Surjective Linear Transformations

Q: How do I find a vector that is not attained?

Reduce the augmented matrix with symbolic constants. Each zero row in the coefficient part leaves a linear expression in the components of $\vec{v}$ that must vanish for consistency. Choose components making that expression non-zero, and the resulting vector provably has an empty preimage.

Q: Does surjectivity mean the preimage is unique?

No, and generally the opposite. Surjectivity asserts existence only. When the domain has larger dimension than the codomain, every preimage is an infinite family with $\dim(U) - \dim(V)$ free parameters, and choosing among them is a separate design decision.

Q: For a square matrix, are surjectivity and injectivity the same condition?

Yes. A square $A$ has full row rank exactly when it has full column rank, which happens exactly when it is nonsingular. So for a transformation between spaces of equal finite dimension, each of the two properties implies the other, and either one alone implies invertibility.

Q: My plant matrix has full row rank but the controller still cannot reach some setpoints. Why?

Full row rank is a statement about exact arithmetic and unbounded inputs. If the smallest singular value is tiny, reaching certain outputs demands inputs far beyond actuator limits, so the direction is mathematically reachable and practically not. Assess authority using the smallest singular value against the actuator saturation limits, not the rank alone.

Overview

A function raises two independent questions about the relationship between its inputs and its outputs. The first is whether anything is lost — that is injectivity. The second is whether anything is missed, and that is surjectivity. A linear transformation $T \colon U \to V$ is surjective when for every $\vec{v}$ in the codomain there exists at least one $\vec{u}$ in the domain with $T(\vec{u}) = \vec{v}$ . Nothing in $V$ is unreachable; the transformation covers its stated codomain completely.

The engineering content of the property is authority. If $T$ maps commands to plant responses, surjectivity says every response the model admits can actually be commanded — the actuator set has enough independent influence to steer the output anywhere in the output space. A non-surjective plant has directions in which it simply cannot be driven, and no controller, however clever, will move it there. Recognising that limitation belongs at design time, not at commissioning.

Proving and disproving surjectivity cost very different amounts of effort, mirroring the situation for injectivity but with the roles of existence and uniqueness swapped. To disprove it, exhibit one vector of the codomain whose defining system is inconsistent: a single leading one in the final column of a reduced augmented matrix settles the matter. To prove it, an argument must supply a preimage for every possible output, which means carrying an entirely symbolic right-hand side through the reduction and demonstrating that the system is consistent whatever values the symbols take.

For a transformation given by a matrix, the criterion collapses to a rank statement. $T(\vec{x}) = A\vec{x}$ from $\mathbb{C}^n$ to $\mathbb{C}^m$ is surjective exactly when $\operatorname{rank}(A) = m$ , that is when $A$ has full row rank and no row of its reduced form is entirely zero. This is the mirror image of the injectivity criterion, which asks for full column rank. Between them the two conditions describe every structural possibility, and a transformation satisfying both is invertible.

Definition

Surjective Linear Transformation

SLT

Let $T \colon U \to V$ be a linear transformation. Then $T$ is surjective if for every vector $\vec{v} \in V$ there exists at least one vector $\vec{u} \in U$ with $T(\vec{u}) = \vec{v}$ . The vector $\vec{u}$ need not be unique; only its existence is required.

The terms onto and surjection are synonyms. Surjectivity is a statement about the codomain as declared, so changing the declared codomain can change the answer without changing any formula.

Linear Transformation

A function $T \colon U \to V$ between vector spaces satisfying $T(\vec{x} + \vec{y}) = T(\vec{x}) + T(\vec{y})$ and $T(\alpha \vec{x}) = \alpha T(\vec{x})$ for all vectors $\vec{x}, \vec{y} \in U$ and all scalars $\alpha$ . Linearity is what makes the existence question a linear system rather than an unstructured search.

Restated because the reduction of surjectivity to a consistency question uses linearity to expand $T(\vec{u})$ into a matrix-vector product or a coefficient-by-coefficient identity.

