Transvections and Elementary Transformations

Introduction

The simplest invertible linear transformations after the scalars are the shears, which fix a hyperplane pointwise and shift every point parallel to a line. In coordinates a shear is an elementary matrix: it differs from the identity in a single off-diagonal entry. Every invertible linear transformation over a field is a product of shears and diagonal scalings, the group generated by the shears alone is the special linear group, and the computation that carries out the factorisation is Gaussian elimination. This article develops the shear, its matrix form, the elimination algorithm it performs, and its place in the classification of unipotent transformations.

Throughout, $F$ is a field, $V$ is a finite-dimensional $F$-vector space of dimension $n$, and the notation of the preceding articles is used: $\operatorname{GL}(V)$, $\operatorname{SL}(V)=\ker\det$, and the elementary matrix $E_{ij}(\lambda)=I+\lambda e_{ij}$. The determinant is not needed to define a transvection, and that is the point of the article: a transvection is a purely linear-algebraic object, defined by a linear functional and a vector, with no inner product, length, angle or orthogonality entering. The last section makes the contrast precise with $LU$ elimination, which is available at this level, and $QR$ decomposition, which is not.

Transvections

Definition

Definition. A transvection, or shear, on $V$ is a linear map $\tau:V \to V$ of the form

$$ \tau(x)=x+f(x)v $$

for a linear functional $f \in V^*$ and a vector $v \in V$ satisfying $f(v)=0$ and $f \neq 0$, $v \neq 0$.

The condition $f(v)=0$ is what makes $\tau$ invertible with an inverse of the same shape.

Proposition. Let $\tau(x)=x+f(x)v$ with $f(v)=0$. Then:

(i) $\tau$ is linear, and the fixed space of $\tau$ is $\ker f$, a hyperplane;

(ii) $\tau(x)-x \in \langle v\rangle$ for every $x$, so $v$ spans the direction of the shift;

(iii) $\tau^{-1}(x)=x-f(x)v$;

(iv) $(\tau-\operatorname{id})(x)=f(x)v$, so $\tau-\operatorname{id}$ has rank $1$ and its square is zero:

$$ (\tau-\operatorname{id})^2(x)=f(x)(\tau-\operatorname{id})(v)=f(x)f(v)v=0 ; $$

(v) $\det\tau=1$, so $\tau \in \operatorname{SL}(V)$.

Proof. (i) Linearity is clear. If $\tau(x)=x$ then $f(x)v=0$ and $v \neq 0$, so $f(x)=0$; conversely $f(x)=0$ gives $\tau(x)=x$. The functional $f$ is nonzero, so $\ker f$ is a hyperplane. (ii) Immediate. (iii) Using $f(v)=0$ twice, $\tau(x-f(x)v)=x-f(x)v+f(x-f(x)v)v=x-f(x)v+f(x)v=x$. (iv) $(\tau-\operatorname{id})(x)=f(x)v$ has image $\langle v\rangle$, of dimension $1$; and $(\tau-\operatorname{id})^2(x)=f(x)(\tau-\operatorname{id})(v)=f(x)\cdot f(v)v=0$. (v) $\tau$ is unipotent with $\tau-\operatorname{id}$ nilpotent of rank $1$, so its only eigenvalue is $1$ and its determinant is a product of eigenvalues, hence $1$.

Property (iv) says that a transvection is exactly a transformation of the form $\operatorname{id}+N$ with $N^2=0$ and $\operatorname{rk}N=1$: the rank-one condition is not automatic, but every rank-one square-zero endomorphism with $\operatorname{im}N \subseteq \ker N$ and $N \neq 0$ is of the form $x\mapsto f(x)v$ with $f(v)=0$.

Coordinates and a Normal Form

Proposition. Let $\tau(x)=x+f(x)v$ be a transvection. Choose $v_1 \in V$ with $f(v_1)=1$, and let $v_2,\dots,v_n$ be a basis of $\ker f$ chosen so that $v_2=v$. Then $(v_1,v_2,\dots,v_n)$ is a basis of $V$, in which $[\tau]=E_{21}(1)$.

