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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.

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 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]

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. This dimension identity is ^[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.

Since $AB=0$, the image of a linear map represented by $B$ lies in the kernel of $A$. The matrix $B$ is nonzero, so this image of a linear map contains a nonzero vector; consequently $\dim \ker A\ge 1$. On the other hand, $A\ne 0$ gives $\mathrm{rank}A\ge 1$. The rank-nullity theorem in dimension two therefore givesThus the image of $A$ is a line through the origin, and $A$ maps onto that line.

Reversing the order of the matrix multiplication need not give zero. For example,gives $AB=0$ but $BA=B\ne 0$. This also shows why information about one composition of linear maps does not establish the corresponding property of the reversed composition.

Write $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. Its rank is one, so its determinant is zero. Choose its adjugate matrix:Direct matrix multiplication gives $AC=CA=(ad-bc)I=0$. Since the entries of $C$ are the entries of the nonzero matrix $A$, rearranged with possible sign changes, $C\ne 0$. This establishes both required annihilation properties.

For a linear map $\alpha$, its kernel of a linear map and image of a linear map areBy the definition of the image, $\alpha x=y$ has a solution exactly when $y\in I$. If $x_0$ is one solution, every other solution is $x_0+k$ for some $k\in K$: subtracting the two equations proves necessity, and linearity proves sufficiency. Thus a solvable equation has a unique solution exactly when $K=\{0\}$.

Write the columns of the given matrix as $c_1=(1,0,1{)}^T$, $c_2=(1,t,t{)}^T$ and $c_3=(t,-2b,0{)}^T$. Its determinant isThe columns $c_1,c_2$ are linearly independent for every real $t$: proportionality would require the proportionality factor to be one from the first coordinate, then simultaneously $t=0$ from the second and $t=1$ from the third. Thus the rank is always at least two.

If $D\ne 0$, the matrix is invertible andIf $D=0$, the rank is exactly two. A direct substitution verifies the nonzero kernel vector $k=(-t^2,t,t-1{)}^T$: its first and third row products vanish identically and its second is $D$. The rank-nullity theorem then givesIn that case the image is the plane spanned by $c_1,c_2$. Their cross product is $(-t,1-t,t{)}^T$, so an equivalent description isThese formulas include the singular case $t=b=0$, giving kernel $\mathbb{R}(0,0,1)$ and image $y_2=0$; no exceptional case was lost by dividing by $t$.

Finally, when $0<b<2$,Therefore the equation has a unique solution for every $t$ and every $y\in {\mathbb{R}}^3$, in particular for every $y\in I$. Outside this parameter range, the singular cases are precisely $t=b\pm \sqrt{b(b-2)}$ when that square root is real.

For any idempotent linear map $\tau$, each vector decomposes asIf $w\in \ker \tau \cap \mathrm{im}\tau$, write $w=\tau v$; then $0=\tau w={\tau }^{2}v=\tau v=w$. Thus the kernel of a linear map and image of a linear map give a direct sum.

For the self-adjoint map in the question, $k\in \ker \tau$ and $w=\tau v$ satisfy $⟨k,w⟩=⟨k,\tau v⟩=⟨\tau k,v⟩=0$. Hence the direct sum is an orthogonal decomposition:

For any linear system $Ax=c$, a solution exists exactly when $c$ belongs to the image of a linear map represented by $A$. If $x_0$ is one solution, all others are exactlysince $A(x-x_0)=0$, and conversely adding a kernel vector preserves the equation. Thus there are no solutions if $c\notin \mathrm{im}A$, one if $c\in \mathrm{im}A$ and $\ker A=\{0\}$, and infinitely many if $c\in \mathrm{im}A$ and the kernel is nontrivial: every real multiple of a nonzero kernel vector gives a distinct solution. Any two solutions differ by a kernel vector. Equivalently, consistency means the augmented matrix and $A$ have equal rank of a matrix; a consistent square linear system is unique exactly when its determinant is nonzero.

For the given matrix, set $d=b-a$ and $s=2a+b$. Perform the column operation $C_1\leftarrow C_1+C_2+C_3$, then the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$. Factoring the first column givesThese operations are determinant-preserving, and the displayed polynomial identity remains valid even when a factored expression vanishes.