Unattainable Vector

A vector $\vec{v} \in V$ for which the equation $T(\vec{u}) = \vec{v}$ has no solution, equivalently whose preimage $T^{-1}(\vec{v})$ is empty. Exhibiting one unattainable vector is a complete disproof of surjectivity; no further argument is needed.

Concepts

Surjectivity is a universal existence statement

Written out, the condition reads: for all $\vec{v} \in V$ , there exists $\vec{u} \in U$ such that $T(\vec{u}) = \vec{v}$ . The outer quantifier is universal and the inner one existential, which fixes how each direction must be argued. A proof must handle an arbitrary, unspecified $\vec{v}$ and produce a formula for $\vec{u}$ in terms of it. A disproof needs only one specific $\vec{v}$ together with a demonstration that no $\vec{u}$ works. As with injectivity, the cheap direction is the negative one, so an engineer suspicious of a model's authority should look for an unreachable output first.

Reduction to consistency with a symbolic right-hand side

For $T \colon \mathbb{C}^n \to \mathbb{C}^m$ with $T(\vec{x}) = A\vec{x}$ , the equation $T(\vec{u}) = \vec{v}$ is the linear system $A\vec{u} = \vec{v}$ . Surjectivity therefore says: the system is consistent for every right-hand side. Row-reduce the augmented matrix $[\, A \mid \vec{v} \,]$ while carrying the symbols $v_1, \dots, v_m$ through untouched. Every zero row produced in the coefficient part leaves behind a linear expression in the $v_i$ that must vanish for consistency. If no zero row appears, no condition is imposed and every $\vec{v}$ is attainable; if a zero row appears, the corresponding expression is a constraint that some $\vec{v}$ will violate.

The matrix criterion: full row rank

The condition ‘no zero rows in the reduced form of $A$ ’ is exactly $\operatorname{rank}(A) = m$ , full row rank. Every row of $A$ contributes a pivot, so no combination of equations degenerates into a constraint on the constants alone. Two consequences follow immediately. A square $A$ gives a surjective $T$ precisely when $A$ is nonsingular, which is also the injectivity condition — for square matrices the two properties coincide. And if $n < m$ the rank cannot reach $m$ , so a transformation from a smaller space into a larger one is never surjective.

Constructing the preimage explicitly

When a proof of surjectivity is required, the cleanest form is constructive: give a formula $\vec{u}(\vec{v})$ and verify $T(\vec{u}(\vec{v})) = \vec{v}$ identically. Two routes produce such a formula. If $A$ is square and nonsingular, take $\vec{u} = A^{-1}\vec{v}$ and the verification is one line. If $A$ is wide, carry the symbolic reduction to completion, set every free variable to zero, and read the pivot variables as linear expressions in the $v_i$ . The resulting formula is one valid choice among infinitely many; surjectivity requires existence only, so the simplest choice is the right one.

Manufacturing an unattainable vector

A disproof does not require guesswork. Carry out the symbolic reduction; each zero row in the coefficient part yields a consistency condition of the form $c_1 v_1 + \cdots + c_m v_m = 0$ . Any $\vec{v}$ violating that equation is unattainable, and constructing one is a matter of choosing values that make the left-hand side non-zero. This is far more reliable than guessing a vector and testing it, and it also explains why the vector is unreachable: it violates a linear law that every output obeys.

Surjectivity is relative to the declared codomain

Unlike injectivity, surjectivity is not a property of the rule alone; it depends on what codomain is declared. The same formula $T(\vec{x}) = A\vec{x}$ is not surjective onto $\mathbb{C}^m$ when the rank is deficient, yet is trivially surjective onto its own range. Restricting the codomain to the range converts any transformation into a surjection without changing a single output value. This is worth stating explicitly in engineering documentation: a claim of full authority is meaningless until the space over which authority is claimed is pinned down.

Deciding whether a linear transformation is surjective

Compare dimensions firstIf

\dim(U) < \dim(V)

T

cannot be surjective. Stop — the formula is irrelevant.

Set $T(\vec{u}) = \vec{v}$ with $\vec{v}$ symbolicDo not choose numeric constants. The whole point is to handle every output at once.

Form the augmented systemEquate components or coefficients to obtain

[\, A \mid \vec{v} \,]

with symbolic entries in the final column.

