Why Equal Dimensions Link Two Different Properties
Injectivity and surjectivity ask different questions about a function. A map is injective if distinct inputs always have distinct outputs. It is surjective if every element of its codomain occurs as an output. Neither definition alone says that the other property must hold. For linear maps between finite-dimensional spaces of the same dimension, however, the Rank-Nullity Theorem connects the two: losing no information in the domain leaves enough image to fill the codomain, and filling the codomain leaves no room for a nonzero kernel.
Throughout, spaces and linear maps are over \(\mathbb{R}\). A space is finite-dimensional if it has a finite basis, and its dimension is the number of vectors in any basis. For a linear map \(T:V\to W\), the kernel is \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\), and the image is \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). The kernel is a subspace of \(V\), and the image is a subspace of \(W\). Their dimensions are called the nullity and rank of \(T\), respectively.
Two definitions translate directly into kernel and image language. The map \(T\) is injective exactly when \(\ker T=\{\mathbf{0}\}\): if \(T(\mathbf{v})=T(\mathbf{w})\), linearity gives \(T(\mathbf{v}-\mathbf{w})=\mathbf{0}\), so a trivial kernel forces \(\mathbf{v}=\mathbf{w}\); conversely, injectivity and \(T(\mathbf{0})=\mathbf{0}\) force every vector in the kernel to be \(\mathbf{0}\). The map is surjective exactly when \(\operatorname{im}T=W\), by the definition of image.
The Equal-Dimensions Theorem
Proof. Suppose first that \(T\) is injective. The kernel criterion just explained gives \(\ker T=\{\mathbf{0}\}\), so \(\dim(\ker T)=0\). Rank-nullity therefore gives $$ \dim(\operatorname{im}T)=\dim V-\dim(\ker T)=n-0=n. $$ Since \(\operatorname{im}T\) is a subspace of \(W\), and it has the same dimension \(n\) as \(W\), it equals \(W\). Indeed, a proper subspace of a finite-dimensional space has smaller dimension: if it had dimension \(n\) inside an \(n\)-dimensional space, a basis for it would already be a basis for the whole space. Thus \(\operatorname{im}T=W\), and \(T\) is surjective. Conversely, suppose that \(T\) is surjective. Then \(\operatorname{im}T=W\), so \(\dim(\operatorname{im}T)=n\). Rank-nullity gives \(\dim(\ker T)=\dim V-\dim(\operatorname{im}T)=n-n=0\). A zero-dimensional subspace contains only \(\mathbf{0}\), so \(\ker T=\{\mathbf{0}\}\), and the kernel criterion gives injectivity. These arguments also cover \(n=0\): both spaces are the zero space, and the same dimension equations apply. \(\square\)
The proof has two distinct steps in each direction. Rank-nullity converts injectivity or surjectivity into a dimension for the image or kernel. Then the equal-dimension assumption identifies that dimension with the one needed to conclude the other property. Without the equality \(\dim V=\dim W\), the dimension calculation would not give these conclusions in this form.
In this setting, saying “injective if and only if surjective” also means that either condition is enough to establish both. Since a function that is both injective and surjective is called bijective, either condition makes \(T\) bijective. This conclusion concerns the map as a whole; it does not say that a particular input has a particular output. Examples still require computing the map, its kernel, or its image.
Using a Kernel to Establish Both Properties
Worked Example: A Two-Dimensional Matrix Map
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T(x,y)=(x+2y,3x+7y)\). Determine whether it is injective and surjective.
To test injectivity, solve \(T(x,y)=(0,0)\). This gives the homogeneous system with coefficient matrix \(\begin{bmatrix}1&2\\3&7\end{bmatrix}\). Subtracting \(3R_1\) from \(R_2\), then subtracting \(2R_2\) from \(R_1\), gives
The only solution is \(x=y=0\), so \(\ker T=\{(0,0)\}\) and \(T\) is injective. Both spaces have dimension \(2\). The theorem therefore says that \(T\) is surjective as well. We can also check surjectivity directly: for an arbitrary \((r,s)\in\mathbb{R}^2\), choose \(x=7r-2s\) and \(y=-3r+s\). Then $$ x+2y=7r-2s-6r+2s=r,\qquad 3x+7y=21r-6s-21r+7s=s. $$ Thus every \((r,s)\) has a preimage. Substituting these expressions into \(T\) recovers \((r,s)\), as required.
Using Surjectivity to Establish Both Properties
Worked Example: A Map from Polynomials to the Plane
Let \(P_1\) be the space of real polynomials of degree at most \(1\), including the zero polynomial. Define \(T:P_1\to\mathbb{R}^2\) by \(T(a+bx)=(a,a+b)\). Determine whether \(T\) is surjective and injective.
