Surjective Transformations, Spanning Sets and Dimension

Overview

Surjectivity is defined by an existence statement over the whole codomain, which is not something that can be checked one target at a time. The structural results on this page replace that statement with a finite construction. Applying a linear transformation to a spanning set of the domain produces a spanning set of the range, so the entire range — an infinite set — is generated by as many vectors as there are members in a spanning set of $U$ . Choose a basis of $U$ and the generation is as economical as possible.

This is the exact counterpart of the way injectivity interacts with linear independence, with the two notions exchanged. An injection preserves independence; a surjection is characterised through spanning. The parallel is not decorative: it explains why the two properties are tested by mirror-image computations and why their dimension consequences point in opposite directions. An injection cannot compress, so $\dim(U) \le \dim(V)$ . A surjection cannot manufacture dimensions it was not given, so $\dim(U) \ge \dim(V)$ .

The practical procedure follows immediately. Take any basis of the domain, apply the transformation to each of its vectors, and ask whether the resulting set spans the codomain. If it does, the transformation is surjective; if it spans only a proper subspace, that subspace is the range and the transformation is not surjective. Because the image set is usually linearly dependent, it must generally be culled before a basis of the range is obtained, and the culling is a routine row reduction.

Two cautions govern the use of these results. The dimension inequality is necessary but never sufficient: a transformation from a large space into a small one can still miss most of the codomain, and only the spanning computation settles the question. And the image of a basis of $U$ is a spanning set of the range, not automatically a basis of it — independence is precisely what a non-injective transformation destroys.

Definition

Spanning Set

SSVS

A set $S$ of vectors in a vector space $W$ spans $W$ when every vector of $W$ can be written as a linear combination of finitely many members of $S$ ; equivalently $\operatorname{span}(S) = W$ . A spanning set need not be linearly independent, and a spanning set that is independent is a basis.

Restated because every theorem on this page is a statement about spanning sets being carried from the domain into the codomain.

Surjective Linear Transformation

SLT

A linear transformation $T \colon U \to V$ is surjective when for every $\vec{v} \in V$ there exists $\vec{u} \in U$ with $T(\vec{u}) = \vec{v}$ . Equivalently, the range of $T$ equals the whole codomain, $\mathcal{R}(T) = V$ .

Surjectivity depends on the declared codomain. Every transformation is surjective onto its own range, so the target space must be named for the claim to have content.

Range of a Linear Transformation

RLT

The set $\mathcal{R}(T) = \{\, T(\vec{u}) \mid \vec{u} \in U \,\}$ of all outputs of $T$ , always a subspace of the codomain. Its dimension is the rank of $T$ .

The results on this page are, in the first instance, statements about how to generate this subspace from finitely many vectors.

Concepts

Images of a spanning set span the range

Let $S = \{\vec{u}_1, \dots, \vec{u}_t\}$ span $U$ and set $R = \{T(\vec{u}_1), \dots, T(\vec{u}_t)\}$ . Claim: $R$ spans $\mathcal{R}(T)$ . Take any $\vec{v} \in \mathcal{R}(T)$ . By definition of the range there is a $\vec{u} \in U$ with $T(\vec{u}) = \vec{v}$ , and because $S$ spans $U$ there are scalars with $\vec{u} = \alpha_1 \vec{u}_1 + \cdots + \alpha_t \vec{u}_t$ . Applying $T$ and using linearity, $\vec{v} = \alpha_1 T(\vec{u}_1) + \cdots + \alpha_t T(\vec{u}_t)$ , which exhibits $\vec{v}$ as a linear combination of $R$ . Every element of the range is therefore reachable from $R$ , so $R$ spans it. Note that $R$ automatically lies inside the range, so no reverse inclusion argument is needed.

Why a basis is the right spanning set to use

The theorem holds for any spanning set of the domain, but the cost of the construction is one evaluation of $T$ per member, so the smallest spanning set is the cheapest. That is a basis, with exactly $\dim(U)$ members. Using a basis also bounds the rank immediately: the range is spanned by $\dim(U)$ vectors, so $\dim(\mathcal{R}(T)) \le \dim(U)$ , whatever the transformation does. For abstract domains — polynomials, matrices, function spaces — the natural basis is usually obvious, which is what makes the method practical where no coefficient matrix is given.

