Codex @codex  0

For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.

Every $g_i\in U^0$ vanishes on $U$, so $U\subseteq {\bigcap }_i\ker g_i$. Conversely, suppose $v\in {\bigcap }_i\ker g_i$. Every $f\in U^0$ is a linear combination ${\sum }_i{a}_i{g}_i$, giving $f(v)=0$. If $v\notin U$, part (b) supplies a functional in $U^0$ with $f(v)\ne 0$, a contradiction. ThereforeFor the dimension statement, take a basis $u_1,\dots ,u_m$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. Relative to the corresponding dual basis, a functional belongs to $U^0$ precisely when its first $m$ coefficients are zero. ThusIf $U=V$, this says $k=0$ and the empty intersection is understood to be $V$.

In the coordinate example choose the linear functionals $g_1(x,y,z)=y$ and $g_2(x,y,z)=z$. Their kernels are the two coordinate planesTheir intersection is the stated axis, and $k=2=3-1$.

If $c(\lambda )=\det (\lambda I-T)=0$, then $\lambda I-T$ has nonzero kernel of a linear map. A nonzero vector $v$ in that kernel of a linear map satisfies $Tv=\lambda v$, proving that $\lambda$ is an eigenvalue. Extend $v$ to a basis. Its first column is $(\lambda ,0,\dots ,0{)}^T$, so the matrix has block formHere $B$ represents the induced operator on $V/⟨v⟩$.

If the algebraic multiplicity is $r$, repeat this argument in that quotient: its characteristic polynomial still has $\lambda$ with multiplicity $r-1$. Induction and lifting a quotient basis gives a basis whose first $r$ columns have the prescribed upper-triangular form. With $U$ the span of those $r$ vectors, the full matrix isThus $U$ is invariant and is contained in $W_{\lambda }$. Conversely, if $(T-\lambda I{)}^{s}v=0$, project to $V/U$. The induced map $C-\lambda I$ is invertible, so its $s$th power cannot kill a nonzero class. Hence $v\in U$. Consequently $\boxed{W_{\lambda }=U,\dim W_{\lambda }=r}$; this proves simultaneously that it is a subspace and $T$-invariant. It also proves $(T-\lambda I{)}^r=0$ on this space, without using Cayley-Hamilton theorem in advance.

To prove independence of the different spaces, take a relation ${\sum }_{\lambda }{v}_{\lambda }=0$ with $v_{\lambda }\in W_{\lambda }$. Fix $\lambda$ and apply $P_{\lambda }(T)={\prod }_{\mu \ne \lambda }(T-\mu I{)}^{r_{\mu }}$. It kills all terms except $v_{\lambda }$. On $W_{\lambda }$, each factor $T-\mu I=(\lambda -\mu )I+N$ is invertible: its inverse is a finite geometric series in the nilpotent $N$. Hence $P_{\lambda }(T)v_{\lambda }=0$ implies $v_{\lambda }=0$. The sum is direct, and its dimension is $\sum r_{\lambda }=\dim V$, soFinally $c(T)={\prod }_{\lambda }(T-\lambda I{)}^{r_{\lambda }}$ kills each summand, hence every vector. This proves $\boxed{c(T)=0}$ without circularly assuming the theorem during the decomposition.

Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne 0$, then $0=R^{a}v={\lambda }^{a}v$, hence $\lambda =0$. The characteristic polynomial splits over $\mathbb{C}$ and has degree $n$, soThere is indeed an eigenvector for the eigenvalue zero: a nonzero vector in the last nonzero image of $R$ lies in its kernel of a linear map. For the bound, choose $v$ with $R^{a-1}v\ne 0$, using minimality. The vectors $v,Rv,\dots ,R^{a-1}v$ are linearly independent. Otherwise choose the least $j$ with nonzero coefficient in a relation and apply $R^{a-1-j}$; only its nonzero multiple of $R^{a-1}v$ survives, a contradiction. Therefore $\boxed{a\le n}$.

