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In the basis $(v,Av)$, the linear map given by $A$ sends $v$ to the second basis vector and sends $Av$ to $A^{2}v=-v$. Equivalently,Multiplying on the left by the matrix inverse $E^{-1}$ givesThis change of basis produces the companion matrix found in part (a)(ii).

The required general linear position condition is that every $n+1$ of the $n+2$ representing vectors are linearly independent. Such an ordered configuration is a projective frame.

Choose representatives $p_0,\dots ,p_n$ of the first $n+1$ projective points; they form a basis. Write a representative of the last point as $p_{n+1}={\sum }_i{a}_i{p}_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to ${\tilde{p}}_i=a_i{p}_i$, so that $p_{n+1}={\sum }_i{\tilde{p}}_i$.

Do the same for the target configuration, obtaining representatives ${\tilde{q}}_i$ with $q_{n+1}={\sum }_i{\tilde{q}}_i$. The unique linear map sending the basis ${\tilde{p}}_i$ to ${\tilde{q}}_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced projective transformation sends every point to its prescribed partner. Thus any two ordered projective frames are projectively equivalent.

The Fréchet derivative at a point $a$ is the unique linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe error divided by $\Vert h\Vert$ tends to zero as $h\to 0$. In standard coordinates this linear map is represented by the Jacobian matrix, but the derivative itself is the map acting on increments, not merely a list of directional rates.

A map $g$ is differentiable at $a$ when a linear map $D{g}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ approximates its increment to first order:This Fréchet derivative is unique, and its matrix in the standard bases is the Jacobian matrix. Differentiability on a domain means differentiability at each point there.

A standard sufficient condition for a zero locus to be a submanifold is the regular level set theorem: if $g$ is $C^1$ on an open domain and $D{g}_a$ is surjective at every $a\in g^{-1}(0)$, thenFor smooth $g$, the submanifold is smooth. The full-rank condition is the relevant constraint qualification.

Under this regularity hypothesis, its tangent space isIndeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the chain rule gives $D{g}_a({\gamma }^{\prime }(0))=0$. Thus all tangent velocities lie in the kernel of a linear map. The rank-nullity theorem makes this kernel $(n-k)$-dimensional, the same dimension as the tangent space, proving equality. In coordinates, solve the homogeneous linear equations given by the Jacobian matrix.

The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional submanifold, but $D{g}_0=0$. Its kernel is all of $\mathbb{R}$, whereas its tangent space is zero. This is the phenomenon that singular defining maps can overestimate tangent spaces.

The map is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatEquivalently, $f(x+h)=f(x)+Lh+o(\Vert h\Vert )$. This linear map is unique and is the Fréchet derivative $D{f}_x$. Its matrix in the standard bases is the Jacobian matrix, whose $(i,j)$ entry is $\partial f_i/\partial x_j$ at $x$. A function is differentiable on its domain if this condition holds at every point.

Choose nonzero vectors $p_i\in {\mathbb{F}}^2$ representing $P_i$, and write $[u,v]=\det (u,v)$. Use the cross-ratio conventionDistinctness makes the denominator nonzero. Every rescaling of a representative occurs once in numerator and denominator, so the value depends only on the projective points. Swapping both pairs simply exchanges the numerator's factors and the denominator's factors:Under an invertible linear map $A$, $[Au,Av]=\det (A)[u,v]$. Two factors of $\det (A)$ occur above and below the fraction, proving projective transformation invariance.

Finally choose homogeneous coordinates making $P_0=[1:0]$, $P_1=[0:1]$, $P_2=[1:1]$. This normalization exists by the allowed theorem about a projective frame. A point distinct from $P_0$ is uniquely $[t:1]$, and the displayed cross-ratio is $t$. Thus equality of the two cross-ratios forces equal coordinates andThis also shows that for fixed first three distinct points the fourth point is determined by its cross-ratio.

A projective transformation of $\mathbb{P}({\mathbb{F}}^{n+1})$ is induced by an invertible linear map $A:{\mathbb{F}}^{n+1}\to {\mathbb{F}}^{n+1}$:This is well defined because replacing $v$ by a nonzero scalar multiple does not change $[Av]$. Multiplying $A$ by a nonzero scalar also leaves $\tau$ unchanged. Conversely these are precisely the transformations in the projective linear group ${\mathrm{PGL}}_{n+1}(\mathbb{F})$. No field automorphism is included in this definition.

The rank-nullity theorem states that a linear map $T:V\to W$ with finite-dimensional $V$ satisfiesChoose a basis $k_1,\dots ,k_r$ of $\ker T$ and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We prove that $T{u}_1,\dots ,T{u}_s$ is a basis of $\mathrm{im}T$.