Culling the image set to a basis

The images of a basis span the range but are generally dependent, because a non-injective transformation collapses independence. To obtain a basis of the range, coordinatise the images in a fixed basis of the codomain, place them as rows of a matrix, row-reduce, and discard the zero rows. The surviving rows form a basis of the same subspace. Alternatively place the images as columns and retain those in pivot positions, which returns a basis consisting of actual images rather than synthetic combinations. Either way the count of surviving vectors is the rank of $T$ .

The basis characterisation of surjectivity

Let $B$ be a basis of $U$ and $C$ the set of its images. Then $T$ is surjective if and only if $C$ spans $V$ . Forward: surjectivity means $\mathcal{R}(T) = V$ , and $C$ spans the range, so $C$ spans $V$ . Backward: assume $C$ spans $V$ and take any $\vec{v} \in V$ . Write $\vec{v} = \beta_1 T(\vec{u}_1) + \cdots + \beta_m T(\vec{u}_m)$ , then define $\vec{u} = \beta_1 \vec{u}_1 + \cdots + \beta_m \vec{u}_m$ . Linearity gives $T(\vec{u}) = \vec{v}$ , so the required input exists and $T$ is surjective. The backward direction is constructive: it does not merely assert a preimage, it builds one from the spanning coefficients.

The dimension inequality and what it does not say

Suppose $T$ is surjective and, for contradiction, $\dim(U) = m < t = \dim(V)$ . A basis of $U$ has $m$ vectors, and their images span $V$ by the characterisation above. So $V$ , a space of dimension $t$ , is spanned by $m$ or fewer vectors with $m < t$ — impossible, since no spanning set can be smaller than the dimension. Hence $\dim(U) \ge \dim(V)$ . Used in contrapositive form this rules out surjectivity from the shapes alone. It is emphatically not sufficient: the transformation on the worked example below maps a four-dimensional domain onto a two-dimensional range inside a three-dimensional codomain, satisfying the inequality comfortably while failing to be surjective.

Range membership, preimages and composition

A vector lies in the range exactly when its preimage is non-empty, which is the bridge between the spanning description of the range and the consistency description of a linear system. Surjectivity is also preserved under composition: if $T \colon U \to V$ and $S \colon V \to W$ are surjective, then given $\vec{w} \in W$ , surjectivity of $S$ supplies $\vec{v}$ with $S(\vec{v}) = \vec{w}$ and surjectivity of $T$ supplies $\vec{u}$ with $T(\vec{u}) = \vec{v}$ , so $(S \circ T)(\vec{u}) = \vec{w}$ . Chaining the dimension inequality along the composition gives $\dim(U) \ge \dim(V) \ge \dim(W)$ , so a pipeline can only lose reach, never gain it.

Testing surjectivity through a spanning set

Compare dimensionsIf

\dim(U) < \dim(V)

, stop:

T

cannot be surjective and the formula is irrelevant.

Choose a basis of the domainAny basis. For

M_{mn}

take the single-entry matrices, for

P_n

the monomials, for

\mathbb{C}^n

the standard unit vectors.

Apply $T$ to each basis vectorThe resulting set spans the range, whatever else is true about

T

Coordinatise and row-reduceExpress each image in a fixed basis of the codomain, stack as rows, and reduce to discard dependent members.

Compare the rank with $\dim(V)$ Equal means the images span the codomain and

T

is surjective. Smaller means the range is a proper subspace.

Describe what is missingEliminate the coordinates to obtain the linear constraints satisfied by every attainable output, and exhibit a vector violating them.

Equations

Images of a spanning set span the range

EQ-SSRLT-01

span ({{\vec{u}}_{1}, \dots, {\vec{u}}_{t}}) = U Longrightarrow span ({T ({\vec{u}}_{1}), \dots, T ({\vec{u}}_{t})}) = R (T)

The construction that generates an infinite subspace from finitely many evaluations of $T$ .

Push-through of a linear combination

EQ-SSRLT-02

Push-through of a linear combination

The identity behind the spanning result: a combination of inputs maps to the same combination of the images, with the coefficients unchanged.