Take an arbitrary target \((r,s)\in\mathbb{R}^2\). To obtain \(T(a+bx)=(r,s)\), we need \(a=r\) and \(a+b=s\). Set \(a=r\) and \(b=s-r\). Then \(p(x)=r+(s-r)x\) belongs to \(P_1\), and $$ T(p)=(r,r+(s-r))=(r,s). $$ This verifies surjectivity for every target. The space \(P_1\) has basis \((1,x)\) and dimension \(2\), while \(\mathbb{R}^2\) also has dimension \(2\). The theorem now implies that \(T\) is injective.
For a direct check, if \(T(a+bx)=(0,0)\), then \(a=0\) and \(a+b=0\), which gives \(b=0\). Thus the only polynomial in the kernel is the zero polynomial. The calculation agrees with the conclusion from equal dimensions. For the preimage constructed above, substitution verifies both coordinates of the target, not just the first.
These examples use opposite starting points. In the first, a kernel calculation establishes injectivity and the theorem supplies surjectivity. In the second, constructing a preimage for an arbitrary target establishes surjectivity and the theorem supplies injectivity. The two routes are equally valid; the convenient one depends on the formula for \(T\).
When the Map Fails One Property
Worked Example: Differentiation on \(P_1\)
Consider \(D:P_1\to P_1\), defined by \(D(p)=p'\). Both domain and codomain have dimension \(2\). Determine whether \(D\) is injective or surjective, and compare the results with the theorem.
Write \(p(x)=a+bx\). Then \(D(p)=b\), a constant polynomial. The equation \(D(p)=0\) holds exactly when \(b=0\), while \(a\) can be any real number. Hence $$ \ker D=\{a:a\in\mathbb{R}\}=\operatorname{span}\{1\},\qquad \dim(\ker D)=1. $$ Since the kernel contains nonzero polynomials, \(D\) is not injective. Rank-nullity gives \(\dim(\operatorname{im}D)=2-1=1\). In fact, every output is constant, and every constant is an output because \(D(cx)=c\). Thus \(\operatorname{im}D=\operatorname{span}\{1\}\), which is not all of \(P_1\); for example, \(x\) is not the derivative of any polynomial in \(P_1\). So \(D\) is not surjective either.
Check: \(D(1)=0\) confirms the nontrivial kernel, and \(D(cx)=c\) confirms that every constant is attained. The theorem predicts that injectivity and surjectivity have the same truth value here, and this calculation shows that both are false.
The theorem does not say every map between equal-dimensional spaces is injective or surjective. It says those properties stand or fall together. In the differentiation example, one dimension of the domain is lost to the kernel, leaving an image of dimension \(1\) rather than the codomain dimension \(2\). The dimension equation accounts for both failures at once.
A Common Mistake: Confusing Rank with Onto-ness
Rank is the dimension of the image, not a label that means “onto.” To be surjective, a map \(T:V\to W\) must have \(\dim(\operatorname{im}T)=\dim W\). In the equal-dimensions theorem, this is the same number as \(\dim V\), but the codomain is what surjectivity asks us to fill.
For a concrete check, define \(S:\mathbb{R}^2\to\mathbb{R}^2\) by \(S(x,y)=(x,0)\). A mistaken argument might notice that the image is nonzero and conclude that \(S\) is onto. But $$ \operatorname{im}S=\{(x,0):x\in\mathbb{R}\} =\operatorname{span}\{(1,0)\}, \qquad \dim(\operatorname{im}S)=1<2. $$ The target \((0,1)\) is not in the image, since \(S(x,y)=(0,1)\) would require \(x=0\) and \(0=1\). The map is not surjective. Its kernel is \(\{(0,y):y\in\mathbb{R}\}\), so it is not injective either, in agreement with the theorem. The correction is to compare the rank with the dimension of the codomain, not merely to observe that the image contains a nonzero vector.
A reliable application of the theorem therefore begins by checking its hypotheses: the map must be linear, both spaces must be finite-dimensional, and their dimensions must be equal. Once these conditions hold, test whichever property is easier. A trivial kernel gives injectivity and hence surjectivity; a full image gives surjectivity and hence injectivity. If the dimensions differ, this particular equivalence cannot be invoked.
Check Your Understanding
For each question, identify the relevant dimension or kernel information before drawing a conclusion.
- Suppose \(T:V\to W\) is linear, both spaces have dimension \(5\), and \(\ker T=\{\mathbf{0}\}\). What can be concluded about \(T\), and which dimension equation justifies it?
- A linear map \(T:V\to W\) is surjective, with \(\dim V=\dim W=3\). Find \(\dim(\ker T)\) and state whether \(T\) is injective.
- For \(D:P_1\to P_1\), \(D(p)=p'\), identify one nonzero vector in the kernel and one element of the codomain that is not in the image.
- Why does a nonzero image alone not establish that a map is surjective? Use \(S(x,y)=(x,0)\) to support your explanation.
- Which hypotheses must be checked before applying the equal-dimensions theorem to a linear map \(T:V\to W\)?