Row-reduce, carrying the symbolsApply operations to the constant column exactly as to the coefficient columns.

Any zero row in the coefficient part?If none, the system is consistent for every

\vec{v}

and

T

is surjective. Read a preimage formula from the pivot rows.

Otherwise build an unattainable vectorEach zero row gives a linear condition on the

v_i

; choose values that violate it and the corresponding

\vec{v}

has empty preimage.

Equations

Defining condition for surjectivity

EQ-SLT-01

\forall \vec{v} \in V \exists \vec{u} \in U T (\vec{u}) = \vec{v}

Every element of the codomain is attained by at least one input. Uniqueness of the input is not required and is generally false.

Negation used for a disproof

EQ-SLT-02

\exists \vec{v} \in V T^{- 1} (\vec{v}) = \emptyset Longrightarrow T non-surjective

One vector with an empty preimage is a complete disproof; no examination of the rest of the codomain is required.

Surjectivity as universal consistency

EQ-SLT-03

T (\vec{x}) = A \vec{x} surjective \Leftrightarrow A \vec{u} = \vec{v} consistent \forall \vec{v} \in ℂ^{m}

The property is exactly the statement that no choice of constants makes the system inconsistent.

Matrix transformation criterion

EQ-SLT-04

T (\vec{x}) = A \vec{x}, A \in M_{m n} Longrightarrow (T surjective \Leftrightarrow rank (A) = m)

Full row rank: every row of the reduced form carries a leading one, so no zero row can generate a constraint on the constants.

Consistency condition from a zero row

EQ-SLT-05

c_{1} v_{1} + c_{2} v_{2} + \dots + c_{m} v_{m} = 0

The residue left in the augmented column by a zero row of the coefficient part. Any $\vec{v}$ violating this equation is unattainable.

Preimage from a square nonsingular matrix

EQ-SLT-06

A nonsingular Longrightarrow \vec{u} = A^{- 1} \vec{v} satisfies T (\vec{u}) = \vec{v}

For square nonsingular coefficient matrices the constructive proof is immediate, and the preimage is unique as well as existent.

Dimension obstruction

EQ-SLT-07

\dim (U) < \dim (V) Longrightarrow T non-surjective

A transformation into a strictly larger space cannot cover it. The defining formula plays no part in this conclusion.

Variable Definitions

Symbols used on this page
Symbol	Name	Meaning	Domain / type
$T$	Linear transformation	The map under test, written $T \colon U \to V$	function from U to V
$U$	Domain	The vector space of available inputs or commands	vector space over C
$V$	Codomain	The declared output space that surjectivity requires $T$ to cover	vector space over C
$\vec{v}$	Target output	An arbitrary element of the codomain, kept symbolic during the test	element of V
$\vec{u}$	Preimage vector	An input satisfying $T(\vec{u}) = \vec{v}$ , constructed as a formula in $\vec{v}$	element of U
$A$	Coefficient matrix	The matrix of $T$ when $U = \mathbb{C}^n$ and $V = \mathbb{C}^m$	m x n matrix
$m$	Codomain dimension	Row count of $A$ ; the rank must reach this value for surjectivity	positive integer
$n$	Domain dimension	Column count of $A$ ; must be at least $m$ for surjectivity to be possible	positive integer
$T^{- 1} (\vec{v})$	Preimage of a target	The set of inputs producing $\vec{v}$ ; non-empty for every $\vec{v}$ exactly when $T$ is surjective	subset of U

Worked Numerical Example

Problem statement

A blending plant has four independently metered feed streams and three controlled product properties. The linearised response of the properties to the feed rates is the matrix $C$ below, giving $T \colon \mathbb{C}^4 \to \mathbb{C}^3$ with $T(\vec{u}) = C\vec{u}$ . Determine whether every combination of the three product properties can be commanded, and if so give a formula for the required feed rates.