Basis characterisation of surjectivity

EQ-SLTB-01

T surjective \Leftrightarrow span ({T ({\vec{u}}_{1}), \dots, T ({\vec{u}}_{m})}) = V

Here $\{\vec{u}_1, \dots, \vec{u}_m\}$ is any basis of the domain $U$ . A finite test for an existence statement quantified over the whole codomain.

Dimension inequality for surjections

EQ-SLTD-01

T : U \to V surjective Longrightarrow \dim (U) \geq \dim (V)

A surjection cannot manufacture dimension. Used in contrapositive form it disposes of a candidate from the shapes alone.

Rank bounds

EQ-SLTD-02

\dim (R (T)) \leq min (\dim (U), \dim (V))

The range is spanned by $\dim(U)$ images and sits inside $V$ , so both dimensions bound the rank. Surjectivity is the case where the second bound is attained.

Range membership and preimages

EQ-RPI-01

\vec{v} \in R (T) \Leftrightarrow T^{- 1} (\vec{v}) \neq \emptyset

Links the spanning description of the range to the consistency description of the corresponding linear system.

Composition of surjections

EQ-CSLTS-01

T : U \to V surjective, S : V \to W surjective Longrightarrow (S \circ T) : U \to W surjective

Variable Definitions

Symbols used on this page
Symbol	Name	Meaning	Domain / type
$T$	Linear transformation	The map $T \colon U \to V$ under test	function from U to V
$S$	Second transformation	A map $S \colon V \to W$ composed after $T$	function from V to W
$B$	Basis of the domain	An independent spanning set of $U$ , with $\dim(U)$ members	finite subset of U
$C$	Image set	The set $T(B)$ ; it spans the range and, when $T$ is surjective, spans $V$	finite subset of V
$R (T)$	Range	The subspace of $V$ generated by $C$	subspace of V
$β_{i}$	Spanning coefficients	Scalars expressing a target $\vec{v}$ as a combination of the images; used to build a preimage	complex scalars
$m$	Domain dimension	$\dim(U)$ ; the number of evaluations of $T$ needed to generate the range	non-negative integer
$t$	Codomain dimension	$\dim(V)$ ; the minimum size of any spanning set of $V$	non-negative integer
$span (S)$	Span of a set	The set of all linear combinations of members of $S$	subspace

Worked Numerical Example

Problem statement

A controller parameterisation converts a $2 \times 2$ block of tuning gains into a quadratic closed-loop numerator polynomial by the linear rule $T \colon M_{22} \to P_2$ below. Determine whether every quadratic numerator can be produced, construct the range from the images of a basis, and identify the structural limitation if surjectivity fails.

