Injective Linear Transformations | KEVOS® Mathematics

Overview

Every function raises the question of whether it loses information. If two different inputs are sent to the same output, the output cannot be traced back to a unique cause, and any attempt to reverse the map is ambiguous. A function that never does this is called injective, or equivalently one-to-one. For a linear transformation $T \colon U \to V$ the condition reads: whenever $T(\vec{x}) = T(\vec{y})$ , it must follow that $\vec{x} = \vec{y}$ .

Stated that way, injectivity looks like an impossible thing to verify. It quantifies over every pair of vectors in an infinite domain. Linearity is what rescues the situation. Because $T(\vec{x}) - T(\vec{y}) = T(\vec{x} - \vec{y})$ , an equality of outputs is exactly the statement that the difference of the inputs is sent to the zero vector. The whole infinite family of pairwise comparisons therefore collapses to a single question about one vector: which inputs does $T$ annihilate? That question is a homogeneous system, and homogeneous systems are settled by one row reduction.

The engineering reading is direct. A linear transformation models a measurement chain, an actuator mixing network, a sensor array or a coordinate change. Injectivity is the property that the model is identifiable: an observed output determines the state that produced it, at most one state can be responsible, and the inverse problem — when it is solvable at all — has a unique answer. A non-injective model has a blind direction, a family of inputs that the instrument cannot tell apart.

Injectivity is one of two structural properties that a linear transformation may possess; the companion property is surjectivity, which asks whether every element of the codomain is actually produced. The two are independent of each other, and a transformation possessing both is invertible. Injectivity is the one allied with linear independence and with dimension counting from below: an injective map cannot compress, so the domain can never be larger than the codomain.

Definition

Injective Linear Transformation

ILT

Let $T \colon U \to V$ be a linear transformation between vector spaces $U$ and $V$ . Then $T$ is injective if, for every pair of vectors $\vec{x}, \vec{y} \in U$ , the equality $T(\vec{x}) = T(\vec{y})$ forces $\vec{x} = \vec{y}$ . Equivalently, distinct inputs always yield distinct outputs.

The terms one-to-one and injection are synonyms in common use. Injectivity is a property of the function on its whole domain, not of any particular input.

Linear Transformation

A function $T \colon U \to V$ between vector spaces that respects both operations: $T(\vec{x} + \vec{y}) = T(\vec{x}) + T(\vec{y})$ for all $\vec{x}, \vec{y} \in U$ , and $T(\alpha \vec{x}) = \alpha T(\vec{x})$ for all scalars $\alpha$ and all $\vec{x} \in U$ . Every statement on this page depends on these two rules and on their immediate consequence $T(\vec{0}) = \vec{0}$ .

Restated here because the argument that reduces injectivity to a homogeneous system uses linearity twice and would fail for an arbitrary function.

Witness Pair for Non-Injectivity

A pair of vectors $\vec{x} \ne \vec{y}$ in the domain with $T(\vec{x}) = T(\vec{y})$ . A single such pair is a complete disproof of injectivity; no further argument is required, and no examination of the rest of the domain is needed.

Concepts

The asymmetry between proving and disproving

Injectivity is a universally quantified statement, so the two directions cost very different amounts of work. To disprove it, exhibit one witness pair: two distinct inputs with identical outputs. That is a finite, checkable object. To prove it, an argument must cover every pair in the domain, which cannot be done by examples. The standard proof therefore starts from the assumed equality $T(\vec{x}) = T(\vec{y})$ with $\vec{x}$ and $\vec{y}$ arbitrary, and derives $\vec{x} = \vec{y}$ by algebra alone. This asymmetry is worth internalising: an engineer testing a model for identifiability should look first for a counterexample, because finding one is cheap and finishes the question.

Linearity converts the definition into a homogeneous system

Assume $T(\vec{x}) = T(\vec{y})$ . Subtract, and use linearity in the form $T(\vec{x}) - T(\vec{y}) = T(\vec{x} - \vec{y})$ . The hypothesis becomes $T(\vec{x} - \vec{y}) = \vec{0}$ . Write $\vec{z} = \vec{x} - \vec{y}$ : the conclusion $\vec{x} = \vec{y}$ is precisely the statement $\vec{z} = \vec{0}$ . So $T$ is injective exactly when the only vector it sends to the zero vector is the zero vector itself. When $U = \mathbb{C}^n$ , $V = \mathbb{C}^m$ and $T(\vec{x}) = A\vec{x}$ , that condition is the requirement that the homogeneous system $A\vec{z} = \vec{0}$ has only the trivial solution — a single row reduction of $A$ settles it.