Every $v\in V$ is a linear combination of the extended basis. Applying $T$ kills the $k_i$ terms, so $Tv$ is a linear combination of $T{u}_1,\dots ,T{u}_s$. This proves the linear span assertion. For linear independence, if ${\sum }_j{b}_{j}T{u}_j=0$, then ${\sum }_j{b}_j{u}_j$ belongs to the kernel of a linear map. Hence ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some scalars $a_i$. The linear independence of the extended basis forces every $a_i$ and $b_j$ to vanish. Consequently $\mathrm{rank}T=s$ and $\mathrm{nullity}T=r$, while $\dim V=r+s$, proving the theorem.

For a linear map $T:V\to W$, its image of a linear map and kernel of a linear map areTheir dimensions define the rank of a linear map and nullity:The image of a linear map measures which output vectors can occur, while the kernel of a linear map measures which input vectors are sent to zero.

Take the linear mapIts kernel of a linear map consists of all $(x,0)$, and its image of a linear map is the same line:Their set intersection is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. ThusThe matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

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 $V$.

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain of a function $V$,Take a basis $k_1,\dots ,k_r$ of the kernel of a linear map and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We show that $T(u_1),\dots ,T(u_s)$ is a basis of the image of a linear map.

Every $v\in V$ is a linear combination of the extended basis, and its $k_i$ terms vanish under $T$. Therefore $T(v)$ is a linear combination of the $T(u_j)$, proving the required linear span property. For linear independence, suppose ${\sum }_j{b}_{j}T(u_j)=0$. Then ${\sum }_j{b}_j{u}_j\in \ker T$, so ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some real scalars $a_i$. The extended basis is linearly independent, forcing every $a_i$ and $b_j$ to be zero. Thus the image of a linear map has dimension $s$, the kernel of a linear map has dimension $r$, and $\dim V=r+s$ proves the result.

A basis provides unique coordinates. DefineThe uniqueness of the coordinates makes this a well-defined function. Adding coordinates and multiplying them by a real scalar shows directly that $\alpha (au+bw)=a\alpha (u)+b\alpha (w)$, so it is a linear map. Substituting $e_i$ gives $\alpha (e_i)=v_i$.

Conversely, every linear map with those three values must send ${\sum }_i{a}_i{e}_i$ to ${\sum }_i{a}_i{v}_i$, by preservation of linear combinations. This determines its value on every element of $V$, proving uniqueness. No linear independence of $v_1,v_2,v_3$ is needed.

For a square matrix $M$, an eigenvalue–eigenvector pair is a scalar $\lambda$ and a nonzero vector $v$ satisfyingThus the linear map represented by $M$ scales this particular direction by $\lambda$. The nonzero condition excludes the zero vector, which would otherwise satisfy the displayed equation for every scalar.

For a square matrix $M$, an eigenvalue–eigenvector pair consists of a scalar $\lambda$ and a nonzero vector $v$ satisfyingThe linear map represented by $M$ therefore scales $v$ by $\lambda$. The exclusion of the zero vector matters: otherwise every scalar would satisfy the equation trivially.

Translate the vertex centroid to the origin, rotate, and translate back:This follows the supplied rotation matrix, so positive $\theta$ rotates clockwise in the usual Cartesian axes. The opposite signs give the conventional counterclockwise matrix. Since the vertex centroid of the centered vertices is zero and a linear map commutes with their average, the final centroid remains c.

A linear map $T:V\to W$ is onto, or surjective, when every $w\in W$ equals $T(v)$ for some $v\in V$; equivalently its image is all of $W$. Here $T$ is not onto, because its image has dimension two while its declared codomain ${\mathbb{R}}^3$ has dimension three. Indeed every image vector satisfies $x+y-3z=0$, so, for example, $(0,0,1)$ is not attained.

The rank-nullity theorem states $\dim V=\dim \ker T+\dim \mathrm{im}T$ for a linear map from a finite-dimensional vector space $V$. Here the domain has dimension four and the image has dimension two. Thus $\dim \ker T=4-2=2$. For example, a kernel basis is $\{(-2,-1,1,0),(-1,-2,0,1)\}$, as follows from the reduced equations.

A linear map is one-to-one, or injective, when equal outputs imply equal inputs. Equivalently its kernel is $\{0\}$: subtracting inputs converts equality of outputs to a zero output. Here $T$ is not one-to-one, since its kernel has dimension two. Concretely, $T(-2,-1,1,0)=T(0)=0$ for distinct inputs.

No. The nonzero vector $v=(2,-9,-3,21{)}^{\top }$ is in the kernel of the linear map, so $T(v)=T(0)=0$ although $v\ne 0$. A linear map is injective exactly when its kernel consists only of zero.