State the parameterisation
The four gains $a, b, c, d$ combine into the three coefficients of a quadratic:
$T ([\begin{matrix} a & b \\ c & d \end{matrix}]) = (a + 2 b - c) + (2 a + 3 b + d) x + (a + b + c + d) x^{2}$
Check the dimension precondition
The domain $M_{22}$ has dimension $4$ and the codomain $P_2$ has dimension $3$ . Since $4 \ge 3$ , the dimension inequality does not rule out surjectivity. This settles nothing on its own; four gains might still act along only two independent directions.
Apply $T$ to a basis of the domain
Take the single-entry basis $\{E_{11}, E_{12}, E_{21}, E_{22}\}$ , which sets one gain to $1$ and the rest to $0$ in turn. The four images span the range:
$T (E_{11}) = 1 + 2 x + x^{2}, T (E_{12}) = 2 + 3 x + x^{2}, T (E_{21}) = - 1 + x^{2}, T (E_{22}) = x + x^{2}$
Coordinatise and row-reduce
Read each image against the basis $\{1,\ x,\ x^{2}\}$ of $P_2$ , stack the four coordinate vectors as rows, and reduce:
$Coordinatise and row-reduce$
Read the rank and the verdict
Only two non-zero rows survive, so $\dim(\mathcal{R}(T)) = 2$ while $\dim(P_2) = 3$ . The images do not span the codomain, so by the basis characterisation $T$ is not surjective — despite four gains acting on three coefficients. The dimension inequality was satisfied and told us nothing.
Cull the image set to a basis
Two of the four images are redundant, and the reduction names which: $T(E_{21}) = 3\,T(E_{11}) - 2\,T(E_{12})$ and $T(E_{22}) = 2\,T(E_{11}) - T(E_{12})$ . Discarding them costs nothing. Either of two bases may then be quoted — the original images $\{1 + 2x + x^{2},\ 2 + 3x + x^{2}\}$ , or the cleaner pair read from the reduced rows, which spans the same subspace:
$R (T) = span ({1 - x^{2}, x + x^{2}})$
Identify the structural constraint
Write a general element as $v_0 + v_1 x + v_2 x^{2} = a(1 - x^{2}) + b(x + x^{2})$ , giving $v_0 = a$ , $v_1 = b$ , $v_2 = -a + b$ . Eliminating $a$ and $b$ leaves one linear law, which is exactly the statement that the polynomial vanishes at $x = -1$ . Every attainable numerator carries a fixed zero at $-1$ , so a target such as $1 + x + x^{2}$ , for which the expression evaluates to $1$ , cannot be produced by any gain setting.
$v_{0} - v_{1} + v_{2} = 0 \Leftrightarrow p (- 1) = 0$
Repair the parameterisation
Let the third gain also influence the linear coefficient, changing the rule to $(a + 2b - c) + (2a + 3b + c + d)x + (a + b + c + d)x^{2}$ . The image of $E_{21}$ becomes $-1 + x + x^{2}$ , and the coordinate array now reduces to rank $3$ . The images span $P_2$ , so the revised parameterisation is surjective and every quadratic numerator is reachable.
$[\begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ - 1 & 1 & 1 \\ 0 & 1 & 1 \end{matrix}] ⟶ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}]$

Result

The original parameterisation offers four tuning gains but only two independent directions of influence, and it silently imposes a fixed closed-loop zero at $x = -1$ on every achievable design. No tuning effort can remove that zero, so an apparently rich four-parameter design space is in fact a two-parameter family — a limitation that surfaces immediately from the spanning test and would otherwise appear as an unexplained floor on achievable performance. Adding one coupling term restores full spanning and with it the ability to place all three numerator coefficients independently.

Applications & Industry Use

Control systems

Achievable closed-loop dynamics from a tuning set

Applying the parameterisation to a basis of the tuning space generates every achievable closed-loop polynomial. If the images fail to span the target polynomial space, some pole or zero locations are structurally unreachable, and no amount of tuning will place them.

Mechanical design

Workspace coverage from actuator directions

The images of the unit actuator commands span the reachable velocity space of a mechanism. Comparing the rank of that image set with the dimension of the task space determines whether the mechanism has full task-space authority in the configuration examined.

Optics & imaging

Reachable wavefront corrections

A deformable mirror maps actuator strokes to a wavefront expressed in a modal basis. Applying the influence map to a basis of actuator commands produces a spanning set of correctable wavefronts; modes outside its span are uncorrectable, which sets a hard floor on residual aberration.

Communications

Coverage of the syndrome space

A parity-check map must be surjective onto the syndrome space for every syndrome pattern to correspond to a real error class. Applying the map to a basis of received-word space and checking that the images span the syndrome space is the design-time verification of that requirement.

Chemical engineering

Independently controllable specifications

Applying the linearised blend model to a basis of feed adjustments generates the reachable specification changes. The rank of that set is the number of specifications that can be moved independently; anything less means certain specifications are locked together by the feed chemistry.

Computer graphics

Expressive range of a blend-shape rig

A character rig maps blend-shape weights linearly to vertex displacements. The images of the individual weights span the achievable deformation space, and its dimension bounds how many independent expressions the rig can represent regardless of how many sliders it exposes.

Design Considerations

Screen by dimension, then compute

The inequality $\dim(U) \ge \dim(V)$ costs nothing and eliminates impossible designs immediately. But a passing dimension check proves nothing at all, as the worked example demonstrates. Treat it strictly as a screen and always follow it with the spanning computation.

