The Equal-Dimension Principle
For a linear map, injectivity and surjectivity describe different things. Injectivity says that distinct inputs cannot have the same output; surjectivity says that every element of the codomain is reached. In general, neither property guarantees the other. When the domain and codomain have the same finite dimension, however, rank-nullity links the size of the kernel to the size of the image and forces the two properties to coincide.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A linear map \(T:V\to W\) is injective if \(T(\mathbf{v}_1)=T(\mathbf{v}_2)\) implies \(\mathbf{v}_1=\mathbf{v}_2\). It is surjective if for every \(\mathbf{w}\in W\), there is some \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\mathbf{w}\). The kernel and image are the subspaces \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\) and \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\), respectively.
Two definitions translate the properties of \(T\) into statements about these subspaces. Injectivity is equivalent to \(\ker T=\{\mathbf{0}\}\): if \(T\) is injective and \(T(\mathbf{v})=\mathbf{0}=T(\mathbf{0})\), then \(\mathbf{v}=\mathbf{0}\); conversely, if the kernel contains only \(\mathbf{0}\) and \(T(\mathbf{v}_1)=T(\mathbf{v}_2)\), linearity gives \(T(\mathbf{v}_1-\mathbf{v}_2)=\mathbf{0}\), so \(\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}\). Surjectivity is equivalent to \(\operatorname{im}T=W\), since the image consists exactly of the outputs attained by \(T\).
A Dimension Fact for Subspaces
Proof. If \(U=W\), their dimensions are equal. For the converse, suppose \(\dim U=\dim W=n\) but \(U\ne W\). Choose \(\mathbf{w}\in W\) with \(\mathbf{w}\notin U\), and let \(\mathbf{u}_1,\dots,\mathbf{u}_n\) be a basis of \(U\). The list \(\mathbf{u}_1,\dots,\mathbf{u}_n,\mathbf{w}\) is linearly independent: in a relation \(\sum_{i=1}^n c_i\mathbf{u}_i+d\mathbf{w}=\mathbf{0}\), if \(d\ne0\), then \(\mathbf{w}=-\sum_{i=1}^n(c_i/d)\mathbf{u}_i\in U\), a contradiction. Thus \(d=0\), and independence of the basis of \(U\) forces every \(c_i=0\). This gives \(n+1\) independent vectors in a space of dimension \(n\), which is impossible. Hence \(U=W\). This argument also covers \(n=0\): the basis of \(U\) is then empty, and any \(\mathbf{w}\notin U\) would give one independent vector in a zero-dimensional space. Finally, if \(U=\{\mathbf{0}\}\), its dimension is zero. If \(\dim U=0\) but \(U\) contained a nonzero vector, that singleton list would be independent, contradicting dimension zero. Therefore \(U=\{\mathbf{0}\}\). \(\square\)
Injectivity Is Equivalent to Surjectivity
Proof. Set \(\dim V=\dim W=n\). First suppose \(T\) is injective. Then \(\ker T=\{\mathbf{0}\}\), so \(\dim(\ker T)=0\). By the Rank-Nullity Theorem, \(0+\dim(\operatorname{im}T)=n\), hence \(\dim(\operatorname{im}T)=n=\dim W\). Since \(\operatorname{im}T\) is a subspace of \(W\), the proposition gives \(\operatorname{im}T=W\), so \(T\) is surjective. Conversely, suppose \(T\) is surjective. Then \(\operatorname{im}T=W\), so \(\dim(\operatorname{im}T)=n\). Rank-nullity gives \(\dim(\ker T)+n=n\), hence \(\dim(\ker T)=0\). The proposition gives \(\ker T=\{\mathbf{0}\}\), which implies that \(T\) is injective. Thus each property implies the other. If \(n=0\), the same equations apply: both spaces consist only of the zero vector, and the conclusion still holds. \(\square\)
The dimension equality is essential to both halves of the argument. In the first direction, a zero-dimensional kernel leaves an image of dimension \(\dim V\); that image fills \(W\) only because \(\dim V=\dim W\). In the reverse direction, an image equal to \(W\) has dimension \(\dim W\); the kernel is forced to have dimension zero only because this equals \(\dim V\). No determinant or square-matrix criterion is needed: the proof is about subspaces and dimensions.
Worked Applications
Worked Example: A Map from \(\mathbb{R}^2\) to \(\mathbb{R}^2\)
Consider \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(2x+y,x+y)\). Show that it is both injective and surjective.
In standard coordinates, the matrix of \(T\) is \(\mathbf{A}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\). Row reduction gives
The matrix has two pivots, so its image has dimension \(2\). Since the domain also has dimension \(2\), rank-nullity gives \(\dim(\ker T)=2-2=0\). Thus \(T\) is injective. Its image has dimension \(2\), equal to the dimension of the codomain, so the subspace \(\operatorname{im}T\) is all of \(\mathbb{R}^2\); \(T\) is surjective as well.
Check by solving: For a target \((r,s)\), the equations \(2x+y=r\) and \(x+y=s\) give \(x=r-s\) and \(y=2s-r\). Substitution yields \(2(r-s)+(2s-r)=r\) and \((r-s)+(2s-r)=s\), so every target is reached. For the zero target, the same formulas give \(x=0\) and \(y=0\), confirming the kernel is trivial.