State the plant model
Each row of $C$ is one product property expressed in terms of the four feed rates $u_1, \dots, u_4$ :
$C = [\begin{matrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 2 \\ 1 & 1 & - 1 & 0 \end{matrix}], T (\vec{u}) = C \vec{u}$
Set the target symbolic
Let $\vec{v} = (v_1, v_2, v_3)$ be an arbitrary commanded property vector. Surjectivity is the claim that $C\vec{u} = \vec{v}$ is consistent for every such $\vec{v}$ . Form the augmented matrix with symbolic constants:
$[\begin{matrix} 1 & 2 & 0 & 1 & v_{1} \\ 0 & 1 & 1 & 2 & v_{2} \\ 1 & 1 & - 1 & 0 & v_{3} \end{matrix}]$
Eliminate in the first column
Entry $(1,1)$ is already a leading one. Apply $R_3 \to R_3 - R_1$ , treating the symbolic constant exactly like any other entry:
$[\begin{matrix} 1 & 2 & 0 & 1 & v_{1} \\ 0 & 1 & 1 & 2 & v_{2} \\ 0 & - 1 & - 1 & - 1 & v_{3} - v_{1} \end{matrix}]$
Eliminate in the second column
Apply $R_3 \to R_3 + R_2$ and $R_1 \to R_1 - 2R_2$ . Row 3 does not collapse: the entry in column 4 survives, and a third leading one appears there.
$[\begin{matrix} 1 & 0 & - 2 & - 3 & v_{1} - 2 v_{2} \\ 0 & 1 & 1 & 2 & v_{2} \\ 0 & 0 & 0 & 1 & v_{2} + v_{3} - v_{1} \end{matrix}]$
Finish the reduction
Clear column 4 above the third leading one with $R_1 \to R_1 + 3R_3$ and $R_2 \to R_2 - 2R_3$ :
$[\begin{matrix} 1 & 0 & - 2 & 0 & - 2 v_{1} + v_{2} + 3 v_{3} \\ 0 & 1 & 1 & 0 & 2 v_{1} - v_{2} - 2 v_{3} \\ 0 & 0 & 0 & 1 & - v_{1} + v_{2} + v_{3} \end{matrix}]$
Read the verdict
There are three leading ones for three rows, so $\operatorname{rank}(C) = 3 = m$ : full row rank. No zero row appears in the coefficient part, so no consistency condition is imposed on $v_1, v_2, v_3$ and the system is solvable for every commanded property vector. By definition, $T$ is surjective. Column 3 carries no leading one, so $u_3$ is a free variable and the preimage is a one-parameter family — existence without uniqueness, exactly what surjectivity asserts.
Extract an explicit preimage formula
Set the free variable $u_3 = 0$ and read the three pivot rows. This is one valid choice; any value of $u_3$ gives another.
$\vec{u} = [\begin{matrix} - 2 v_{1} + v_{2} + 3 v_{3} \\ 2 v_{1} - v_{2} - 2 v_{3} \\ 0 \\ - v_{1} + v_{2} + v_{3} \end{matrix}]$
Verify the formula identically
Multiply out $C\vec{u}$ symbolically. Row 1: $(-2v_1 + v_2 + 3v_3) + 2(2v_1 - v_2 - 2v_3) + (-v_1 + v_2 + v_3) = v_1$ . Row 2: $(2v_1 - v_2 - 2v_3) + 2(-v_1 + v_2 + v_3) = v_2$ . Row 3: $(-2v_1 + v_2 + 3v_3) + (2v_1 - v_2 - 2v_3) = v_3$ . A numeric spot check with $\vec{v} = (3, -2, 5)$ gives $\vec{u} = (7, -2, 0, 0)$ , and $C\vec{u} = (3, -2, 5)$ as required.

Result

The plant has full actuation authority over its three controlled properties: any commanded property vector is reachable, and the required feed rates are given in closed form by a linear expression in the setpoints. The free variable $u_3$ is one degree of blending freedom that does not affect the product properties at all, and it is available for a secondary objective such as minimising the cost of the most expensive stream. Had a zero row appeared in the coefficient part, the residual expression in that row would have named the linear law that every achievable property vector must obey.

Applications & Industry Use

Control systems

Actuation authority and reachability

The map from actuator commands to output rates must be surjective onto the space of demanded output rates, or some demands are physically unreachable. Full row rank of the input matrix is the design condition, and a rank-deficient row set identifies precisely which combination of outputs cannot be driven independently.