To find all kernels and images of a linear map, putIf $J$ is the all-ones matrix, then $A=aJ+dP$. The direct sum ${\mathbb{R}}^3=\mathbb{R}u\oplus H$ is invariant under $A$: on $\mathbb{R}u$ it acts by multiplication by $s$, and on $H$ by $dP$. Since $P^3=I$, its restriction to $H$ is invertible.

For $d\ne 0$ and $s\ne 0$, the kernel is $\{0\}$ and the image of a linear map is all of ${\mathbb{R}}^3$. Decompose the right-hand side asApplying the inverse on each invariant summand gives the unique solutionFor $d\ne 0$ and $s=0$, the kernel is $\mathbb{R}u$ and the image of a linear map is $H$. A solution exists exactly when $c+2=0$. For $c=-2$, all solutions areFor $d=0$ and $a\ne 0$, so $a=b$, the kernel is $H$ and the image of a linear map is $\mathbb{R}u$. A solution exists exactly when $c=1$. All such solutions areFinally, for $a=b=0$, the kernel is ${\mathbb{R}}^3$ and the image of a linear map is $\{0\}$. The right-hand side has first entry one, so there is no solution for any $c$. These four cases exhaust all parameter values.

The ascending chain $\ker T\subseteq \ker T^2\subseteq \cdots$ must stabilize because $V$ is finite-dimensional. Once $\ker T^l=\ker T^{l+1}$, it stays stable: if $T^{l+2}v=0$, then $Tv\in \ker T^{l+1}=\ker T^l$, so $v\in \ker T^{l+1}=\ker T^l$; induction repeats the argument. Choose a positive such $l$, giving $\ker T^{2l}=\ker T^l$.

If $v\in \ker T^l\cap \mathrm{im}{T}^l$, write $v=T^{l}w$. Then $T^{2l}w=0$, so $w\in \ker T^l$ and $v=0$. By the rank-nullity theorem, the dimensions of those two subspaces sum to $\dim V$. HenceThis proves the finite-dimensional form of the Fitting lemma.

If $\det T=0$, the kernel contains a nonzero vector $u$, and $\mathrm{im}T$ is a proper vector subspace. Extend a basis of that image of a linear map to a basis of $V$ and choose a nonzero linear functional $\lambda$ which vanishes on the former basis. Define $S(v)=\lambda (v)u$. It is a nonzero endomorphism, since $\lambda$ takes a nonzero value somewhere, andThus a singular endomorphism has a nonzero two-sided annihilator, as required.

For an idempotent linear map $T$, the decomposition $v=(v-Tv)+Tv$ shows $V=\ker T\oplus \mathrm{im}T$: the first term is killed by $T$, the second is fixed, and a vector both fixed and killed is zero. An adapted basis makes its matrix $\mathrm{diag}(0,I_r)$, with $r=\mathrm{rank}T$. Consequently equal ranks give identical adapted matrices and therefore a change of basis relating $T_1,T_2$. Conversely, similar matrices have the same rank because pre- and postmultiplication by invertible matrices preserve rank. Hence $T_1$ and $T_2$ are similar exactly when their ranks agree, proving the similarity classification of idempotent linear maps.

The monomials $x^i{y}^j$ with $i,j\ge 0$ and $i+j\le n$ form a basis of the given vector space. There are $k+1$ of total degree $k$, soOn that basis the linear operator is diagonal:Both summands of the eigenvalue are nonnegative integers. Their sum is zero precisely when $i,j\in \{0,1\}$. Thus the kernel is the span of those members of $1,x,y,xy$ whose total degree is at most $n$. In particular it is $\mathrm{span}\{1,x,y\}$ for $n=1$ and $\mathrm{span}\{1,x,y,xy\}$ for $n\ge 2$; for $n=0$ it is the constants. The image of a linear map is the span of all the other permitted monomials, since each has a nonzero diagonal eigenvalue and hence is the image of its reciprocal multiple. No excluded monomial can occur in an image because its diagonal coefficient is zero.

