Injective Transformations, Independence and Dimension

Overview

Injectivity is defined by a condition on pairs of vectors, but its real force is structural. An injective linear transformation does not merely avoid collisions between individual inputs; it preserves the entire independence structure of the domain. Whenever a set of vectors in $U$ admits only the trivial relation of linear dependence, the same is true of the set of their images in $V$ . Nothing is flattened, nothing is folded, and the geometry of independence is transported intact.

That observation converts an infinite verification into a finite one. To decide whether $T$ is injective, take any one basis of the domain, apply $T$ to each of its vectors, and test the resulting set for linear independence. If the images are independent, $T$ is injective; if they are dependent, $T$ is not. Because a basis is finite for any finite-dimensional space, this reduces the question to a single rank computation on a matrix built from the images. It also explains why injectivity is so cheap to test in practice even for transformations whose domain is a space of polynomials or matrices, where no coefficient matrix is handed to you.

The dimension consequence follows immediately and is the most-used result of the group. If $T$ is injective and $B$ is a basis of $U$ with $\dim(U)$ vectors, then the images of $B$ form an independent set of the same size inside $V$ . A vector space cannot contain more independent vectors than its dimension, so $\dim(U) \le \dim(V)$ . Contrapositively, a transformation from a larger space into a smaller one is never injective, whatever formula defines it. This is a one-line test that disposes of entire classes of models before any arithmetic is done.

Together these results establish the correct mental picture: an injective transformation embeds a faithful copy of the domain inside the codomain. The image of a basis of $U$ is a basis of the range of $T$ , so the range is a subspace of $V$ with exactly $\dim(U)$ dimensions. Everything about $U$ — independence, spanning within the range, dimension — is reproduced there. What injectivity does not deliver is coverage: the embedded copy may be a proper subspace, and filling out the codomain is the separate business of surjectivity.

Definition

Relation of Linear Dependence

RLD

For a set $S = \{\vec{u}_1, \vec{u}_2, \dots, \vec{u}_t\}$ in a vector space, a relation of linear dependence is an equation $\alpha_1 \vec{u}_1 + \alpha_2 \vec{u}_2 + \cdots + \alpha_t \vec{u}_t = \vec{0}$ with specified scalars $\alpha_i$ . The relation is trivial when every $\alpha_i = 0$ . The set $S$ is linearly independent when the trivial relation is the only one.

Restated here because every proof on this page begins by writing down an arbitrary relation of linear dependence on the image set and pulling it back through the transformation.

Basis

A set of vectors in a vector space $U$ that is both linearly independent and spans $U$ . Every finite-dimensional vector space has a basis, all its bases have the same number of vectors, and that number is $\dim(U)$ .

The basis test for injectivity holds for any basis of the domain. There is no need to choose a distinguished one, which is what makes the test practical for abstract spaces.

Embedding

An injective linear transformation, viewed as identifying the domain with a subspace of the codomain. The image of the domain under an embedding is a subspace of $V$ whose dimension equals $\dim(U)$ , and on which $T$ is a one-to-one correspondence.

Concepts

Injections preserve linear independence

Let $T \colon U \to V$ be injective and let $S = \{\vec{u}_1, \dots, \vec{u}_t\}$ be a linearly independent subset of $U$ . Claim: $R = \{T(\vec{u}_1), \dots, T(\vec{u}_t)\}$ is linearly independent in $V$ . Begin with an arbitrary relation $\alpha_1 T(\vec{u}_1) + \cdots + \alpha_t T(\vec{u}_t) = \vec{0}$ . By linearity the left-hand side equals $T(\alpha_1 \vec{u}_1 + \cdots + \alpha_t \vec{u}_t)$ , so the combination $\alpha_1 \vec{u}_1 + \cdots + \alpha_t \vec{u}_t$ lies in the kernel of $T$ . Injectivity makes the kernel trivial, so that combination equals $\vec{0}$ — which is a relation of linear dependence on the independent set $S$ , forcing every $\alpha_i = 0$ . The only relation on $R$ is the trivial one, so $R$ is independent. Note where each hypothesis is used: linearity to move the scalars inside, injectivity to make the kernel trivial, independence of $S$ to finish.