Proof. The vector $v_1$ is not in $\ker f$, so $v_1,v_2,\dots,v_n$ is a basis. For $j \ge 2$, $f(v_j)=0$ and hence $\tau(v_j)=v_j$. For $v_1$, $\tau(v_1)=v_1+f(v_1)v=v_1+v_2$. So the first column of the matrix is $(1,1,0,\dots,0)^{\mathsf{T}}$ and all other columns are standard basis vectors, that is, the matrix is $I+e_{21}=E_{21}(1)$.

Note that the direction vector $v$ in the formula is fixed by $\tau$, because $f(v)=0$; the vectors that are moved are those $x$ with $f(x) \neq 0$, and each is moved parallel to $v$ by the amount $f(x)$. The hyperplane $\ker f$ is fixed pointwise, and the line $\langle v\rangle$ lies inside it and is the image of $\tau-\operatorname{id}$.

Corollary. Every transvection is conjugate in $\operatorname{GL}(V)$ to every other: for each $n$, all transvections lie in a single conjugacy class, represented by $E_{21}(1)$.

Proof. The normal form is independent of $\tau$ and of the field.

This is in sharp contrast with the orthogonal and symplectic cases, where the transvections present are constrained by the form and split into several classes.

Elementary Matrices

The Elementary Matrices are Transvections

Definition. For $n \ge 2$, $i \neq j$ and $\lambda \in F$, the elementary matrix is $E_{ij}(\lambda)=I+\lambda e_{ij}$, where $e_{ij}$ is the matrix with a single $1$ in position $(i,j)$.

Proposition. In the standard basis of $F^n$, the matrix $E_{ij}(\lambda)$ is the transvection $x \mapsto x+\lambda x_j e_i$, where $x_j$ is the $j$-th coordinate functional and $e_i$ the $i$-th basis vector. It satisfies $E_{ij}(\lambda)E_{ij}(\mu)=E_{ij}(\lambda+\mu)$ and $E_{ij}(\lambda)^{-1}=E_{ij}(-\lambda)$, so the matrices $E_{ij}(\lambda)$, $\lambda \in F$, form a subgroup isomorphic to the additive group $(F,+)$.

Proof. The matrix $e_{ij}$ sends $x$ to $x_je_i$, so $(I+\lambda e_{ij})x=x+\lambda x_je_i$. The functional $f(x)=\lambda x_j$ and the vector $v=e_i$ satisfy $f(v)=\lambda \cdot 0=0$ for $i \neq j$. Products: $(I+\lambda e_{ij})(I+\mu e_{ij})=I+(\lambda+\mu)e_{ij}$, since $e_{ij}^2=0$.

The elementary column operations are the same matrices applied on the right, $A \mapsto AE_{ij}(\lambda)$; they replace column $j$ by column $j$ plus $\lambda$ times column $i$. The row operations are the left multiplications $A \mapsto E_{ij}(\lambda)A$, which replace row $i$ by row $i$ plus $\lambda$ times row $j$. Row and column operations generate the same groups of matrices, but acting on the two sides of a matrix; and for an operator on $V$, a change of basis realises these operations as conjugation.

Interrelations

Proposition. For distinct indices $i,j,k$ one has the commutator identity

$$ E_{ij}(\lambda)E_{jk}(\mu)E_{ij}(-\lambda)E_{jk}(-\mu)=E_{ik}(\lambda\mu), $$

and for $t \in F^{\times}$ and the diagonal matrix $D=\operatorname{diag}(1,\dots,t,\dots,t^{-1},\dots,1)$ with $t$ and $t^{-1}$ in positions $i$ and $j$,

$$ D E_{ij}(\lambda) D^{-1}=E_{ij}(t^2\lambda), \qquad D E_{ji}(\lambda) D^{-1}=E_{ji}(t^{-2}\lambda). $$

Proof. The first is the standard elementary computation $[E_{ij}(\lambda),E_{jk}(\mu)]=E_{ik}(\lambda\mu)$ for $i,j,k$ distinct, obtained by multiplying out $2 \times 2$ blocks. The second is conjugation of the matrix with a single off-diagonal entry, which scales that entry by $t\cdot t$ when it lies above the diagonal and by $t^{-1}\cdot t^{-1}$ when it lies below it.