The matrix criterion: full column rank

For $T(\vec{x}) = A\vec{x}$ with $A$ of size $m \times n$ , reduce $A$ and count pivot columns. If every one of the $n$ columns is a pivot column, there are no free variables, the only solution of $A\vec{z} = \vec{0}$ is trivial, and $T$ is injective. If any column is free, each free variable generates a non-zero $\vec{z}$ that $T$ annihilates, and the map is not injective. Two corollaries follow at once. A square $A$ gives an injective $T$ precisely when $A$ is nonsingular. And if $n > m$ the rank cannot reach $n$ , so a transformation from a larger space to a smaller one is never injective.

Manufacturing a witness pair on demand

When a transformation fails to be injective, the failure is never isolated — it is systematic. Let $\vec{z} \ne \vec{0}$ satisfy $T(\vec{z}) = \vec{0}$ . Then for any $\vec{x}$ in the domain, $T(\vec{x} + \vec{z}) = T(\vec{x}) + T(\vec{z}) = T(\vec{x})$ , so $\vec{x}$ and $\vec{x} + \vec{z}$ form a witness pair. A single annihilated direction therefore produces an entire parallel family of collisions, one for every point of the domain. This is why non-injectivity of a physical model is not a rare edge case but a permanent, everywhere-present ambiguity.

Abstract domains behave identically

Nothing in the argument requires the domain to be a space of column vectors. For a transformation $T \colon P_3 \to M_{22}$ from cubic polynomials to $2 \times 2$ matrices, the same subtraction produces $T(p - q) = O$ , and equating coefficients turns that single equation into a homogeneous linear system in the differences of the coefficients. If the resulting coefficient matrix is nonsingular, only the trivial solution survives and the two polynomials must be identical. The lesson generalises: injectivity of a transformation between any pair of finite-dimensional spaces reduces to a rank computation once bases are fixed.

Why the property is called identifiability in practice

Interpret $T$ as an instrument: $\vec{x}$ is a physical state and $T(\vec{x})$ the recorded data. Injectivity says the data determine the state uniquely, so estimating $\vec{x}$ from $T(\vec{x})$ is a well-posed question. Failure of injectivity is failure of identifiability, and no amount of noise-free data, extra averaging or better estimation algorithm can recover the lost information — the ambiguity is in the map, not the measurement. The only remedies are structural: add an independent sensing channel, or add prior constraints that exclude all but one member of each ambiguous family.

Deciding whether a linear transformation is injective

Compare dimensions firstIf

\dim(U) > \dim(V)

T

cannot be injective. Stop — no computation with the formula is needed.

Assume $T(\vec{x}) = T(\vec{y})$ Subtract and apply linearity to reach

T(\vec{x} - \vec{y}) = \vec{0}

Convert to a homogeneous systemEquate components or coefficients to obtain

A\vec{z} = \vec{0}

in the differences

\vec{z} = \vec{x} - \vec{y}

Row-reduce the coefficient matrixCount pivot columns. Full column rank means no free variables.

Trivial solution only?If yes,

\vec{z} = \vec{0}

, hence

\vec{x} = \vec{y}

and

T

is injective.

Otherwise build a witnessTake any non-trivial

\vec{z}

, pick any

\vec{x}

, and report the pair

\vec{x}

and

\vec{x} + \vec{z}

Equations

Defining condition for injectivity

EQ-ILT-01

T (\vec{x}) = T (\vec{y}) \Rightarrow \vec{x} = \vec{y} \forall \vec{x}, \vec{y} \in U

Equal outputs can only arise from equal inputs. The quantifier ranges over the entire domain, which is why examples cannot establish the property.

Contrapositive form

EQ-ILT-02

\vec{x} \neq \vec{y} \Rightarrow T (\vec{x}) \neq T (\vec{y})

Distinct inputs give distinct outputs. Logically identical to the definition and often the more intuitive reading.

Reduction to a single annihilated vector

EQ-ILT-03

T (\vec{x}) = T (\vec{y}) \Leftrightarrow T (\vec{x} - \vec{y}) = \vec{0}

The consequence of linearity that turns a two-vector condition into a one-vector condition. Everything computational on this page follows from it.

Trivial-annihilator criterion

EQ-ILT-04

Trivial-annihilator criterion

The practical test. The set of vectors satisfying $T(\vec{z}) = \vec{0}$ is the kernel of $T$ , so this states that injectivity is equivalent to a trivial kernel.