Suppose $S^b=0$. Use the increasing chain of vector subspacesChoose a basis successively, extending a basis of each kernel of a linear map to one of the next kernel of a linear map. If a basis vector first enters at level $j$, its image under $S$ belongs to $\ker S^{j-1}$, spanned entirely by earlier basis vectors. Thus every matrix column has nonzero entries only above its diagonal: the matrix is strictly upper triangular.

Conversely, for such a matrix and its ordered basis $e_1,\dots ,e_n$, one has $S⟨e_1,\dots ,e_j⟩\subseteq ⟨e_1,\dots ,e_{j-1}⟩$. Iterating $n$ times kills every vector. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.

The polynomials $t^c$ and $(t-1{)}^d$ are coprime. Bézout's identity gives polynomials $u,v$ with $u(t)t^c+v(t)(t-1{)}^d=1$. SetFor every $x$, the decompositionhas first summand in $V_0$ and second in $V_1$, since $T^c(T-I{)}^d=0$ and polynomial expressions in $T$ commute. If $x$ belongs to both kernels, the same Bézout identity gives $x=0$. Thus $V=V_0\oplus V_1$. Both vector subspaces are invariant under $T$.

On $V_0$, the restriction of $T$ is nilpotent; on $V_1$, the restriction of $T-I$ is nilpotent. Apply (b) separately, then concatenate the two bases. This yieldswhere $A,N$ are strictly upper triangular. Empty blocks are allowed if one subspace is zero. In particular $C=I+N$ has precisely the required diagonal.

For a normal operator, (a) gives $\Vert Tx\Vert =\Vert T^{\ast }x\Vert$, so one side is zero exactly when the other is. Hence $\boxed{\ker T=\ker T^{\ast }}$.

For real $\lambda$, the adjoint of $T-\lambda I$ is $T^{\ast }-\lambda I$. Expanding both products shows that their difference is $T^{\ast }T-T{T}^{\ast }=0$. Thus $T-\lambda I$ is normal, and applying the kernel of a linear map conclusion to it givesThis proves equality of each real eigenspace, hence the same eigenvalues and eigenvectors on the stated real vector space. If neither has real eigenvalues, the conclusion is vacuous. Upon complexification one instead pairs $\lambda$ with $\overline{\lambda }$ for the adjoint; no claim of the same complex eigenvalues is needed here.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.

Every $g_i\in U^0$ vanishes on $U$, so $U\subseteq {\bigcap }_i\ker g_i$. Conversely, suppose $v\in {\bigcap }_i\ker g_i$. Every $f\in U^0$ is a linear combination ${\sum }_i{a}_i{g}_i$, giving $f(v)=0$. If $v\notin U$, part (b) supplies a functional in $U^0$ with $f(v)\ne 0$, a contradiction. ThereforeFor the dimension statement, take a basis $u_1,\dots ,u_m$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. Relative to the corresponding dual basis, a functional belongs to $U^0$ precisely when its first $m$ coefficients are zero. ThusIf $U=V$, this says $k=0$ and the empty intersection is understood to be $V$.

In the coordinate example choose the linear functionals $g_1(x,y,z)=y$ and $g_2(x,y,z)=z$. Their kernels are the two coordinate planesTheir intersection is the stated axis, and $k=2=3-1$.

The homogeneous solution is $A\cos x+B\sin x$. The left boundary condition removes $A$, and the right boundary condition gives $B\sin l=0$. Thus the kernel of a linear map associated with this boundary value problem isThe Fredholm alternative now gives the complete classification:For necessity at $l=m\pi$, multiply by the homogeneous solution $\sin x$ and use the Lagrange identity for a Sturm-Liouville operator to obtain ${\int }_0^{l}f\sin xdx=0$. For sufficiency, the particular solutionsatisfies $z(0)=0$ and $z^{\prime \prime }+z=f$. At the other endpoint,which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed Lipschitz continuity of $f$ in particular makes $f$ continuous, so the displayed integral defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this boundary value problem.