These identities are exactly what the commutator theorems of the preceding article use: the first produces an elementary matrix of a new index from two of the others, and the second rescales the parameter.

Gaussian Elimination

The Algorithm

Definition. An elementary row operation on a matrix is one of the following: interchange two rows; add a multiple of one row to another; multiply a row by a nonzero scalar. The first two are implemented by transvections and permutation matrices, the third by a diagonal matrix.

Theorem (Gaussian elimination). Every matrix over a field can be brought to reduced row-echelon form by finitely many elementary row operations. If the leading principal minors of $A$ are all nonzero, the operations can be taken to be additions of multiples of rows to later rows, and the result is the $LU$ factorisation $A=LU$ with $L$ unit lower triangular and $U$ upper triangular.

Proof. The elimination step clears the entries below a pivot by subtracting a multiple of the pivot row from each later row; each such step is left multiplication by a lower elementary matrix. The pivot at stage $k$ is the $k$-th leading principal minor of the partially reduced matrix divided by its predecessor, so it is nonzero exactly when the corresponding leading principal minor of $A$ is nonzero. Collecting the inverse elementary matrices gives $L$; the reduced matrix is $U$.

With row interchanges allowed, every invertible matrix has a factorisation $PA=LU$ with $P$ a permutation matrix, obtained by partial pivoting. This is the standard computational form.

Example. The matrix $A=\begin{pmatrix}2&1\\4&5\end{pmatrix}$ has $U=E_{21}(-2)A=\begin{pmatrix}2&1\\0&3\end{pmatrix}$ and $L=E_{21}(2)=\begin{pmatrix}1&0\\2&1\end{pmatrix}$, so $A=LU$. The leading principal minors of $A$ are $2$ and $2 \cdot 5-1\cdot4=6$, both nonzero, so no interchange is needed; the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ needs one, and its $LU$ factorisation exists only after the interchange.

Why Nothing Here Needs a Metric

The operations above are linear combinations of the rows with coefficients in $F$. They involve only addition and multiplication, not length, distance, angle or orthogonality. Two consequences follow:

(i) Elimination is available over any field, and over any commutative ring in the weaker Smith form. The $LU$ and row-echelon forms are defined by the linear structure alone, and they furnish the generation of $\operatorname{GL}_n(F)$ by elementary and diagonal matrices without any notion of size.

(ii) The $QR$ decomposition needs an inner product. Gram–Schmidt orthogonalisation produces an orthogonal matrix $Q$ and an upper triangular $R$ with $A=QR$; but the construction projects onto the orthogonal complement of the span of earlier columns, a notion requiring an inner product, and it normalises vectors, requiring square roots. Over an arbitrary field, or over a real space with a degenerate or indefinite form, the decomposition is not available in general. This is the exact boundary: $LU$ is a statement about linear structure and filtration by coordinate subspaces, while $QR$ is a statement about Euclidean geometry, and it is not a statement about vector spaces over an arbitrary field at all.

The same contrast appears in the menu of generation statements: the elementary matrices generate the special linear group because they are defined by incidence relations among basis vectors, whereas the orthogonal group is defined by the preservation of a form and is generated by reflections and form-constrained transvections, a different and metric-dependent story.

Generation of the Special Linear Group

Theorem. For $n \ge 2$ and any field $F$, $\operatorname{SL}_n(F)$ is generated by the elementary matrices $E_{ij}(\lambda)$.