The basis test is an equivalence

Let $B = \{\vec{u}_1, \dots, \vec{u}_m\}$ be a basis of $U$ and let $C = \{T(\vec{u}_1), \dots, T(\vec{u}_m)\}$ . Then $T$ is injective if and only if $C$ is linearly independent. Forward: a basis is independent, so the preservation result applies directly. Backward: assume $C$ is independent and take $\vec{u} \in \mathcal{K}(T)$ . Write $\vec{u} = \alpha_1 \vec{u}_1 + \cdots + \alpha_m \vec{u}_m$ , which is possible because $B$ spans $U$ . Then $\vec{0} = T(\vec{u}) = \alpha_1 T(\vec{u}_1) + \cdots + \alpha_m T(\vec{u}_m)$ is a relation of linear dependence on the independent set $C$ , so every $\alpha_i = 0$ and therefore $\vec{u} = \vec{0}$ . The kernel is trivial and $T$ is injective. The backward direction is where the spanning half of the basis property is used; independence alone would not suffice.

The dimension inequality and its contrapositive

Suppose $T \colon U \to V$ is injective and, for contradiction, that $\dim(U) = m > t = \dim(V)$ . Take a basis $B$ of $U$ ; it has $m$ vectors, and their images form a linearly independent set of $m$ distinct vectors inside $V$ . But a vector space of dimension $t$ cannot contain a linearly independent set of more than $t$ vectors. With $m > t$ this is impossible, so the assumption fails and $\dim(U) \le \dim(V)$ . Used in contrapositive form this is the cheapest test in the subject: if the domain has strictly larger dimension than the codomain, no injective linear transformation between them exists. A map from $M_{23}$ to $\mathbb{C}^4$ , for example, has $6 > 4$ and is non-injective regardless of its entries.

Necessary, not sufficient

The inequality $\dim(U) \le \dim(V)$ is a necessary condition for injectivity, never a sufficient one. The zero transformation from $\mathbb{C}^2$ to $\mathbb{C}^5$ satisfies the inequality comfortably and is as far from injective as a map can be. Passing the dimension test only means the question has not yet been settled and the independence computation must still be done. Confusing the two directions is the single most common error in applying these results.

The image of a basis is a basis of the range

If $T$ is injective and $B$ is a basis of $U$ , then $C = T(B)$ is independent by the preservation result and spans the range of $T$ , because every output is $T$ of some linear combination of $B$ and therefore the matching combination of $C$ . So $C$ is a basis of the range, and $\dim(\mathcal{R}(T)) = \dim(U)$ . The rank of an injective transformation equals the dimension of its domain, and its nullity is zero. This is the precise sense in which an injection places an undistorted copy of $U$ inside $V$ : the copy is the range, and it has exactly the right dimension.

Injectivity survives composition

If $T \colon U \to V$ and $S \colon V \to W$ are both injective, so is $S \circ T$ . Assume $(S \circ T)(\vec{x}) = (S \circ T)(\vec{y})$ , that is $S(T(\vec{x})) = S(T(\vec{y}))$ . Injectivity of $S$ gives $T(\vec{x}) = T(\vec{y})$ , and injectivity of $T$ then gives $\vec{x} = \vec{y}$ . Chaining the dimension inequality along the composition yields $\dim(U) \le \dim(V) \le \dim(W)$ . The converse is only partial: if $S \circ T$ is injective then $T$ must be injective, but $S$ need not be — $S$ is only constrained on the range of $T$ .

Testing injectivity through a basis

Compare dimensionsIf

\dim(U) > \dim(V)

, stop:

T

cannot be injective and no computation is needed.

Choose any basis of the domainAny basis works. For

\mathbb{C}^n

take the standard unit vectors; for

P_n

take the monomials; for

M_{mn}

take the single-entry matrices.

Apply $T$ to each basis vectorEvaluate the defining formula once per basis vector to obtain the image set

C

Coordinatise the imagesExpress each image in a fixed basis of the codomain and stack the coordinate vectors as the rows or columns of a matrix.

Row-reduce and countIf the rank equals the number of basis vectors,

C

is independent and

T

is injective. Otherwise it is not.

Read the by-productsThe rank is

\dim(\mathcal{R}(T))

, and the dependent columns identify a relation that produces an explicit kernel vector.

Equations

Preservation of linear independence

EQ-ILTLI-01

T injective, S = {{\vec{u}}_{1}, \dots, {\vec{u}}_{t}} independent \Rightarrow {T ({\vec{u}}_{1}), \dots, T ({\vec{u}}_{t})} independent

The image of an independent set under an injection is independent, with the same number of distinct vectors.