More parameters is not more authority

A design exposing many tuning parameters may still act along very few independent directions. The number that matters is the rank of the image set, not the parameter count. Reporting the rank alongside the parameter count prevents a false impression of design freedom.

Spanning is preserved, independence is not

The images of a basis always span the range, but they are a basis of the range only when $T$ is injective. Assuming otherwise leads to inflated rank claims and to bases that contain redundant members. Cull the image set explicitly rather than assuming it is already independent.

Report the missing directions, not just the failure

When the images do not span the codomain, eliminating the coordinates yields the linear laws every achievable output satisfies. Those laws are the actionable output: they name the structural limitation, and a single vector violating them is a concrete demonstration for a design review.

The backward proof is a construction

The characterisation does not merely assert that a preimage exists when the images span $V$ ; the spanning coefficients of a target are directly the coordinates of a preimage in the chosen basis. Implementations should exploit this, obtaining the required input from the same solve that establishes spanning rather than from a second computation.

Rank near the boundary is fragile

A design whose image set spans the codomain only marginally — with a tiny smallest singular value — is surjective in exact arithmetic and useless in practice, because reaching some outputs demands unbounded parameter values. Report the smallest singular value against the parameter limits, not the rank alone.

Standards & Codes

Notation, interchange and numerical standards that govern how this material is written down, stored and computed in production systems.

Applicable standards, conventions and reference implementations
Reference	Title	Relevance to this topic
`ISO 80000-2`	Quantities and units — Part 2: Mathematics	Fixes the notation for spans, mappings, composition and dimension used throughout, including the arrow form $T \colon U \to V$ and the composition symbol in $S \circ T$ .
`LAPACK / BLAS reference`	Linear Algebra PACKage reference implementation	Deciding whether an image set spans the codomain is a rank determination, delivered in production by the singular value decomposition (`xGESVD`) or a rank-revealing QR (`xGEQP3`).
`IEEE 754-2019`	IEEE Standard for Floating-Point Arithmetic	Governs the arithmetic in which the image coordinates and their rank are computed, and why a marginal spanning verdict must be reported with the singular value that supports it.
`ISO 24998`	Optics and photonics — Adaptive optics terminology	Adaptive optics systems are specified by the modal space their actuators can reach; the spanning test on this page is the algebraic form of the correctable-mode requirement.
`ISO/IEC 40314`	Mathematical Markup Language (MathML) Version 3.0	Encodes the span expressions, quantified statements and coordinate arrays on this page as semantic markup rather than images, keeping them searchable and accessible.

Material Selection

For a mathematical topic, "material" is the numeric representation: the scalar field, storage format and precision the computation is built from.

Representation and precision selection
Representation	Select when	Trade-off
Exact rational arithmetic	Design-stage parameterisations with exact coefficients, where the structural constraint on the range is the deliverable.	Gives exact constraint equations and an unambiguous spanning verdict, but coefficient growth limits the practical scale.
IEEE 754 binary64 with SVD rank test	Image sets computed from identified or measured models, the normal engineering case.	Robust, and the singular values grade how nearly spanning fails; costs several times a plain elimination.
IEEE 754 binary64 with pivoted QR	Large basis sets where a full decomposition is uneconomical but the redundant images must be named.	Cheaper and identifies which images are redundant, but the rank boundary is less sharply resolved.
Original-image basis from column reduction	Applications where each image corresponds to a physical parameter and the basis must consist of real parameter effects.	Maximum interpretability of which parameters matter, but the basis may be poorly conditioned for numerical projection.
Constraint (implicit) representation of the range	Runtime feasibility testing of targets against a fixed parameterisation.	Reduces a feasibility check to evaluating a few linear forms, but constructing a parameter setting from constraints requires a separate solve.
Finite field arithmetic	Parity-check and syndrome coverage questions, where the codomain is a space over a finite field.	Exact with bounded operand size; a set may span over the rationals yet fail to span over the field in use.

Manufacturing Notes

Implementation notes — how the result is actually produced by hand, by algorithm and by library, including cost and numerical behaviour.