Matrix transformation criterion

EQ-ILT-05

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

Injectivity of a matrix transformation is full column rank: every column of $A$ is a pivot column, so no free variables exist.

Systematic construction of a witness pair

EQ-ILT-06

T (\vec{z}) = \vec{0}, \vec{z} \neq \vec{0} Longrightarrow T (\vec{x} + \vec{z}) = T (\vec{x}) \forall \vec{x} \in U

One annihilated direction produces a collision at every point of the domain, not merely at one.

Dimension obstruction

EQ-ILT-07

\dim (U) > \dim (V) Longrightarrow T non-injective

A transformation into a strictly smaller space must collide. The formula defining $T$ 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 admissible inputs	vector space over C
$V$	Codomain	The vector space in which outputs are recorded	vector space over C
$\vec{x}$	First input vector	An arbitrary element of the domain	element of U
$\vec{y}$	Second input vector	A second arbitrary element of the domain, compared with $\vec{x}$	element of U
$\vec{z}$	Difference vector	$\vec{z} = \vec{x} - \vec{y}$ ; the vector whose annihilation decides injectivity	element of U
$A$	Coefficient matrix	The matrix representing $T$ when $U = \mathbb{C}^n$ and $V = \mathbb{C}^m$	m x n matrix
$n$	Domain dimension	Number of input coordinates, equal to the column count of $A$	positive integer
$m$	Codomain dimension	Number of output coordinates, equal to the row count of $A$	positive integer

Worked Numerical Example

Problem statement

An actuator bank drives three independent commands into a structure, and four strain gauges record the response. The response model is linear, $T \colon \mathbb{C}^3 \to \mathbb{C}^4$ with $T(\vec{x}) = A\vec{x}$ . Determine whether the gauge readings identify the command vector uniquely, and if not, exhibit two command vectors the instrument cannot distinguish.

State the model
The four gauge readings are linear in the three commands $x_1, x_2, x_3$ . Collect the coefficients into a $4 \times 3$ matrix:
$A = [\begin{matrix} 1 & 2 & - 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \\ 3 & 5 & 0 \end{matrix}], T (\vec{x}) = A \vec{x}$
Set up the injectivity test
Suppose two command vectors give identical gauge readings, $T(\vec{x}) = T(\vec{y})$ . Subtracting and applying linearity gives $T(\vec{x} - \vec{y}) = \vec{0}$ , that is $A\vec{z} = \vec{0}$ with $\vec{z} = \vec{x} - \vec{y}$ . Injectivity holds exactly when this homogeneous system has only the trivial solution.
Row-reduce the coefficient matrix
Apply $R_2 \to R_2 - 2R_1$ , $R_3 \to R_3 - R_1$ , $R_4 \to R_4 - 3R_1$ , then scale and clear the second column. The reduced row-echelon form is:
$[\begin{matrix} 1 & 0 & 5 \\ 0 & 1 & - 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}]$
Read the pivot pattern
Pivot columns are 1 and 2, so $\operatorname{rank}(A) = 2$ . Column 3 carries no leading one, so $z_3$ is a free variable and the system has non-trivial solutions. By the full-column-rank criterion, $T$ is not injective: the three columns of $A$ are linearly dependent.
Extract an annihilated direction
Setting $z_3 = 1$ and back-reading the two pivot rows gives $z_1 = -5$ and $z_2 = 3$ . Verify directly against the original matrix — the check must be made on $A$ , not on its reduced form:
$\vec{z} = [\begin{matrix} - 5 \\ 3 \\ 1 \end{matrix}], A \vec{z} = [\begin{matrix} - 5 + 6 - 1 \\ - 10 + 9 + 1 \\ - 5 + 3 + 2 \\ - 15 + 15 + 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \end{matrix}]$
Build the witness pair
Choose any command vector, say $\vec{x} = (1, 0, 2)$ , and set $\vec{y} = \vec{x} + \vec{z} = (-4, 3, 3)$ . These are distinct commands. Evaluating both:
$T (\vec{x}) = [\begin{matrix} - 1 \\ 4 \\ 5 \\ 3 \end{matrix}] = T (\vec{y})$
Repair the instrument
The dependence is a property of the sensing geometry, not of the data. Relocating the fourth gauge so that its response row becomes $(3, 5, 1)$ changes the matrix to $A'$ . Reducing $A'$ gives leading ones in all three columns, so $\operatorname{rank}(A') = 3$ , the homogeneous system has only the trivial solution, and the modified instrument is injective.
$A' = [\begin{matrix} 1 & 2 & - 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \\ 3 & 5 & 1 \end{matrix}] ⟶ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}]$

Result

The original four-gauge array is blind along the command direction $(-5, 3, 1)$ : any two commands differing by a multiple of that vector produce byte-identical readings, so the array cannot identify the applied command. Adding four sensors did not help because only two of the four rows were independent. Changing one gauge location to break the dependence restores full column rank and with it a unique inverse reading.