Since ${\lambda }_1=-4/5$, the kernel of a linear map $L+\frac{4}{5}I$ is the span of $\psi (x)=\cos (3x/2)$. The self-adjoint Fredholm alternative requires $⟨f,\psi ⟩=0$ for solvability. If it holds, an arbitrary multiple of $\psi$ can be added, so a solution is not unique.

For $f=\cos x$,so case (A) has no solution. For $f=\sin x\sin (3x/2)$,by sine-function orthogonality. Thus case (B) has solutions, and they are nonunique.

If $\varphi =T^{\prime }\psi$ and $v\in \ker T$, then $\varphi (v)=\psi (Tv)=0$. Thus $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^{\circ }$, using the annihilator of a vector subspace.

Conversely, let $\varphi$ vanish on the kernel of a linear map $\ker T$. Define a linear functional on $\mathrm{im}T$ by ${\psi }_0(Tv)=\varphi (v)$. This is well-defined: $Tv=Tw$ implies $v-w\in \ker T$, so $\varphi (v)=\varphi (w)$. Extend ${\psi }_0$ to a linear functional $\psi$ on $W$, using the extension argument from part (i). Then $T^{\prime }\psi =\varphi$, provingThe dimension formula from part (i) and the rank-nullity theorem now giveTo deduce the eigenvalue statement, apply this equality of rank of a linear map to $T-\lambda I$. Its dual map is $T^{\prime }-\lambda I$, so their kernel of a linear map dimensions agree. Thus the same $\lambda$ are eigenvalues, with the same geometric multiplicities.

For algebraic multiplicities as well, part (a)(iii) givesso equality of rank applies to every power. HenceAfter extending to a field where the characteristic polynomials split, take $r\ge \dim V$. These kernel of a linear map spaces are the generalized eigenspaces; the primary decomposition theorem for linear maps identifies their dimensions with the algebraic multiplicities. The same proof works over the original field for every eigenvalue in it. We obtainIn fact the argument also includes the zero eigenvalue.

The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.

Induction from $M^2=M$ gives $M^j=M$ for every integer $j\ge 1$. Separate the zeroth term of the matrix exponential series:Therefore the exponential of an idempotent linear map isThe last form shows its action directly: it multiplies the image of a linear map of $M$ by $e^{\alpha }$ and fixes its kernel of a linear map. No Hermiticity assumption on the idempotent matrix is needed.

Differentiation satisfies $D(af+bg)=aD(f)+bD(g)$ for constant scalars $a,b$, so it is a linear map on suitable vector spaces of differentiable functions. An arbitrary set $S$ need not even be a vector space, and an endomorphism interpretation also requires $D(S)\subseteq S$.

For a noninvertible example, take $S=\mathrm{span}\{1,x\}$. Then $D(a+bx)=b$ and the kernel of a linear map is the nonzero linear span of $1$, so $D:S\to S$ is not injective. The matrix in this basis is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

For an invertible example, take $S=\mathrm{span}\{e^x,e^{2x}\}$. In this basis, $D$ has matrix $\mathrm{diag}(1,2)$, andThus differentiation is not always invertible; the domain and codomain determine the answer.

To find the kernel of a linear map, perform the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-(1+\alpha )R_1$, followed by $R_3\leftarrow R_3-R_2$. The equivalent homogeneous equations areThe resulting determinant is ${\alpha }^2(3-\alpha )$. If $\alpha \notin \{0,3\}$ all coordinates vanish. At $\alpha =0$ only $x-y+z=0$ remains; at $\alpha =3$ the equations give $y=2z$, $x=-5z$. Therefore kernel bases areThe empty family is the basis of the zero vector subspace.