Cost of the spanning test

Evaluating $T$ on $m$ basis vectors costs $m$ applications of the defining rule. Reducing the resulting $m \times t$ coordinate array costs about $2mt\min(m,t) - \tfrac{2}{3}\min(m,t)^3$ operations. Since surjectivity requires $m \ge t$ , the dominant term is $O(mt^2)$ , negligible at the dimensions typical of design-time checks.

Hand procedure

Place the coordinatised images as the rows of a matrix and row-reduce. The number of non-zero rows is the rank, and comparison against $\dim(V)$ gives the verdict directly. Rows that reduce to zero exhibit the redundant images, and the operations that produced them are the explicit dependence relations.

Building a preimage from the spanning coefficients

When the images span $V$ , solving for the coefficients $\beta_i$ that express a target as a combination of the images yields a preimage immediately: $\vec{u} = \beta_1 \vec{u}_1 + \cdots + \beta_m \vec{u}_m$ in terms of the original basis. When $m > t$ the coefficient system is underdetermined and a minimum-norm solution is usually the sensible choice.

Library routes

numpy.linalg.matrix_rank compared against

\dim(V)

answers the spanning question numerically; scipy.linalg.orth returns an orthonormal basis of the span. Symbolic systems accept the images directly and expose Matrix.rank() and Matrix.rref() over the rationals, which suits abstract domains where no coefficient matrix exists in advance.

Verification discipline

Two checks catch most errors. Confirm that every original image satisfies every derived constraint on the range, and confirm that the rank plus the number of independent constraints equals $\dim(V)$ . When the verdict is surjective, spot-check by choosing a random target, computing the spanning coefficients, building the preimage, and evaluating $T$ on it.

Failure Modes & Common Mistakes

Failure modes, root causes and prevention
Failure mode / mistake	Impact	Root cause	Prevention & detection
Reading the dimension inequality as sufficient	high	Concluding surjectivity from $\dim(U) \ge \dim(V)$ without computing the span of the images.	State the inequality as necessary only, and always follow a passing check with the spanning computation.
Treating the image of a basis as a basis of the range	high	Assuming the images inherit independence as well as spanning.	Only injective transformations preserve independence. Cull the image set by row reduction before quoting a basis or a rank.
Counting parameters instead of rank	medium	Reporting design freedom from the number of tuning parameters rather than the dimension of their image.	Compute the rank of the image set. Redundant parameters inflate the count without adding reachable directions.
Applying the test to a non-spanning set	medium	Using an independent but non-spanning subset of the domain in place of a basis.	The spanning half of the basis property is essential. A non-spanning set generates only part of the range and understates the rank.
Inconsistent coordinatisation of the images	medium	Reading one image against one basis ordering of the codomain and another against a different ordering.	Fix a single ordered basis of the codomain at the outset and apply it identically to every image.
Exact-zero rank test on floating-point images	high	Declaring that the images span the codomain because no computed pivot is exactly zero.	Use a tolerance scaled by the norm of the coordinate array, and report the smallest singular value with the verdict.
Assuming surjectivity of $S$ from surjectivity of $S \circ T$	medium	Reversing the composition result, which constrains only the combined reach.	From a surjective composition conclude that $S$ is surjective; nothing follows about $T$ , which may miss part of $V$ .
Discarding the constraint equations	low	Reporting only that the images fail to span, without eliminating coordinates to name the missing directions.	Derive the constraints from the reduced array; they cost nothing extra and are the actionable diagnosis.

FAQs

Why do the images of a spanning set automatically span the range?

Because linearity carries a linear combination of inputs to the same linear combination of their images. Every element of the range is $T$ of some input, that input is a combination of the spanning set, and pushing the combination through $T$ writes the output as a combination of the images with identical coefficients.

Do the images of a basis form a basis of the range?

They span the range but need not be independent. Independence survives only when $T$ is injective, so for a general transformation the image set must be culled by row reduction before it can be quoted as a basis. Its size after culling is the rank.

If the domain has larger dimension than the codomain, is the transformation surjective?

Not necessarily. The inequality $\dim(U) \ge \dim(V)$ is necessary for surjectivity but never sufficient. A transformation from a four-dimensional space into a three-dimensional one can easily have a two-dimensional range, as the worked example on this page shows.

How is this different from the injective case?