Process engineering

Blend feasibility over a specification space

A blending unit with several feed streams must reach any point of a specification envelope. Surjectivity of the linearised response guarantees a feasible recipe exists for every target, and the free variables in the preimage give the slack that a cost-minimising optimiser can exploit.

Computer graphics

Colour gamut coverage

Rendering a colour from primaries is a linear map from primary intensities to a tristimulus vector. Surjectivity onto the target colour space is the algebraic form of full gamut coverage; when it fails, the unattainable directions are exactly the out-of-gamut colours that must be handled by a mapping policy.

Structural engineering

Load paths and applied force distributions

The equilibrium relation maps internal member forces to resultant nodal loads. Surjectivity means the structure can carry any applied load pattern, while a deficiency identifies load directions with no equilibrating internal state — a mechanism, and a design fault.

Communications

Syndrome coverage in error detection

A parity-check matrix maps received words to syndromes. Surjectivity onto the syndrome space is what guarantees every syndrome pattern corresponds to a real error class, so no decoder table entry is wasted and the check matrix carries its full designed rank.

Power systems

Dispatch feasibility across a network

The linear map from generator injections to nodal power flows must cover the required set of flow patterns for a dispatch to be realisable. Rank deficiency in the network incidence relations reveals flow patterns that no combination of injections can produce.

Design Considerations

Declare the codomain before claiming surjectivity

The property depends entirely on what space is being covered. The same transformation is non-surjective onto $\mathbb{C}^m$ and trivially surjective onto its range. Engineering statements about authority or coverage must name the target space explicitly, otherwise the claim cannot be checked or falsified.

Screen by dimension before computing

If $\dim(U) < \dim(V)$ the transformation cannot be surjective, whatever its coefficients. That check is free. The converse is not available: $\dim(U) \ge \dim(V)$ permits surjectivity but does not deliver it, and the rank must still be computed.

Keep the right-hand side symbolic

Testing with a handful of numeric targets can only reveal failure, never establish coverage. The reduction must be carried out with symbolic constants so that the consistency conditions, if any, appear explicitly as linear expressions in the components of $\vec{v}$ .

Report the consistency conditions, not just the verdict

When surjectivity fails, the residual expressions left by zero rows are the useful output: they are the linear laws every achievable output obeys, and they define the range as a solution set. Reporting only that the map is non-surjective discards the description of what is achievable.

Free variables in the preimage are exploitable slack

A surjective transformation from a larger domain has free variables in every preimage. Those parameters cost nothing to the primary objective and can be spent on secondary criteria — minimum energy, minimum cost, actuator wear. Setting them all to zero, as a bare reduction does, is a choice and should be a conscious one.

Near-deficiency is worse than the rank suggests

A matrix with full row rank but a very small smallest singular value is technically surjective while requiring enormous inputs to reach some outputs. The practical statement of authority is not the rank but the smallest singular value scaled against actuator limits. Report both.

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	Defines the quantifier symbols and mapping notation used in the formal statement of surjectivity, and fixes the typography for the transformation arrow and for set membership.
`IEC 61131-3`	Programmable controllers — Part 3: Programming languages	Control logic implemented under this standard depends on the assumption that commanded setpoints are reachable; a non-surjective plant model invalidates that assumption and produces persistent setpoint offsets.
`LAPACK / BLAS reference`	Linear Algebra PACKage reference implementation	Full row rank is determined in production by `xGESVD` or by an LQ-style factorisation; the minimum-norm preimage of a wide system is produced by `xGELSD`.
`IEEE 754-2019`	IEEE Standard for Floating-Point Arithmetic	Governs the arithmetic of the symbolic-constant reduction when it is carried out numerically, and explains why a near-zero pivot must be judged against a tolerance rather than against zero.
`CIE 15`	Colorimetry	Tristimulus reproduction from a set of primaries is a linear map; the colorimetric notion of gamut coverage is the surjectivity of that map onto the target colour space.