Worked Example: Evaluation of a Linear Polynomial
Let \(P_1(\mathbb{R})\) be the vector space of real polynomials of degree at most \(1\), including the zero polynomial. Define \(S:P_1(\mathbb{R})\to\mathbb{R}^2\) by \(S(p)=(p(0),p(1))\). Show that surjectivity forces injectivity here.
Every polynomial has the form \(p(x)=a+bx\), so \(\dim P_1(\mathbb{R})=2\), with basis \((1,x)\). The codomain also has dimension \(2\). To reach any target \((r,s)\), choose \(p(x)=r+(s-r)x\). Then \(p(0)=r\), while \(p(1)=r+(s-r)=s\). Thus \(S\) is surjective.
In the basis \((1,x)\) and the standard basis of \(\mathbb{R}^2\), its matrix is \(\begin{bmatrix}1&0\\1&1\end{bmatrix}\), since \(S(1)=(1,1)\) and \(S(x)=(0,1)\). Reducing gives
There are two pivots, so \(\dim(\operatorname{im}S)=2\). Rank-nullity gives \(\dim(\ker S)=2-2=0\), and therefore \(\ker S=\{0\}\). This proves injectivity. Check: the constructed polynomial gives \(S(p)=(r,s)\) exactly, and if \(S(a+bx)=(0,0)\), then \(a=0\) and \(a+b=0\), so \(b=0\) as well.
Why Equal Dimensions Cannot Be Dropped
Worked Example: Two Maps with Different Dimension Counts
Consider the inclusion \(J:\mathbb{R}^2\to\mathbb{R}^3\), \(J(x,y)=(x,y,0)\), and the projection \(Q:\mathbb{R}^3\to\mathbb{R}^2\), \(Q(x,y,z)=(x,y)\). These maps show that the equivalence does not extend to unequal dimensions.
For \(J\), if \(J(x,y)=(0,0,0)\), then \(x=0\) and \(y=0\); hence \(J\) is injective. But \((0,0,1)\) is not in its image, since every output has third coordinate \(0\). Thus \(J\) is not surjective. Its image is \(\{(x,y,0):x,y\in\mathbb{R}\}\), which has dimension \(2\), while the codomain has dimension \(3\). Rank-nullity agrees: \(\dim(\ker J)+\dim(\operatorname{im}J)=0+2=2\), the dimension of its domain.
For \(Q\), every \((a,b)\in\mathbb{R}^2\) is reached because \(Q(a,b,0)=(a,b)\), so \(Q\) is surjective. It is not injective: \(Q(0,0,0)=Q(0,0,1)=(0,0)\), even though the inputs differ. In fact \(\ker Q=\{(0,0,z):z\in\mathbb{R}\}\) has dimension \(1\), and \(\operatorname{im}Q=\mathbb{R}^2\) has dimension \(2\). The sum \(1+2=3\) is the domain dimension, as rank-nullity requires.
Check: direct substitution verifies both claims: \(J(1,2)=(1,2,0)\), whereas no input maps to \((0,0,1)\); and \(Q(4,-1,0)=(4,-1)\), while \(Q(4,-1,0)=Q(4,-1,7)\).
Common Pitfall and Practical Use
A common error is to use the codomain dimension in place of the domain dimension in rank-nullity. For the inclusion \(J:\mathbb{R}^2\to\mathbb{R}^3\), a mistaken calculation might say, “The kernel is zero, so the image has dimension \(3\), the codomain dimension, and \(J\) is onto.” The correction is \(\dim(\ker J)+\dim(\operatorname{im}J)=\dim\mathbb{R}^2=2\), so \(\dim(\operatorname{im}J)=2\), not \(3\). The image cannot be all of \(\mathbb{R}^3\). Rank-nullity always uses the domain dimension.
For maps between finite-dimensional spaces of equal dimension, it is enough to establish either injectivity or surjectivity; the theorem supplies the other property. For example, one may prove injectivity by showing the kernel is trivial, or prove surjectivity by showing the image is the entire codomain. This shortcut is valid only after checking all hypotheses: the map must be linear, both spaces must be finite-dimensional, and their dimensions must be equal. Without those conditions, the two properties must be tested separately.
Check Your Understanding
Use the definitions and rank-nullity carefully; in each problem, identify the dimensions involved.
- If \(T:V\to W\) is linear, \(\dim V=\dim W=5\), and \(\dim(\ker T)=0\), what is \(\dim(\operatorname{im}T)\), and what property of \(T\) follows?
- A linear map \(S:V\to W\) is surjective, with \(\dim V=\dim W=4\). Determine \(\dim(\ker S)\) and state the resulting property.
- Explain why a subspace \(U\) of a finite-dimensional space \(W\) must equal \(W\) when their dimensions are equal.
- For \(J:\mathbb{R}^2\to\mathbb{R}^3\), \(J(x,y)=(x,y,0)\), state which of injectivity and surjectivity holds and give a vector witnessing failure of the other.
- Can the equal-dimension theorem be applied to a linear map \(T:\mathbb{R}^3\to\mathbb{R}^2\)? Give the hypothesis that prevents its use.