The two are mirror images. Injectivity preserves linear independence and forces $\dim(U) \le \dim(V)$ ; surjectivity is characterised by spanning and forces $\dim(U) \ge \dim(V)$ . Together, in the case of equal dimensions, either property alone implies the other and hence invertibility.

What does it mean when the images span only a proper subspace?

The transformation has a structural limitation: eliminating coordinates from the span produces linear equations that every achievable output satisfies. Those equations describe exactly what cannot be reached, and any vector violating them is a concrete unattainable target.

How do I actually build a preimage once I know the images span the codomain?

Express the target as a linear combination of the images; the coefficients obtained are the coordinates of a preimage in the basis you started with. When the domain dimension exceeds the codomain dimension the coefficient system is underdetermined, so choose the minimum-norm solution or fix the free parameters against a secondary objective.

Does surjectivity survive composition of transformations?

Yes. If both stages are surjective the composition is, and the argument works backwards through the chain: a target in the final space is hit by an intermediate vector, which is in turn hit by an input. The dimensions must be non-increasing along the chain, so a pipeline can only lose reach.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section SLT, subsections SSSLT, SLTD and CSLT. Licensed under the GNU Free Documentation License v1.2.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics. International Organization for Standardization.
Halmos, P. R. Finite-Dimensional Vector Spaces, 2nd edition. Springer, 1974.
Golub, G. H. and Van Loan, C. F. Matrix Computations, 4th edition. Johns Hopkins University Press, 2013.
Skogestad, S. and Postlethwaite, I. Multivariable Feedback Control: Analysis and Design, 2nd edition. Wiley, 2005.

AI Suggested Questions

Build a transformation from a 5-dimensional space to a 3-dimensional one that satisfies the dimension inequality but has a 2-dimensional range, and show the culled image set.
Show how the spanning coefficients of a target vector translate directly into a preimage, with a worked underdetermined example.
Why does an injective transformation preserve independence while a surjective one is characterised by spanning? Explain the duality precisely.
For a deformable mirror influence matrix, how do I determine which wavefront modes are uncorrectable?
Prove that if S∘T is surjective then S is surjective, and give a concrete T that is not surjective in such a composition.
If my image coordinate array has singular values 8, 2 and 1e-11, should I report the transformation as surjective?

Related Calculators

Spanning Set Tester

Check whether a set of vectors spans a target space, cull it to a basis, and report the rank with a selectable tolerance.

Surjectivity Tester

Apply a transformation to a chosen basis, test whether the images span the codomain, and report the constraints on the range.

Dimension Compatibility Checker

Compare domain and codomain dimensions to screen for impossible injections and surjections before any computation.

Overview

Definition

Spanning Set

Surjective Linear Transformation

Range of a Linear Transformation

Concepts

Images of a spanning set span the range

Why a basis is the right spanning set to use

Culling the image set to a basis

The basis characterisation of surjectivity

The dimension inequality and what it does not say

Range membership, preimages and composition

Testing surjectivity through a spanning set

Equations

Variable Definitions

Worked Numerical Example

Problem statement

State the parameterisation

Check the dimension precondition

Apply T to a basis of the domain

Coordinatise and row-reduce

Read the rank and the verdict

Cull the image set to a basis

Identify the structural constraint

Repair the parameterisation

Applications &amp; Industry Use

Achievable closed-loop dynamics from a tuning set

Workspace coverage from actuator directions

Reachable wavefront corrections

Coverage of the syndrome space

Independently controllable specifications

Expressive range of a blend-shape rig

Design Considerations

Screen by dimension, then compute

More parameters is not more authority

Spanning is preserved, independence is not

Report the missing directions, not just the failure

The backward proof is a construction

Rank near the boundary is fragile

Standards &amp; Codes

Material Selection

Manufacturing Notes

Cost of the spanning test

Hand procedure

Building a preimage from the spanning coefficients

Library routes

Verification discipline

Failure Modes &amp; Common Mistakes

FAQs

Related Engineering Topics

References

AI Suggested Questions

Related Calculators

Continue learning

Apply $T$ to a basis of the domain

Applications & Industry Use

Standards & Codes

Failure Modes & Common Mistakes