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 with symbolic constants	Structural design questions where the consistency conditions themselves are the deliverable.	Yields exact linear laws describing the range, but symbolic reduction is expensive and the expressions grow quickly with the row count.
IEEE 754 binary64 with SVD rank test	Numerical plant models identified from data, the normal engineering case.	Robust and graded by the singular values, but full row rank must be judged against a declared tolerance rather than exactly.
IEEE 754 binary64 with LQ or pivoted QR	Wide matrices where the question is only whether the rows are independent.	Considerably cheaper than an SVD and identifies dependent rows, but the rank estimate is less reliable when singular values decay smoothly.
Minimum-norm least squares solver	A preimage is needed numerically for a wide system and the free variables should be resolved sensibly rather than zeroed.	Returns the smallest-norm preimage, which is usually the physically sensible one, at the cost of a decomposition rather than a plain elimination.
Fixed-point arithmetic	Embedded controllers computing preimages online from a fixed plant matrix.	Deterministic timing and no rounding surprises within range, but scaling must be designed so intermediate products cannot overflow.
Finite field arithmetic	Parity-check and syndrome computations, where the codomain is a space over a finite field.	Exact with bounded operand size; row rank over the field can differ from row rank of the same integer matrix over the rationals.

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 decision

Deciding surjectivity of $T(\vec{x}) = A\vec{x}$ for $A$ of size $m \times n$ with $n \ge m$ requires one elimination, about $2m^2 n - \tfrac{2}{3}m^3$ floating-point operations. Carrying a symbolic right-hand side multiplies the bookkeeping but not the asymptotic cost. An SVD is roughly an order of magnitude more expensive and returns the margin by which full row rank is achieved.

Hand procedure

Augment with a column of symbols $v_1, \dots, v_m$ , not with numbers, and apply every row operation to that column exactly as to the others. Work to fully reduced form so the pivot variables are read off without back-substitution. Then set every free variable to zero to obtain the simplest preimage formula.

Verifying a claimed preimage formula

Substitute the formula into the original transformation and expand symbolically; each output component must simplify to the corresponding $v_i$ identically, with all other symbols cancelling. A numeric spot check with one arbitrary target is a useful secondary test but does not replace the symbolic verification.

Library routes

numpy.linalg.matrix_rank(A) compared against the row count answers the question numerically. numpy.linalg.lstsq and scipy.linalg.lstsq return the minimum-norm preimage for a wide system, which is preferable to the zero-free-variable choice for physical actuation. Symbolic systems handle the fully symbolic right-hand side directly through linsolve and report the consistency conditions as part of the answer.

Surjectivity under composition

If $T \colon U \to V$ and $S \colon V \to W$ are both surjective then $S \circ T$ is surjective: given $\vec{w} \in W$ , surjectivity of $S$ supplies $\vec{v}$ with $S(\vec{v}) = \vec{w}$ , and surjectivity of $T$ then supplies $\vec{u}$ with $T(\vec{u}) = \vec{v}$ . Chaining stages therefore preserves authority, and a single non-surjective stage caps the coverage of the whole chain.

Failure Modes & Common Mistakes

Failure modes, root causes and prevention
Failure mode / mistake	Impact	Root cause	Prevention & detection
Testing with numeric targets only	high	Solving $T(\vec{u}) = \vec{v}$ for a few chosen $\vec{v}$ , succeeding, and declaring the map onto.	Carry a symbolic right-hand side through the reduction. Numeric targets can disprove surjectivity but can never establish it.
Confusing surjective with injective	high	Treating both as general statements that the transformation is well behaved.	Surjectivity is full row rank and concerns existence; injectivity is full column rank and concerns uniqueness. For non-square shapes at most one can hold.
Claiming surjectivity without naming the codomain	medium	Asserting coverage while leaving the target space implicit.	State the codomain explicitly. Every transformation is surjective onto its own range, so an unqualified claim carries no information.
Assuming more inputs guarantees coverage	medium	Believing that $n > m$ , more actuators than outputs, is sufficient.	Only the row rank matters. Redundant actuators that act along existing directions add columns without adding rank.
Dropping the symbolic column during reduction	medium	Reducing the coefficient matrix alone and inferring consistency from its shape.	The consistency conditions live in the augmented column. Reducing $A$ alone answers a rank question but discards the explicit description of the range.
Exact-zero pivot test on floating-point data	high	Accepting a pivot of magnitude $10^{-16}$ as evidence of full row rank.	Apply a tolerance scaled by the matrix norm, or decide the rank from singular values and report the smallest one.
Reading surjectivity as invertibility	medium	Writing $T^{-1}$ after establishing only that every output is attained.	Preimages of a surjective non-injective map are infinite sets, not single vectors. An inverse function requires injectivity as well.
Zeroing free variables without thinking	low	Accepting the reduction's default preimage as the physical command.	Choose the free variables against a secondary objective, or take the minimum-norm solution, rather than an arbitrary corner of the solution set.

