What Injectivity Asks
A map is injective when distinct inputs always produce distinct outputs. For a linear map, this question can be reduced to a particularly simple equation: determine which inputs are sent to the zero vector. If the only such input is the zero vector, then no two distinct inputs can share an output. If a nonzero input is sent to zero, that input and the zero vector already have the same output.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) may be \(\mathbb{R}\) or \(\mathbb{C}\). A map \(T:V\to W\) is linear if \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and all scalars \(a,b\in\mathbb{F}\). It is injective if \(T(\mathbf{u})=T(\mathbf{v})\) implies \(\mathbf{u}=\mathbf{v}\) for all \(\mathbf{u},\mathbf{v}\in V\). Write \(\mathbf{0}_V\) and \(\mathbf{0}_W\) for the zero vectors in the two spaces, to keep their roles distinct.
Linearity ensures that \(\mathbf{0}_V\) is always in the kernel. Indeed, \(T(\mathbf{0}_V)=T(\mathbf{0}_V+\mathbf{0}_V)=T(\mathbf{0}_V)+T(\mathbf{0}_V)\); subtracting \(T(\mathbf{0}_V)\) from both sides gives \(T(\mathbf{0}_V)=\mathbf{0}_W\). Also, linearity gives \(T(-\mathbf{v})=-T(\mathbf{v})\), since \(T(-\mathbf{v})=(-1)T(\mathbf{v})\). In particular, for any \(\mathbf{u},\mathbf{v}\in V\), \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})\).
The Kernel Criterion
Proof. First suppose \(T\) is injective. Let \(\mathbf{v}\in\ker T\). By the definition of the kernel, \(T(\mathbf{v})=\mathbf{0}_W\). We also know \(T(\mathbf{0}_V)=\mathbf{0}_W\). Therefore \(T(\mathbf{v})=T(\mathbf{0}_V)\), and injectivity implies \(\mathbf{v}=\mathbf{0}_V\). Since the zero vector is in the kernel, this proves \(\ker T=\{\mathbf{0}_V\}\). Conversely, suppose \(\ker T=\{\mathbf{0}_V\}\), and take any \(\mathbf{u},\mathbf{v}\in V\) such that \(T(\mathbf{u})=T(\mathbf{v})\). Linearity gives \(T(\mathbf{u}-\mathbf{v})=T(\mathbf{u})-T(\mathbf{v})=\mathbf{0}_W\), so \(\mathbf{u}-\mathbf{v}\in\ker T\). The kernel assumption then gives \(\mathbf{u}-\mathbf{v}=\mathbf{0}_V\), which means \(\mathbf{u}=\mathbf{v}\). This is injectivity. \(\square\)
The two directions suggest two useful ways to apply the result. To prove injectivity, start with an arbitrary vector \(\mathbf{v}\) satisfying \(T(\mathbf{v})=\mathbf{0}_W\) and show that \(\mathbf{v}=\mathbf{0}_V\). To prove failure of injectivity, it is enough to find one nonzero \(\mathbf{v}\) with \(T(\mathbf{v})=\mathbf{0}_W\). Such a vector is a concrete certificate that the kernel is not trivial.
Worked Example: A Map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\)
Consider the linear map \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(2x+y,\;x+y)\). Test whether \(T\) is injective using its kernel.
Set the output equal to the zero vector and solve both coordinate equations:
Subtract the second equation from the first to obtain \(x=0\). Substituting into \(x+y=0\) gives \(y=0\). Thus the only input mapped to \((0,0)\) is \((0,0)\), so \(\ker T=\{(0,0)\}\). The kernel criterion proves that \(T\) is injective.
Check: Substitution gives \(T(0,0)=(2\cdot0+0,\;0+0)=(0,0)\), as required. The equations also show that any input in the kernel must have both coordinates zero.
When a Nonzero Kernel Vector Is Found
For a map that is not injective, the decisive calculation is not merely that some outputs look alike. It is enough to exhibit a nonzero kernel vector: linearity ensures that this vector and the zero vector have the same output. Conversely, finding no such vector by testing only a few selected inputs does not establish that the kernel is trivial; the full equation must be solved or otherwise justified.
Worked Example: Differentiation on Quadratic Polynomials
Let \(P_2(\mathbb{R})\) be the vector space of real polynomials of degree at most \(2\), including the zero polynomial. Consider \(D:P_2(\mathbb{R})\to P_2(\mathbb{R})\), defined by \(D(p)=p'\). Test whether \(D\) is injective.