For the change of basis, let $B_0,C_0$ be the matrices whose columns are the two ordered bases in standard coordinates. Then $v=B_0[v{]}_B=C_0[v{]}_C$, so $[v{]}_B=S[v{]}_C$ with $S=B_0^{-1}{C}_0$. If $A$ and $A^{\prime }$ represent the same linear map in the two bases, $[\Phi (v){]}_B=A[v{]}_B$ and also $[\Phi (v){]}_B=S{A}^{\prime }[v{]}_C$. Therefore $S{A}^{\prime }=AS$, andFor the concrete bases, their vectors satisfy $c_1=b_1+b_2$, $c_2=b_2+b_3$, $c_3=b_1+b_2+b_3$. These coordinate columns giveAt $\alpha =0$, every row of $A_0$ is $(1,-1,1)$, so every row of $A_{0}S$ is $(0,0,1)$. Multiplying on the left by the displayed inverse givesThe original PDF asks for this full product; its final factor $S$ is missing from the converted TeX.

The linearization of this reduction of the FitzHugh-Nagumo model at its trivial equilibrium point isA double zero eigenvalue requires both the constant and linear coefficients to vanish. Hence the codimension-two point isThere the polynomial is $m^2(m+3/2)$. The kernel of a linear map of $L$ is one-dimensional, so its zero eigenvalue has a nontrivial size-two block in Jordan normal form. This is a reflection-symmetric double-zero point of Bogdanov–Takens bifurcation type, with one stable direction; the nonlinear coefficients are computed below rather than assumed.

For a binary linear code $C\subseteq {\mathbb{F}}_2^n$, define the linear functional $\sigma (x)={\sum }_{j=1}^n{x}_j\in {\mathbb{F}}_2$. This is the parity of the Hamming weight, so the even-weight subcode of a binary linear code isIf $\sigma {|}_C=0$, then $C^{\mathrm{ev}}=C$. Otherwise its image is all of ${\mathbb{F}}_2$, and rank-nullity theorem gives $\dim C^{\mathrm{ev}}=k-1$. The kernel of a linear map is a linear subspace, proving the required dichotomy including closure under addition.

Let $G$ be a full-rank $k\times n$ generator matrix. Messages $a\in {\mathbb{F}}_2^k$ correspond bijectively to codewords $aG$. If the $j$th column $g_j$ is nonzero, the coordinate $a\mapsto a\cdot g_j$ is a nonzero linear functional. Choose $a_0$ with $a_0\cdot g_j=1$. Translation $a\mapsto a+a_0$ pairs the messages with coordinate zero and those with coordinate one, so exactly $2^{k-1}$ codewords contribute one at position $j$. Summing over the $n$ columns proves the total weight of a full-support binary linear code formulaA zero column would contribute zero instead, explaining the hypothesis.

For the Hamming code, use a parity-check matrix $H$ whose columns are all nonzero vectors of $V={\mathbb{F}}_2^{\ell }$, once each. Then $C_{H,\ell }=\ker H$ and $n=2^{\ell }-1$. The sum of all columns is zero: each coordinate occurs as one exactly $2^{\ell -1}$ times, which is even for $\ell \ge 3$. ConsequentlyUnder this column labelling, the support of a vector of a codeword is identified with a set of distinct nonzero columns whose sum is zero. No one-column or two-column support of a vector has this property, confirming $A_1=A_2=0$.

We can count the low-weight coefficients of a binary Hamming code directly. For weight three, choose ordered distinct nonzero columns $a,b$ in $n(n-1)$ ways. The third must be $a+b$, which is nonzero and different from both. Every unordered support of a vector has $3!$ orderings, so $A_3=n(n-1)/3!$.

For weight four, the first two columns $a,b$ are again distinct and nonzero. The third $c$ must avoid $a,b,a+b$, giving $n-3$ possibilities. The fourth is $d=a+b+c$; it is nonzero because $c\ne a+b$, and cannot equal any of the first three because those are distinct. Conversely every ordered zero-sum four-tuple arises in this way. Dividing by $4!$ gives $A_4=n(n-1)(n-3)/4!$.