FAQs

What is the difference between surjective and injective?

Injectivity constrains inputs: no two distinct inputs share an output, so nothing is lost. Surjectivity constrains outputs: nothing in the codomain is missed. In matrix terms injectivity is full column rank and surjectivity is full row rank, so for a non-square matrix at most one of the two can hold.

Why do I have to keep the right-hand side symbolic?

Because the claim is about every possible target simultaneously. A numeric target that happens to be solvable says nothing about the rest of the codomain. Carrying symbols through the reduction produces, in one pass, either a proof that no consistency condition exists or the explicit conditions that some targets will violate.

Can a transformation from a smaller space to a larger one be surjective?

No. If $\dim(U) < \dim(V)$ then the range has dimension at most $\dim(U)$ , which is strictly less than $\dim(V)$ , so the range is a proper subspace and some vectors are unattainable. The obstruction is purely dimensional and independent of the coefficients.

How do I find a vector that is not attained?

Reduce the augmented matrix with symbolic constants. Each zero row in the coefficient part leaves a linear expression in the components of $\vec{v}$ that must vanish for consistency. Choose components making that expression non-zero, and the resulting vector provably has an empty preimage.

Does surjectivity mean the preimage is unique?

No, and generally the opposite. Surjectivity asserts existence only. When the domain has larger dimension than the codomain, every preimage is an infinite family with $\dim(U) - \dim(V)$ free parameters, and choosing among them is a separate design decision.

For a square matrix, are surjectivity and injectivity the same condition?

Yes. A square $A$ has full row rank exactly when it has full column rank, which happens exactly when it is nonsingular. So for a transformation between spaces of equal finite dimension, each of the two properties implies the other, and either one alone implies invertibility.

My plant matrix has full row rank but the controller still cannot reach some setpoints. Why?

Full row rank is a statement about exact arithmetic and unbounded inputs. If the smallest singular value is tiny, reaching certain outputs demands inputs far beyond actuator limits, so the direction is mathematically reachable and practically not. Assess authority using the smallest singular value against the actuator saturation limits, not the rank alone.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section SLT, subsection ESLT. Licensed under the GNU Free Documentation License v1.2.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics. International Organization for Standardization.
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.
Anderson, E. et al. LAPACK Users' Guide, 3rd edition. Society for Industrial and Applied Mathematics, 1999.

AI Suggested Questions

Take a 3x5 matrix of my choosing and carry out the symbolic-constant reduction, showing the consistency conditions explicitly.
Construct a 4x4 transformation that is injective but not surjective onto a larger declared codomain, and explain the bookkeeping.
How do I convert the residual expressions from zero rows into a set of linear equations describing the range?
Why is the minimum-norm preimage usually preferable to the zero-free-variable preimage in an actuation problem?
If a plant matrix has full row rank with singular values 12, 3 and 0.002, what does that imply about reachable setpoints under actuator saturation?
Show that the composition of two surjective linear transformations is surjective, and give an example where the composition is surjective but the first stage is not.

Related Calculators

Surjectivity Tester

Enter a matrix and get a full-row-rank verdict, the consistency conditions from any zero rows, and a sample unattainable vector.

Symbolic Right-Hand Side Solver

Reduce an augmented matrix with symbolic constants and return the preimage formula plus any conditions the constants must satisfy.

Matrix Rank & Nullity Calculator

Report rank, nullity and the pivot column set with a selectable numerical tolerance.