Write an arbitrary input as \(p(x)=a+bx+cx^2\), where \(a,b,c\in\mathbb{R}\). Its derivative is \(D(p)(x)=b+2cx\). To belong to the kernel, this derivative must be the zero polynomial. A polynomial is the zero polynomial exactly when all its coefficients are zero, so
The coefficient \(a\) is unrestricted. Consequently, the kernel consists of all constant polynomials, \(\ker D=\{a:a\in\mathbb{R}\}\), and this kernel is not trivial. In particular, the constant polynomial \(p(x)=1\) is nonzero and satisfies \(D(p)=0\). By the kernel criterion, \(D\) is not injective.
Check: \(D(1)=0\), while \(1\) is not the zero polynomial. More generally, every constant \(a\) has derivative \(0\), and the coefficient calculation shows that no polynomial with \(b\ne0\) or \(c\ne0\) is in the kernel.
Worked Example: Multiplication by \(x+1\)
Define \(S:P_1(\mathbb{R})\to P_2(\mathbb{R})\) by \(S(p)(x)=(x+1)p(x)\), where \(P_1(\mathbb{R})\) consists of real polynomials of degree at most \(1\). Test whether \(S\) is injective.
Take an arbitrary \(p(x)=a+bx\). Expanding the output gives
For this output to be the zero polynomial, each coefficient must vanish. The constant, linear, and quadratic coefficients give, respectively, \(a=0\), \(a+b=0\), and \(b=0\). In particular, the first equation gives \(a=0\), and then the second gives \(b=0\); the third is satisfied as well. Therefore the only polynomial in \(\ker S\) is the zero polynomial, so \(S\) is injective.
Check: Substituting \(p(x)=0+0x\) gives \(S(p)(x)=(x+1)0=0\). The coefficient equations prove that every polynomial sent to zero has exactly these zero coefficients.
Why the Criterion Does Not Depend on Dimension
The proof of the kernel criterion uses only the defining properties of a linear map and the definitions of injectivity and kernel. It does not count vectors, compare dimensions, or require a matrix representation. The method therefore applies to maps on polynomial spaces and other vector spaces just as it applies to coordinate vectors. When a map is given by a formula, the practical task is to set the formula equal to the codomain’s zero vector and solve for every possible input.
For coordinate maps \(T:\mathbb{R}^n\to\mathbb{R}^m\), this means solving \(T(\mathbf{x})=\mathbf{0}\) for an arbitrary column vector \(\mathbf{x}\). For maps on polynomial spaces, it means equating every coefficient of the output polynomial to zero. In either setting, one must retain all solutions: a kernel can contain infinitely many vectors or polynomials, and finding one nonzero member is enough to rule out injectivity.
Proof. Suppose \(\mathbf{v}_1\) and \(\mathbf{v}_2\) both satisfy \(T(\mathbf{v})=\mathbf{w}\). Then \(T(\mathbf{v}_1)=\mathbf{w}=T(\mathbf{v}_2)\). Since \(T\) is injective, equality of these outputs implies \(\mathbf{v}_1=\mathbf{v}_2\). Thus there cannot be two distinct solutions. This conclusion is “at most one”: it does not assert that a solution exists for every \(\mathbf{w}\in W\). \(\square\)
Common Mistake: Testing Only One Input
A nonzero output for one nonzero input does not prove that a map is injective. For example, let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x,0)\). A mistaken argument might calculate \(T(1,0)=(1,0)\ne(0,0)\) and conclude that the kernel is trivial. This calculation checks only one input; it does not rule out other nonzero inputs in the kernel.
The correction is to solve the full kernel equation. Here \(T(x,y)=(0,0)\) means \(x=0\), while \(y\) is unrestricted. Thus \(\ker T=\{(0,y):y\in\mathbb{R}\}\), which contains \((0,1)\ne(0,0)\). Indeed, \(T(0,1)=(0,0)=T(0,0)\), so the map is not injective. The criterion asks whether every vector sent to zero is the zero vector, not whether a few selected nonzero vectors have nonzero outputs.
Check Your Understanding
For each map, use its kernel to decide whether it is injective. When it is not, give a nonzero vector or polynomial in the kernel.
- Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x-y,\;2x-y)\). Solve \(T(x,y)=(0,0)\).
- Let \(U:P_2(\mathbb{R})\to P_2(\mathbb{R})\) be \(U(p)=p'\). Find all polynomials in \(\ker U\).
- Let \(S:\mathbb{R}^3\to\mathbb{R}^2\) be \(S(x,y,z)=(x+z,\;y-z)\). Determine whether its kernel is trivial.
- State why a single calculation \(T(\mathbf{v})\ne\mathbf{0}\) for one chosen nonzero \(\mathbf{v}\) is not enough to prove injectivity.
- Prove that if an injective map \(T\) satisfies \(T(\mathbf{u})=T(\mathbf{v})\), then \(\mathbf{u}=\mathbf{v}\), and identify the definition being used.