The trace is the sum of these eigenvalues. At degrees zero and one all contributions vanish. At degree two the contributions are $2,0,2$, summing to four; at degree three they are $6,2,2,6$, summing to sixteen; at degree four they are $12,6,4,6,12$, summing to forty. ThereforeThis is the diagonal monomial action of a second-order Euler operator.

Put $p=1-R$, $b=\mu -r1$, $A=\sigma {\sigma }^T$ and $q=R+\epsilon$. First assume the covariance matrix $A$ is invertible and initial portfolio wealth is positive. Introduce the accumulated fee processThe Itô formula and the self-financing portfolio equation giveThis converts the path-dependent payment into a terminal-utility problem with state $y$. The volatility-penalized terminal fee changes its drift as well as its spot volatility.

Let $J(t,y)$ be the manager's value function. Its Hamilton-Jacobi-Bellman equation, with terminal condition $J(T,y)=y^p/p$, isThe homogeneity of CRRA utility suggests $J(t,y)=f(t)y^p/p$. Since $y{J}_y=f(t)y^p$ and $y^2{J}_{yy}=-Rf(t)y^p$, the portfolio-dependent expression isIts coefficient is positive even when $p<0$. Completing the square givesThus the stock fractions are constant. Substitution into the Hamilton-Jacobi-Bellman equation yieldsIn particular, multiplying the fee by $a$ does not affect the optimal investment portfolio.

To prove global optimality rather than merely solve a formal Hamilton-Jacobi-Bellman equation, write $v_t={\sigma }^T{\pi }_t$, $v^{\ast }={\sigma }^T{\pi }^{\ast }$ and let $y^{\ast }$ be the fee under the proposed constant policy. Then $b^T\pi =q{v}^{\ast }\cdot v$. Define a probability $Q$ byThis density is a true martingale because $v^{\ast }$ is constant. By the Girsanov theorem, $W_t^Q=W_t-p{v}^{\ast }t$ is a Brownian motion under $Q$. The geometric Brownian motion formula for the fee ratio givesFor admissible, locally square-integrable exposures with finite accumulated quadratic variation, setThis stochastic exponential is a positive local martingale, so $E_Q{Z}_T\le 1$. Moreover,For $p>0$, the exponent $p/(1+\epsilon )$ lies between zero and one. The Jensen inequality for this concave function gives $E_Q[Z_T^{p/(1+\epsilon )}]\le 1$. For $p<0$, the power is convex and decreasing, so the Jensen inequality gives $E_Q[Z_T^{p/(1+\epsilon )}]\ge (E_Q{Z}_T{)}^{p/(1+\epsilon )}\ge 1$; an infinite moment only makes this inequality stronger. Dividing by $p$, with its corresponding sign, proves $E[U(y_T)]\le E[U(y_T^{\ast })]$ in both cases. Equality is achieved by the constant policy.

For an ordinary terminal-wealth agent with relative risk aversion $q$, the same Hamilton-Jacobi-Bellman equation calculation, with no fee penalty, maximizes $b^T\pi -q{\pi }^{T}A\pi /2$. It therefore gives exactly the same optimal investment portfolio. For $q\ne 1$, the preceding stochastic exponential and Jensen inequality verification applies with penalty zero and exponent $1-q$, proving global optimality for this agent too. The spot volatility penalty adds $\epsilon$ to effective relative risk aversion. If $q=1$, the printed expression for $U_0$ has a zero denominator; its appropriate limit, after subtracting an irrelevant constant, is $U_0(x)=\log x$, which still gives ${\pi }^{\ast }=A^{-1}b$. In this logarithmic case, the terminal log-wealth difference from the candidate equals $\log Z_T$ for a positive stochastic exponential $Z$, so $E[\log Z_T]\le \log E[Z_T]\le 0$ proves its optimality.

If $A$ is singular, the same conclusion holds in the absence of arbitrage, provided $b$ lies in the image of a linear map $x\mapsto Ax$: replace $A^{-1}$ by the Moore-Penrose inverse, and add any vector in the kernel of $A$ to ${\pi }^{\ast }$. These additions change neither returns nor Brownian portfolio exposures. If $b$ has a nonzero component in that kernel, a zero-volatility position has a nonzero excess return. Scaling its profitable sign gives arbitrage, so a finite attained optimum as claimed requires this degeneracy to be excluded.