Proof (standard). This is the classical generation theorem of the theory of the linear groups, quoted here as standard; its two verifiable ends are recorded. Since $\det A=1$ the first column of $A$ is nonzero, and if its first entry vanishes then adding a suitable multiple of a row with nonzero first-column entry to the first row makes the pivot nonzero. The row transvections $E_{k1}(\lambda)$ with $\lambda=-a_{k1}/a_{11}$ then clear the first column below the pivot, and the column transvections $E_{1j}(\mu)$ with $\mu=-a_{1j}/a_{11}$ clear the first row to its right, leaving a matrix $\operatorname{diag}(c,A'')$ with $c\det A''=1$. The subgroup generated by the elementary matrices is normal in $\operatorname{SL}_n(F)$, and the classical theorem of Dieudonné identifies it with $\operatorname{SL}_n(F)$; the ingredients are that $\operatorname{diag}(c,c^{-1},1,\dots,1) \in E$ for $n \ge 3$, together with the identities $E_{ij}(\lambda)=[E_{ik}(\lambda),E_{kj}(1)]$ for distinct $i,j,k$ and $DE_{12}(\lambda)D^{-1}=E_{12}(t^2\lambda)$, proved above, which handle $n \ge 3$ and $n=2$ with $|F|>3$ respectively. The two small fields are settled by direct enumeration of the six and the twenty-four elements: $E_{12}(1)$ and $E_{21}(1)$ generate $\operatorname{SL}_2(\mathbb{F}_2)\cong S_3$ and $\operatorname{SL}_2(\mathbb{F}_3)$.

Corollary. $\operatorname{GL}_n(F)$ is generated by the elementary matrices together with the diagonal matrices.

Proof. Every invertible matrix can be reduced to a diagonal matrix by row operations, by the elimination theorem; the elementary operations are elementary matrices and interchanges, and an interchange is a product of elementary matrices and a diagonal sign change over a field.

Unipotent Transformations

Definition and Basic Properties

Definition. A linear map $u:V \to V$ is unipotent if $u=\operatorname{id}+N$ with $N$ nilpotent. Equivalently, $u$ has characteristic polynomial $(x-1)^n$, so its only eigenvalue is $1$ over an algebraically closed field.

Proposition. (i) A unipotent map is invertible, with $u^{-1}=\operatorname{id}-N+N^2-\cdots$ (a finite sum), and $\det u=1$. (ii) If $u$ is unipotent and $S$ is any invertible map then $SuS^{-1}$ is unipotent with nilpotent part $SNS^{-1}$. (iii) The product of commuting unipotent maps is unipotent. (iv) A transvection is unipotent with $N^2=0$ and $\operatorname{rk}N=1$.

Proof. (i) The displayed sum telescopes with $\operatorname{id}+N$ because $N$ is nilpotent; the determinant is the product of the eigenvalues, all equal to $1$. (ii) $S(\operatorname{id}+N)S^{-1}=\operatorname{id}+SNS^{-1}$ and nilpotence is preserved by conjugation. (iii) $(\operatorname{id}+N)(\operatorname{id}+M)=\operatorname{id}+(N+M+NM)$ and $N+M+NM$ is nilpotent if $N,M$ commute, since then $N+M$ and $NM$ are commuting nilpotent elements and their sum is nilpotent. (iv) is the computation of the transvection section.

Jordan Decomposition of a Unipotent Map

Theorem. Let $u$ be unipotent and suppose $c_u$ splits, which is automatic when $F$ is algebraically closed. Then $u$ is similar, over $F$, to a direct sum of Jordan blocks $J_e(1)=\operatorname{id}_e+N_e$ with $N_e$ the superdiagonal shift, and the multiset of block sizes is determined by $u$. In particular $u$ is a transvection exactly when it has a single Jordan block of size $2$ and all other blocks of size $1$.