Pull-back of a relation of linear dependence

EQ-ILTLI-02

Pull-back of a relation of linear dependence

The identity that makes the proof work: a relation among the images is the image of a combination of the originals, which must therefore lie in the kernel.

Basis characterisation of injectivity

EQ-ILTB-01

T injective \Leftrightarrow T (B) = {T ({\vec{u}}_{1}), \dots, T ({\vec{u}}_{m})} independent

Here $B = \{\vec{u}_1, \dots, \vec{u}_m\}$ is any basis of the domain $U$ . A finite test for an infinitely quantified property; it holds for every basis, so any convenient one may be chosen.

Dimension inequality for injections

EQ-ILTD-01

T : U \to V injective Longrightarrow \dim (U) \leq \dim (V)

An injection cannot compress. Used in contrapositive form it rules out injectivity from the shapes alone.

Rank and nullity of an injection

EQ-ILTD-02

\dim (R (T)) = \dim (U), \dim (K (T)) = 0

An injective transformation has zero nullity, so its rank consumes the entire dimension of the domain.

Composition of injections

EQ-CILTI-01

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

Injectivity is preserved by chaining, so a pipeline of identifiable stages is identifiable end to end.

Dimension chain along a composition

EQ-ILTD-03

\dim (U) \leq \dim (V) \leq \dim (W)

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 $\dim(U)$ vectors	finite subset of U
$C$	Image of the basis	The set $T(B)$ , tested for linear independence in $V$	finite subset of V
$α_{i}$	Relation coefficients	Scalars in a relation of linear dependence on the image set	complex scalars
$m$	Domain dimension	$\dim(U)$ ; also the number of vectors in $B$ and in $C$	non-negative integer
$t$	Codomain dimension	$\dim(V)$ ; the maximum size of any independent set in $V$	non-negative integer
$K (T)$	Kernel	Trivial exactly when $T$ is injective; the object the pull-back argument passes through	subspace of U
$R (T)$	Range	The embedded copy of $U$ inside $V$ , with $C$ as a basis when $T$ is injective	subspace of V

Worked Numerical Example

Problem statement

A quadratic strain profile across a beam section, $p(x) = a + bx + cx^{2}$ , is encoded for transmission as a $2 \times 2$ block of derived quantities by the linear rule $T \colon P_2 \to M_{22}$ . Decide whether the encoding is injective using the basis test, and confirm that the dimension inequality is not violated.

State the encoding
The encoder forms two sums and two differences of the profile coefficients:
$T (a + b x + c x^{2}) = [\begin{matrix} a + b & a - 2 c \\ c & b - c \end{matrix}]$
Check the dimension precondition
The domain $P_2$ of quadratic polynomials has dimension $3$ ; the codomain $M_{22}$ has dimension $4$ . Since $3 \le 4$ , the dimension inequality does not rule out injectivity. This settles nothing on its own — it only means the computation is worth doing.
Apply $T$ to a basis of the domain
Take the monomial basis $B = \{1,\ x,\ x^{2}\}$ , corresponding to $(a,b,c) = (1,0,0)$ , $(0,1,0)$ and $(0,0,1)$ in turn:
$T (1) = [\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}], T (x) = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], T (x^{2}) = [\begin{matrix} 0 & - 2 \\ 1 & - 1 \end{matrix}]$
Coordinatise the images
Read each matrix in the order (top-left, top-right, bottom-left, bottom-right) to obtain coordinate vectors in $\mathbb{C}^4$ , and stack them as rows:
$[\begin{matrix} 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & - 2 & 1 & - 1 \end{matrix}]$
Row-reduce and count
Apply $R_2 \to R_2 - R_1$ , scale by $-1$ , then clear column 2 from rows 1 and 3:
$Row-reduce and count$
Draw the conclusion
Three non-zero rows for three image vectors, so the rank is $3$ and the images of the basis are linearly independent. By the basis characterisation, $T$ is injective: distinct strain profiles always produce distinct encoded blocks. The range has dimension $3$ inside the four-dimensional $M_{22}$ , so the encoding is not surjective.
Use preservation on a different independent set
Independence is preserved for every independent set, not just for the chosen basis. Take $\{1 + x,\ x - x^{2}\}$ , which is independent in $P_2$ . Its images must be independent in $M_{22}$ , and indeed neither is a scalar multiple of the other:
$T (1 + x) = [\begin{matrix} 2 & 1 \\ 0 & 1 \end{matrix}], T (x - x^{2}) = [\begin{matrix} 1 & 2 \\ - 1 & 2 \end{matrix}]$
Contrast with a shape that cannot work
Suppose instead the encoder were required to compress a $2 \times 3$ sensor block into four numbers, $T \colon M_{23} \to \mathbb{C}^4$ . Here $\dim(M_{23}) = 6 > 4$ , so the dimension inequality is violated and no such encoder is injective, whatever coefficients are chosen. Some pair of distinct sensor blocks must always collide.
$\dim (M_{23}) = 6 > 4 = \dim (ℂ^{4}) Longrightarrow T non-injective$