Applications & Industry Use

Instrumentation & metrology

Sensor array identifiability

A sensor array measures a linear functional of the physical state. Injectivity of the sensing map is the condition that the recorded vector determines the state. Testing rank of the sensing matrix at design time is far cheaper than discovering during commissioning that two fault conditions produce the same signature.

Communications

Uniquely decodable linear codes

A linear encoder maps a message block to a longer codeword. Injectivity of the encoder is the minimum requirement for decodability: two distinct messages sharing a codeword make error-free decoding impossible even on a noiseless channel. Generator matrices are therefore always chosen with full column rank.

Computer graphics

Degenerate transformation detection

Modelling transforms composed from scales, shears and projections may silently become rank-deficient. A non-injective transform flattens geometry, so distinct vertices coincide, surface normals vanish and shading fails. Checking injectivity of the linear part detects the collapse before it reaches the rasteriser.

Chemical process engineering

Distinguishing feed compositions

Online analysers report a small number of linear combinations of species concentrations. If the analyser map is not injective, two different feed compositions produce identical readings and control action based on those readings is directed at an ambiguous state. Rank analysis of the analyser response matrix identifies which species pairs are indistinguishable.

Structural health monitoring

Damage localisation

Damage is modelled as a change in a stiffness parameter vector, which maps linearly to a change in measured modal response for small perturbations. Non-injectivity means two damage locations produce the same response change, so the monitoring system can detect damage but not localise it. Sensor placement is chosen precisely to restore injectivity.

Data science

Identifiable regression models

In a linear model the design matrix maps parameters to fitted values. When it lacks full column rank, infinitely many parameter vectors give an identical fit and individual coefficients are meaningless, even though predictions remain well defined. Detecting this before interpretation prevents confident statements about unidentifiable quantities.

Design Considerations

Check dimensions before touching the formula

If $\dim(U) > \dim(V)$ the transformation cannot be injective, whatever its coefficients. This one-line test disposes of many cases immediately and should always precede a row reduction. The converse is not available: $\dim(U) \le \dim(V)$ permits injectivity but does not deliver it.

Injectivity is a property of the map, not of the data

It is tempting to conclude from a data set in which all observed outputs happen to be distinct that the map is injective. It is not evidence. Injectivity is decided by the rank of the transformation, and a rank-deficient map will produce distinct outputs for most randomly chosen input pairs while still being ambiguous along a specific direction.

Rank deficiency is a knife edge in exact arithmetic and a gradient in floating point

Over the rationals a matrix either has full column rank or it does not. In floating point, near-dependence between columns is the practical concern: the map is technically injective but numerically close to a map that is not, and inverting it amplifies noise enormously. Report the smallest singular value or a condition number alongside the rank, never the rank alone.

Adding rows is not the same as adding information

Extra output channels increase the row count of $A$ but only help if they increase the rank. A sensor that measures a linear combination of existing channels adds a row of zeros to the reduced form and leaves the ambiguity untouched. Design sensor placement against the rank of the assembled matrix, not against the sensor count.

Prefer a constructive disproof

When a transformation fails the test, report the annihilated direction $\vec{z}$ rather than only the fact of failure. The direction is physically meaningful — it names the combination of inputs the instrument cannot see — and it immediately generates witness pairs for validation and regression tests.

Injectivity does not imply the map can be undone

An injective transformation can be reversed on its range, but if the range is a proper subset of the codomain there is no function defined on all of $V$ that inverts it. Full invertibility requires surjectivity as well. Do not read injectivity as a licence to write $T^{-1}$ without checking the second property.

