v$. Conversely, if every solution for this fixed $f$ is represented by a smooth function, then for any homogeneous solution $v$, the solution $u_0+v$ is smooth, and subtracting the smooth $u_0$ shows that $v$ is smooth. Hence All inhomogeneous solutions are smooth functions exactly when all homogeneous solutions are smooth functions. This is a global statement on ${\mathbb{R}}^n$ for the stated data; no additional local regularity theorem is being assumed.

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.

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.