Result

The three-into-four encoding is lossless: the encoded block determines the strain profile uniquely, and a decoder exists on the range. The range is a three-dimensional subspace of $M_{22}$ , so one linear constraint is satisfied by every legitimate encoded block — a free integrity check for the receiver. The six-into-four variant is provably lossy before a single coefficient is chosen, which is the kind of conclusion the dimension inequality delivers at zero cost.

Applications & Industry Use

Communications

Generator matrices for block codes

A linear block encoder maps $k$ message symbols to $n > k$ channel symbols. Injectivity is unique decodability, and the basis test is exactly the requirement that the $k$ columns of the generator matrix be independent. The dimension inequality is the reason no useful code compresses: $k \le n$ is forced.

Aerospace engineering

Sensor suite sizing

Before selecting instruments, the count of independent measurements is compared with the number of state variables to be resolved. If the state space has more dimensions than the measurement vector, no linear estimator can identify the state and the suite must be expanded. This screening uses only dimensions, no sensor models.

Computer graphics

Preserving the independence of a frame

A modelling transform applied to a set of independent basis vectors must return an independent set, or the local frame collapses and normals, tangents and bitangents become degenerate. Testing the images of the three axis vectors is the standard cheap check for a degenerate transform.

Systems biology

Identifiability screening of reaction networks

Parameter vectors are mapped linearly to observable time-course features in a local analysis. Comparing the number of parameters against the number of independent observables gives an immediate upper bound on identifiability, before any expensive profile-likelihood computation is attempted.

Manufacturing metrology

Fixture and probe layout

A coordinate measuring routine samples a part at a fixed set of points and infers a small number of form parameters. The map from parameters to probe readings must be injective for the fit to be meaningful, and the basis test applied to the parameter basis is the design-time criterion for probe layout.

Cryptography

Linear layers in block ciphers

The diffusion layer of a block cipher is a linear transformation over a finite field that must be invertible, hence in particular injective. The design check is that images of the standard basis remain independent over that field, which is a rank computation in modular arithmetic.

Design Considerations

Run the dimension test first, every time

It costs nothing, requires no coefficients, and eliminates a large fraction of candidate designs. Only when $\dim(U) \le \dim(V)$ does the independence computation become worth performing. Reversing the order wastes effort on models that were doomed by their shapes.

Never read the inequality backwards

$\dim(U) \le \dim(V)$ does not imply injectivity. The zero transformation satisfies it for every pair of spaces and is maximally non-injective. Treat the inequality strictly as a necessary condition and state it as such in any written analysis.

Any basis will do — choose the cheap one

The basis characterisation holds for every basis of the domain, so pick the one that makes evaluating $T$ trivial: standard unit vectors for $\mathbb{C}^n$ , monomials for polynomial spaces, single-entry matrices for matrix spaces. Choosing an awkward basis multiplies the arithmetic without changing the verdict.

Independence is preserved; spanning is not

An injection carries independent sets to independent sets, but it does not carry spanning sets of $U$ to spanning sets of $V$ — only to spanning sets of the range. Assuming otherwise silently conflates injectivity with surjectivity and produces claims about coverage that the hypothesis does not support.

The rank is the by-product worth keeping

The same reduction that answers the injectivity question returns $\dim(\mathcal{R}(T))$ and, when the test fails, an explicit dependence relation among the images. That relation pulls back to a non-zero kernel vector, which is the constructive form of the failure. Discarding it and reporting only pass or fail throws away the diagnosis.

Chain injectivity deliberately across a pipeline

Because compositions of injections are injective, a processing chain is identifiable end to end if each stage is. The corollary is that a single non-injective stage destroys identifiability for the whole chain irrecoverably, so the weakest stage should be identified and fixed rather than compensated downstream.