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	Item 2-8.4 fixes the notation for mappings, and the standard prescribes the arrow form $T \colon U \to V$ together with the terms injective, surjective and bijective used throughout this page.
`IEEE 754-2019`	IEEE Standard for Floating-Point Arithmetic	Governs the arithmetic in which the rank test is executed. Exact cancellation to zero is not reproducible in binary64, so an injectivity decision taken by comparing a pivot against zero requires an explicit tolerance.
`LAPACK / BLAS reference`	Linear Algebra PACKage reference implementation	Column-pivoted QR (`xGEQP3`) and the singular value decomposition (`xGESVD`) are the production instruments for deciding full column rank, and therefore for deciding injectivity of a matrix transformation.
`ISO/IEC 40314`	Mathematical Markup Language (MathML) Version 3.0	Encodes the transformation notation and matrices on this page as semantic markup rather than images, keeping them searchable and machine-readable.
`IEC 60751`	Industrial platinum resistance thermometers and sensors	A representative instrumentation standard: calibration relations of this kind are linearised into sensing matrices whose column rank determines whether a multi-probe installation can identify a temperature field uniquely.

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	Small symbolic models where the question is structural: does a designed sensing geometry lose a direction?	Gives an unambiguous yes or no with no tolerance, but coefficient growth makes it impractical above a few hundred columns.
IEEE 754 binary64 with an SVD rank test	Measured or estimated coefficient matrices, the normal engineering case.	Robust and standard, at the cost of $O(mn^2)$ work and a tolerance choice that must be documented alongside the answer.
IEEE 754 binary64 with pivoted QR	Large systems where a full SVD is too expensive but a rank estimate with a permutation is enough.	Several times cheaper than SVD and reveals which columns are the dependent ones, but the rank estimate is less reliable for matrices with a gradual singular value decay.
IEEE 754 binary32	Embedded or GPU pipelines where memory bandwidth dominates and the sensing matrix is known to be well conditioned.	Roughly seven significant digits leaves little margin between a genuinely dependent column and a merely correlated one.
Fixed-point integer arithmetic	Fixed-geometry sensing matrices baked into an embedded controller with known integer coefficients.	Deterministic and exact for the stated geometry, but overflow during elimination must be bounded a priori and any recalibration invalidates the analysis.
Finite field arithmetic	Linear codes and cryptographic maps, where injectivity of the encoder is the decodability condition.	Exact and fast with no growth in operand size; note that column rank over a finite field can differ from rank over the rationals for the same integer matrix.

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 injectivity of $T(\vec{x}) = A\vec{x}$ for $A$ of size $m \times n$ with $m \ge n$ costs one elimination, about $2mn^2 - \tfrac{2}{3}n^3$ floating-point operations. A rank-revealing QR is a small constant factor more; a singular value decomposition is roughly an order of magnitude more but returns the margin by which the test passes or fails.

Doing it by hand

Reduce the coefficient matrix only — there is no augmented column, because the right-hand side is zero and stays zero under every row operation. Count leading ones and compare against the column count. Carrying a column of zeros through the reduction is a common waste of effort and an occasional source of arithmetic error.

Library routines

numpy.linalg.matrix_rank(A) answers the question for a numeric matrix via the SVD, with a tolerance derived from the largest singular value and the machine epsilon. scipy.linalg.null_space(A) returns an orthonormal basis for the annihilated directions, which is the constructive form of the answer. Symbolic systems expose Matrix.nullspace() and Matrix.rank() over the rationals.

Verifying a claimed witness pair

Substitute both vectors into the original transformation, not into any reduced or factored form. An arithmetic slip during elimination produces a vector that annihilates the reduced matrix but not the original one, and the check on $A$ catches it immediately.

Injectivity under composition

If $T \colon U \to V$ and $S \colon V \to W$ are both injective, the composition $S \circ T$ is injective: equal outputs of the composition force equal outputs of $T$ by injectivity of $S$ , and then equal inputs by injectivity of $T$ . In matrix terms, a product of full-column-rank matrices with compatible shapes retains full column rank. This makes injectivity a property that survives pipelining of processing stages.