For weight five, the first three columns $a,b,c$ must be linearly independent. If $a+b+c=0$, the sum condition would force the fourth and fifth to coincide. Thus there are $n(n-1)(n-3)$ choices for the first three. Put $t=a+b+c$. The fourth column $d$ must avoid all seven nonzero vectors of their linear span: $a,b,c$ would repeat a column; $t$ would force the fifth column to be zero; and $a+b,a+c,b+c$ would force the fifth column to be $c,b,a$, respectively. Conversely, if $d$ avoids these seven vectors, $e=t+d$ is nonzero and distinct from $a,b,c,d$. This gives $n-7$ choices for $d$ and determines $e$. Dividing by $5!$ yieldsAt $\ell =3$, the last count is zero, as the argument also shows.

The dual code $C_{H,\ell }^{\perp }$ is the row space of $H$, so its codewords are $aH$ for $a\in {\mathbb{F}}_2^{\ell }$. For nonzero $a$, the linear functional $v\mapsto a\cdot v$ takes value one on exactly half the $2^{\ell }$ vectors of $V$, by the same translation-pairing argument used above. Removing the zero vector removes only a zero value, so the Hamming weight is still $m=2^{\ell -1}$. Since $H$ has matrix rank $\ell$, distinct $a$ give distinct codewords. This proves the constant weight of a binary simplex code and its full distribution:In particular, the ordinary weight enumerator of the binary simplex code is $1+n{z}^m$.

Apply the MacWilliams identity to this dual code, whose size is $2^{\ell }=n+1$. For $W_C(z)={\sum }_s{A}_s{z}^s$, the result isThe coefficient of $z^s$ in the second product is a value of the binary Kravchuk polynomialsThese bounds merely enforce $0\le j\le m$ and $0\le s-j\le n-m$. Since $n-m=2^{\ell -1}-1$, they agree with the bounds in the PDF. Therefore the requested formula isFor a further check on the low-weight counts, $n=2m-1$ givesThus $K_3(m)=m-1$, $K_4(m)=\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$, and $K_5(m)=-\left(\genfrac{}{}{0ex}{}{m-1}{2}\right)$. Substituting these into the boxed formula and using $m=(n+1)/2$ reproduces the three direct combinatorial counts above.

Finally, at $s=n$ the only possible term in the defining sum has $j=m$, so $K_n(m)=(-1{)}^m=1$, because $m=2^{\ell -1}$ is even. ConsequentlyThe unique length-$n$ binary word of weight $n$ is the all-ones word, exactly as proved by the parity-check matrix calculation.

The two diagonal $2\times 2$ blocks allow the characteristic polynomial to factor immediately:Thus the characteristic equation isSolving the two block eigenvalue equations gives the following nonzero eigenvectors and corresponding eigenvalues:For example, the lower block maps $(1,i{)}^T$ to $(2i,-2{)}^T=2i(1,i{)}^T$, and maps $(1,-i{)}^T$ to zero. The four eigenvalues are distinct. Alternatively, the determinant of the matrix with columns $a,b,c,d$ is $(-2)(-2i)=4i\ne 0$. Hence they form a basis and span the complex vector space ${\mathbb{C}}^4$. Their nonzero scalar multiples are all the corresponding eigenvectors.

For any eigenvector $v$ with eigenvalue ${\lambda }_v$, its one-dimensional vector subspace satisfies $A(sv)={\lambda }_{v}sv$. If ${\lambda }_v\ne 0$, every $tv$ has preimage $(t/{\lambda }_v)v$, so the restricted linear map is onto that same subspace. ConsequentlyThe last line is the kernel of a linear map direction: multiplication by $A$ collapses that line to the zero-dimensional subspace.

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.