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 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	The independence test on the image set is a rank determination, delivered in production by column-pivoted QR (`xGEQP3`) or the singular value decomposition (`xGESVD`).
`IEEE 754-2019`	IEEE Standard for Floating-Point Arithmetic	Determines the arithmetic in which the image coordinates are computed and reduced, and therefore why near-dependence rather than exact dependence is the practical failure mode.
`ISO/IEC 18033-3`	Information technology — Security techniques — Encryption algorithms — Block ciphers	Block cipher diffusion layers are specified as invertible linear maps over finite fields; the independence-of-basis-images criterion on this page is the design-time check for that requirement.
`ISO/IEC 40314`	Mathematical Markup Language (MathML) Version 3.0	Encodes the set notation, matrices and quantified statements on this page as structured 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	Symbolic transformations, textbook-scale designs, and any claim of injectivity that must hold exactly rather than numerically.	Delivers an unambiguous independence verdict, but the intermediate fractions grow quickly and the approach does not scale to large image sets.
IEEE 754 binary64 with an SVD rank test	Image vectors computed from measured or floating-point coefficients, which is the normal engineering case.	Robust, and the singular values grade how close the images are to dependence; costs several times a plain elimination.
IEEE 754 binary64 with pivoted QR	Large basis sets where a full decomposition is uneconomical but the dependent members must still be identified.	Cheaper and names the dependent images, but the rank estimate degrades when the singular values decay gradually.
Finite field arithmetic modulo a prime	Coding theory and cipher design, where independence of the images is required over the field the code actually uses.	Exact, fast and with bounded operand size; independence over a finite field can differ from independence over the rationals for the same integer data.
Integer arithmetic with a determinant or Smith normal form	Square cases with integer entries, where an exact non-zero determinant certifies independence outright.	A single decisive number, but only applicable when the number of images equals the codomain dimension, and determinants overflow quickly.
Dimension counting alone	Early screening, before any coefficients are available.	Free and conclusive when it fails, but silent when it passes; it can never establish injectivity.

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 basis test

Evaluating $T$ on $m$ basis vectors costs $m$ applications of the defining formula. Reducing the resulting $m \times t$ coordinate array costs about $2mt\min(m,t) - \tfrac{2}{3}\min(m,t)^3$ operations. Since $m \le t$ is required for injectivity anyway, this is dominated by $O(m^2 t)$ and is negligible for the modest dimensions typical of design-time checks.

Doing it by hand

Write the coordinate vectors as the rows of a matrix rather than the columns. Row reduction then answers the independence question by counting non-zero rows, and any row that reduces to zero exhibits a dependence relation directly in terms of the operations applied. Placing them as columns answers the same question but makes the relation harder to read off.

Choosing between exact and numerical verdicts

In exact arithmetic the images are independent or they are not. In floating point the meaningful quantity is the smallest singular value of the coordinate array relative to the largest. Report that ratio alongside the verdict: a transformation whose images are independent by a margin of $10^{-14}$ is, for engineering purposes, not injective.

Library route for abstract domains

Symbolic systems accept the images directly and expose Matrix.rank() and Matrix.rref() over the rationals, which is the natural route when the domain is a polynomial or matrix space. For numerical work, build the coordinate array explicitly and call numpy.linalg.matrix_rank; there is no need to construct an abstract representation of $T$ first.

Extracting the kernel vector from a failed test

If the images are dependent, the reduction produces scalars $\alpha_i$ , not all zero, with $\sum \alpha_i T(\vec{u}_i) = \vec{0}$ . By linearity the vector $\sum \alpha_i \vec{u}_i$ is a non-zero element of the kernel, expressed in the domain's own terms. This pull-back turns an abstract failure into a concrete counterexample and costs no extra computation.