Failure Modes & Common Mistakes

Failure modes, root causes and prevention
Failure mode / mistake	Impact	Root cause	Prevention & detection
Concluding injectivity from examples	high	Testing a handful of input pairs, observing distinct outputs, and declaring the map one-to-one.	Prove it by reducing to the homogeneous system. Examples can only disprove injectivity, never establish it.
Confusing injective with surjective	high	Reading both as vague statements about the map being well behaved, or assuming one implies the other.	Injective constrains inputs (no collisions); surjective constrains outputs (nothing missed). For non-square shapes at most one of them can hold.
Testing the wrong homogeneous system	medium	Row-reducing an augmented matrix with a non-zero right-hand side while investigating injectivity.	Injectivity depends only on the coefficient matrix. Reduce $A$ alone; a right-hand side belongs to a consistency question, not this one.
Assuming more outputs guarantees injectivity	medium	Believing that $m > n$ , more measurements than unknowns, is sufficient.	Only the rank matters. Check $\operatorname{rank}(A) = n$ explicitly; redundant sensors add rows without adding rank.
Exact-zero pivot test on floating-point data	high	Comparing a computed pivot against zero, so rounding noise of order $10^{-16}$ is accepted as a genuine pivot.	Use a scaled tolerance relative to the matrix norm, or decide rank from singular values and report the margin.
Ignoring near-dependence	medium	Reporting full column rank for a matrix whose smallest singular value is many orders below the largest.	Report the condition number with the rank. A technically injective but ill-conditioned map is unusable for inverse problems.
Treating injectivity as invertibility	medium	Writing $T^{-1}$ after establishing only that distinct inputs give distinct outputs.	Confirm surjectivity as well, or restrict the codomain to the range before speaking of an inverse.
Applying the property to a non-linear map	low	Using the difference argument on a function that is not a linear transformation.	Verify additivity and scalar-multiple preservation first. Without them, $T(\vec{x}) - T(\vec{y}) = T(\vec{x} - \vec{y})$ is false and the whole reduction collapses.

FAQs

How is injectivity different from a matrix being nonsingular?

Nonsingularity is defined only for square matrices; injectivity is defined for any linear transformation. For a square matrix the two coincide: $T(\vec{x}) = A\vec{x}$ is injective exactly when $A$ is nonsingular. For a rectangular $A$ the corresponding condition is full column rank, which is the natural generalisation.

Can a linear transformation from a larger space to a smaller one ever be injective?

No. If $\dim(U) > \dim(V)$ then any transformation from $U$ to $V$ must send some non-zero vector to zero, and that produces collisions everywhere. This is a purely dimensional obstruction and holds regardless of the coefficients defining the map.

Why is it enough to check whether only the zero vector is annihilated?

Because linearity converts an equality of outputs into an equality of a difference with zero: $T(\vec{x}) = T(\vec{y})$ is equivalent to $T(\vec{x} - \vec{y}) = \vec{0}$ . If the only vector sent to zero is the zero vector, the difference must vanish and the inputs must agree.

Does injectivity mean I can compute the input from the output?

It means at most one input corresponds to a given output, so the answer is unique when it exists. It does not guarantee existence: an output outside the range has no preimage at all. Practical inversion also requires the map to be well conditioned, otherwise the unique answer is numerically unreliable.

How does injectivity behave under composition of transformations?

It is preserved. If $T$ and $S$ are both injective then so is $S \circ T$ , because a collision in the composition forces a collision in $S$ , hence in $T$ . The converse is only partial: if $S \circ T$ is injective then $T$ must be injective, but $S$ need not be.

What is the fastest reliable way to test injectivity numerically?

Compute the singular values of the coefficient matrix and compare the smallest against a tolerance scaled by the largest and by the machine epsilon. This is what a library rank function does. A pivoted QR factorisation is a cheaper alternative when the matrix is large and the singular value gap is clear.

If a transformation is not injective, how much information is lost?

Exactly the dimension of the set of annihilated vectors. If that set has dimension $k$ , then every output has a $k$ -parameter family of inputs mapping to it, and the outputs determine the input only up to that family. The number $k$ is the nullity of the coefficient matrix.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section ILT, subsection EILT. 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.
Anderson, E. et al. LAPACK Users' Guide, 3rd edition. Society for Industrial and Applied Mathematics, 1999.
IEEE 754-2019, IEEE Standard for Floating-Point Arithmetic. Institute of Electrical and Electronics Engineers.

AI Suggested Questions

Give me a 5x3 matrix whose transformation is injective and a 5x3 matrix whose transformation is not, and show the row reduction that distinguishes them.
If a sensing matrix is technically full column rank but has condition number 10^9, what does that mean physically for identifiability?
Prove that if the composition S∘T is injective then T must be injective, and give an example where S is not.
How do I choose a numerical tolerance for a rank test so that the injectivity decision is defensible in an engineering report?
Show how an injective linear transformation between abstract polynomial spaces reduces to a matrix rank calculation once bases are chosen.
What is the relationship between injectivity of a linear transformation and the number of degrees of freedom in its inverse problem?

Related Calculators

Injectivity Tester

Enter a coefficient matrix and get a verdict on injectivity, the pivot pattern, and an explicit annihilated direction when the test fails.

Matrix Rank & Nullity Calculator

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

Null Space Basis Calculator

Produce a basis for the set of vectors a matrix annihilates, in exact rational or floating-point arithmetic.