Proof. This is the Jordan decomposition applied to the unique eigenvalue $1$; the rank-one condition forces exactly one block of size $2$.

A unipotent map with $N^2=0$ but $\operatorname{rk}N=r \ge 2$ is a product of commuting transvections in suitable coordinates: $N$ is then a sum of $r$ rank-one square-zero pieces with disjoint images and kernels, and each piece is a transvection, the pieces commuting. This is the sense in which transvections are the elementary unipotents.

Proposition (orders). In characteristic $0$ a nontrivial unipotent map has infinite order. In characteristic $p>0$, if $N$ is nilpotent then $(\operatorname{id}+N)^p=\operatorname{id}+N^p$ by the binomial theorem, because the intermediate binomial coefficients vanish in characteristic $p$; consequently $\operatorname{id}+N$ has order a power of $p$. For a transvection, $N^2=0$, and $\operatorname{id}+N$ has order dividing $p$: the order is exactly $p$ when $p$ is odd, and is $2$ when $p=2$, since then $(\operatorname{id}+N)^2=\operatorname{id}$.

Proof. The characteristic $p$ identity is the binomial theorem, and if $N^e=0$ then $(\operatorname{id}+N)^{p^m}=\operatorname{id}+N^{p^m}=\operatorname{id}$ for $p^m \ge e$, so the order divides $p^m$ and is a power of $p$. For a transvection, $N^2=0$ and $(\operatorname{id}+N)^k=\operatorname{id}+kN$ for $0 \le k \le p-1$, which is $\operatorname{id}$ only for $k=0$ when $p$ is odd, so the order is $p$; for $p=2$ the order is $2$. In characteristic $0$ the same formula $(\operatorname{id}+N)^k=\operatorname{id}+kN+\cdots$ has the nonzero term $kN$, so no positive power is the identity.

Dilatations and the Centre

Definition. A dilatation is a linear map $\delta$ that fixes a hyperplane $H$ pointwise and acts on a complementary line $\langle v\rangle$ by $\delta(v)=\lambda v$ with $\lambda \in F^{\times}$. A homothety is a scalar map $\lambda\operatorname{id}_V$.

Proposition. Each dilatation lies in $\operatorname{GL}(V)$, its determinant is $\lambda$, and the homotheties are exactly the dilatations whose fixed hyperplane can be chosen arbitrarily and whose scalar is independent of $H$.

Proof. In a basis adapted to the decomposition $V=H \oplus \langle v\rangle$ the matrix is diagonal with one entry $\lambda$ and the rest $1$, so it is invertible with determinant $\lambda$. A scalar map $\lambda\operatorname{id}$ fixes every hyperplane; conversely if a map fixes every hyperplane pointwise it fixes every vector, and if it fixes a hyperplane $H$ pointwise and acts on a complementary line by $\lambda$, its fixed hyperplane is exactly $H$ unless $\lambda=1$.

Theorem. Over any field, $\operatorname{GL}(V)$ is generated by transvections and dilatations.

Proof. By Gaussian elimination, $\operatorname{GL}(V)$ is generated by the elementary matrices and the diagonal matrices in any basis. Each elementary matrix $E_{ij}(\lambda)$ is a transvection in that basis. A diagonal matrix $\operatorname{diag}(d_1,\dots,d_n)$ is the product over $i$ of the dilatations that fix the coordinate hyperplane $\{x_i=0\}$ and scale the coordinate line $\langle e_i\rangle$ by $d_i$; each factor is a dilatation with fixed hyperplane the complementary coordinate hyperplane. Hence the transvections and dilatations generate $\operatorname{GL}(V)$.

Example. The identity

$$ \begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&0\\-1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix} $$

exhibits the quarter-turn $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ of determinant $1$ as a product of transvections: the outer factors are the transvection $E_{12}(1)$ and the inner factor is the inverse of the transvection $E_{21}(1)$. It also shows that a product of transvections need not be triangular in any basis: the quarter turn has eigenvalues $\pm i$, which lie outside $\mathbb{R}$, so over $\mathbb{R}$ it is irreducible and in particular not conjugate to an elementary matrix.