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 that $\dim(U) \le \dim(V)$ makes $T$ injective.	State the inequality only as a necessary condition, and always follow a passing dimension check with the independence computation.
Testing a spanning set instead of a basis	medium	Applying the characterisation to a spanning set that is not independent, whose images are then dependent for trivial reasons.	The test requires a basis. Reduce a spanning set to a basis first, or the dependent verdict carries no information about $T$ .
Concluding the images span the codomain	high	Assuming that an injection carries a basis of $U$ to a basis of $V$ .	The images form a basis of the range, a subspace of $V$ of dimension $\dim(U)$ . Spanning $V$ is surjectivity and needs a separate argument.
Coordinatising images in inconsistent orders	medium	Reading matrix entries row-wise for one image and column-wise for another when building coordinate vectors.	Fix one ordering of the codomain basis at the start and apply it identically to every image.
Exact-zero rank test on floating-point images	high	Declaring independence because no computed pivot is exactly zero.	Use a tolerance scaled by the norm of the coordinate array, and report the margin with the verdict.
Assuming injectivity of $S$ from injectivity of $S \circ T$	medium	Reversing the composition result, which only constrains $S$ on the range of $T$ .	From an injective composition conclude only that $T$ is injective; test $S$ separately on its whole domain.
Ignoring the pull-back of a dependence relation	low	Reporting only that the test failed, without converting the relation into an explicit kernel vector.	Record the coefficients from the reduction and form $\sum \alpha_i \vec{u}_i$ ; the diagnosis is free once the reduction is done.
Applying the results to a non-linear map	low	Using the pull-back identity for a function that does not preserve addition and scalar multiplication.	Verify linearity first; every proof on this page moves scalars through $T$ and fails immediately without it.

FAQs

Why is testing the images of one basis enough to decide injectivity?

Because a basis both spans the domain and is independent. Spanning lets every kernel candidate be written as a combination of basis vectors, and independence of the images then forces all its coefficients to vanish, so the kernel is trivial. Both halves of the basis property are used, which is why an arbitrary independent set will not do.

Does an injective transformation carry a basis of the domain to a basis of the codomain?

Only to a basis of its range. The images are independent and span the range, so they form a basis of that subspace, but the range may be a proper subspace of the codomain. Reaching all of $V$ is surjectivity, a separate property.

If the dimension inequality holds, is my transformation injective?

Not necessarily. The inequality is only necessary. The zero transformation from $\mathbb{C}^2$ into $\mathbb{C}^5$ satisfies it and sends everything to the zero vector. After a passing dimension check the independence of the basis images must still be computed.

What happens to injectivity when I compose transformations?

It is preserved: the composition of two injections is injective. In the other direction, an injective composition $S \circ T$ forces $T$ to be injective, but says nothing about $S$ outside the range of $T$ . Practically, one non-injective stage in a chain destroys identifiability for the whole chain.

How do these results relate to rank and nullity?

An injective transformation has nullity zero, so its rank equals the dimension of its domain. Since the rank can never exceed the dimension of the codomain, the inequality $\dim(U) \le \dim(V)$ falls straight out of the rank-nullity accounting.

Can an injective transformation exist between spaces of equal dimension?

Yes, and in that case it is automatically surjective as well, hence invertible. When $\dim(U) = \dim(V)$ the range is a subspace of $V$ with the full dimension of $V$ , which forces it to be all of $V$ . Equal dimensions are what make the two properties collapse into one.

What do I do with the dependence relation when the test fails?

Pull it back. If $\sum \alpha_i T(\vec{u}_i) = \vec{0}$ with the $\alpha_i$ not all zero, then $\sum \alpha_i \vec{u}_i$ is a non-zero kernel vector written in the domain's own language. It generates witness pairs and names the input combination the transformation cannot see.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section ILT, subsections ILTLI, ILTD and CILT. 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.
MacWilliams, F. J. and Sloane, N. J. A. The Theory of Error-Correcting Codes. North-Holland, 1977.

AI Suggested Questions

Construct a linear transformation from P3 into M22 that satisfies the dimension inequality but is still not injective, and show the dependent basis images.
Prove that if S∘T is injective then T is injective, and give a concrete S that is not injective in such a composition.
For an injective transformation, why is the image of a basis a basis of the range but not of the codomain?
Show how a dependence relation among basis images pulls back to an explicit kernel vector, with a worked 4x3 example.
How does the independence-of-basis-images criterion translate into the design rule for a generator matrix of a linear block code?
If the coordinate array of my basis images has singular values 5, 2 and 1e-13, is the transformation injective in any useful sense?

Related Calculators

Linear Independence Tester

Enter a set of vectors or coordinatised images and get an independence verdict, the rank, and any dependence relation found.

Injectivity Tester

Apply a transformation to a chosen basis, test the images, and report injectivity together with the rank of the range.

Dimension Compatibility Checker

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