Remark. The centre of $\operatorname{GL}(V)$ consists of the homotheties, and the homotheties are the intersection of the family of all dilatations in the following weak sense: a homothety is a dilatation for every choice of complementary hyperplane and line, whereas a non-scalar dilatation is a dilatation only for the specific pair $(H,\langle v\rangle)$. The centre meets $\operatorname{SL}(V)$ in the $n$-th roots of unity, as in the preceding articles.

Summary

A transvection on a finite-dimensional vector space over a field is a map $\tau(x)=x+f(x)v$ with $f(v)=0$, $f \neq 0$ and $v \neq 0$. It fixes the hyperplane $\ker f$ pointwise, translates each vector along $\langle v\rangle$ by the amount $f(x)$, is invertible with $\tau^{-1}(x)=x-f(x)v$, has rank-one nilpotent part $\tau-\operatorname{id}=v \otimes f$ with square zero, and lies in the special linear group with $\det\tau=1$. All transvections on a space of dimension $n$ are conjugate in $\operatorname{GL}(V)$ and are represented by the elementary matrix $E_{21}(1)$; the matrices $E_{ij}(\lambda)=I+\lambda e_{ij}$ are exactly the transvections in the standard basis, and the matrices $E_{ij}(\lambda)$ for fixed $i \neq j$ form a copy of the additive group. The identities $[E_{ij}(\lambda),E_{jk}(\mu)]=E_{ik}(\lambda\mu)$ and $D E_{ij}(\lambda)D^{-1}=E_{ij}(t^2\lambda)$ drive the commutator theory.

Row and column operations are left and right multiplication by these matrices, and Gaussian elimination is the computation that implements them; it yields the $LU$ factorisation when the leading principal minors are nonzero and the general row-echelon form with partial pivoting, and it requires no metric. Consequently the elementary matrices generate $\operatorname{SL}_n(F)$ for $n \ge 2$ and, together with the diagonal matrices, generate $\operatorname{GL}_n(F)$; in contrast the $QR$ decomposition requires an inner product and square roots and is not a statement about an arbitrary field.

Transvections are the elementary unipotent transformations: $u$ is unipotent when $u=\operatorname{id}+N$ with $N$ nilpotent, such a $u$ has determinant $1$ and inverse $\operatorname{id}-N+N^2-\cdots$, and a unipotent map is a transvection exactly when its Jordan form is a single block of size $2$ and singleton blocks. A transvection generates an infinite cyclic group in characteristic zero and a group of order the characteristic in positive characteristic. Finally, dilatations fix a hyperplane and scale a complementary line, they generate $\operatorname{GL}(V)$ together with the transvections, and the scalars among them are the centre of $\operatorname{GL}(V)$.

Summary of Notation

Symbol	Meaning
$F$	a field
$V$	$F$-vector space of dimension $n$
$\tau(x)=x+f(x)v$, $f(v)=0$	transvection, with $f \in V^*$, $v \in V$
$\ker f$	fixed hyperplane of the transvection
$\langle v\rangle$	direction of the shift
$E_{ij}(\lambda)=I+\lambda e_{ij}$	elementary matrix, a transvection
$e_{ij}$	matrix unit
$L$, $U$, $P$	lower triangular, upper triangular, permutation factors in $PA=LU$
$u=\operatorname{id}+N$	unipotent transformation, $N$ nilpotent
$N_e$	nilpotent superdiagonal shift
$J_e(1)=\operatorname{id}_e+N_e$	Jordan block of a unipotent map
$\delta$	dilatation fixing a hyperplane and scaling a complementary line
$\lambda\operatorname{id}_V$	homothety
$\operatorname{rk}N$	rank of an endomorphism
$V^*$	dual space