Surjectivity as a Question About Dimension
A linear map is surjective when every vector in its codomain is the output of at least one input. For a map \(T:V\to W\), the set of all outputs is its image, \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). Thus surjectivity asks whether the image fills \(W\). If \(W\) is finite-dimensional, this question can be answered by comparing two dimensions: the dimension of the image and the dimension of the codomain.
Throughout this tutorial, \(V\) and \(W\) are vector spaces over \(\mathbb{R}\), and \(W\) is finite-dimensional. The domain \(V\) need not be finite-dimensional. Linearity ensures that the image is a subspace of \(W\): \(T(\mathbf{0})=\mathbf{0}\), and if \(T(\mathbf{u}),T(\mathbf{v})\in\operatorname{im}T\), then \(aT(\mathbf{u})+bT(\mathbf{v})=T(a\mathbf{u}+b\mathbf{v})\) is also in the image.
The rank records how many independent directions in the codomain the map can produce. It is not, by definition, the dimension of the domain or the number of inputs one happens to use. The relevant comparison is between the image and the codomain, because surjectivity concerns whether every codomain vector is reached.
When a Subspace Fills Its Ambient Space
The dimension test rests on a fact about subspaces. If a subspace is proper—that is, if it does not equal the whole space—then it must have strictly smaller dimension. Here is a proof that includes why the subspace itself has a finite basis.
Proof. Write \(n=\dim W\). First, \(U\) has a finite basis. Start with the empty list in \(U\). Whenever the vectors chosen so far do not span \(U\), choose another vector of \(U\) outside their span. The resulting list remains linearly independent: a relation with a nonzero coefficient on the new vector would express that vector as a combination of the previous ones. An independent list in an \(n\)-dimensional space has at most \(n\) vectors; this follows by writing its vectors in coordinates relative to a basis of \(W\) and applying the pivot test for independence. Therefore this selection process must stop after at most \(n\) choices, and when it stops the list spans \(U\). It is a basis of \(U\).
If \(U=W\), their dimensions are equal because they are the same space. Conversely, suppose \(U\ne W\), and let \(\mathbf{u}_1,\dots,\mathbf{u}_k\) be a basis of \(U\). Choose \(\mathbf{w}\in W\setminus U\). The list \(\mathbf{u}_1,\dots,\mathbf{u}_k,\mathbf{w}\) is independent: if \(a_1\mathbf{u}_1+\cdots+a_k\mathbf{u}_k+c\mathbf{w}=\mathbf{0}\) and \(c\ne0\), then \(\mathbf{w}=-\frac{1}{c}\sum_{i=1}^k a_i\mathbf{u}_i\in U\), a contradiction. Hence \(c=0\), and independence of the basis of \(U\) forces every \(a_i=0\). Since an independent list in \(W\) has at most \(n\) vectors, \(k+1\le n\). Thus \(\dim U=k<n=\dim W\). A proper subspace cannot have the same dimension as \(W\), proving the converse. \(\square\)
The finite-dimensional hypothesis matters: it gives a finite upper bound on the length of an independent list and makes dimension comparison decisive. In this tutorial, that hypothesis is placed on the codomain, so it also guarantees that \(\operatorname{im}T\), as a subspace of \(W\), has a finite dimension.
The Dimension Test for Surjectivity
Proof. By definition, \(\operatorname{rank}T=\dim(\operatorname{im}T)\), and linearity makes \(\operatorname{im}T\) a subspace of \(W\). If \(T\) is surjective, then \(\operatorname{im}T=W\), so \(\operatorname{rank}T=\dim W\). Conversely, if \(\operatorname{rank}T=\dim W\), then \(\dim(\operatorname{im}T)=\dim W\). The lemma applied to the subspace \(\operatorname{im}T\subseteq W\) gives \(\operatorname{im}T=W\), which is exactly surjectivity. Both directions use the finite-dimensionality of \(W\) through the lemma. \(\square\)
The theorem gives a test, not merely a necessary condition. A rank smaller than \(\dim W\) rules out surjectivity. A rank equal to \(\dim W\) guarantees that no codomain directions are missing. The test makes no requirement that \(V\) be finite-dimensional; it is the codomain dimension that sets the comparison.
Applying the Test
Worked Example: A Map Whose Image Misses a Direction
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(x,y,0)\). Decide whether \(T\) is surjective using its rank, then verify the conclusion directly.
The outputs have the form \((x,y,0)=x(1,0,0)+y(0,1,0)\). Therefore \(\operatorname{im}T=\operatorname{span}\{(1,0,0),(0,1,0)\}\). These two vectors are independent: \(a(1,0,0)+b(0,1,0)=(a,b,0)=(0,0,0)\) forces \(a=b=0\). They are a basis of the image, so \(\operatorname{rank}T=2\). The codomain has dimension \(3\), since its standard basis has three vectors. Thus \(2<3\), and the dimension test says \(T\) is not surjective.
Direct check: the vector \((0,0,1)\) is in the codomain but is not an output of \(T\), because every output has third coordinate \(0\). This confirms the conclusion from the rank comparison.
Worked Example: Differentiation Onto a Polynomial Space
Let \(P_2\) be the real polynomials of degree at most \(2\), including the zero polynomial, and \(P_1\) the real polynomials of degree at most \(1\). Define \(D:P_2\to P_1\) by \(D(p)=p'\). Determine its rank and whether it is surjective.
Write an arbitrary input as \(p(x)=a+bx+cx^2\), where \(a,b,c\in\mathbb{R}\). Then \(D(p)=b+2cx\), so every image lies in \(P_1\). The image is exactly \(P_1\): given any \(q(x)=\alpha+\beta x\in P_1\), choose \(p(x)=\alpha x+\frac{\beta}{2}x^2\in P_2\). Differentiating gives \(D(p)=\alpha+\beta x=q(x)\).
Equivalently, \(D(1)=0\), \(D(x)=1\), and \(D(x^2)=2x\), so the image is spanned by \(1\) and \(2x\). These are independent: if \(a\cdot 1+b\cdot 2x\) is the zero polynomial, its constant coefficient gives \(a=0\) and its coefficient of \(x\) gives \(2b=0\), hence \(b=0\). Therefore \(\operatorname{rank}D=2\). The polynomials \(1,x\) form a basis of \(P_1\), so \(\dim P_1=2\). The rank equals the codomain dimension, and \(D\) is surjective.
Check: for the chosen preimage, \(p'(x)=\left(\alpha x+\frac{\beta}{2}x^2\right)'=\alpha+\beta x=q(x)\). Thus the construction reaches an arbitrary target polynomial, as claimed.
Worked Example: A Coordinate Map with Full Rank
Define \(S:\mathbb{R}^3\to\mathbb{R}^2\) by \(S(x,y,z)=(x+z,y+z)\). Find its image and rank, then decide whether it is surjective.
Using the standard basis vectors of \(\mathbb{R}^3\), compute \(S(1,0,0)=(1,0)\), \(S(0,1,0)=(0,1)\), and \(S(0,0,1)=(1,1)\). Every input \((x,y,z)\) gives \(S(x,y,z)=x(1,0)+y(0,1)+z(1,1)\), so the image is spanned by these outputs. Since it contains \((1,0)\) and \((0,1)\), which form a basis of \(\mathbb{R}^2\), the image is all of \(\mathbb{R}^2\) and has dimension \(2\). Thus \(\operatorname{rank}S=2=\dim\mathbb{R}^2\), and the test gives surjectivity.
Check: for any target \((a,b)\in\mathbb{R}^2\), take the input \((a,b,0)\). Then \(S(a,b,0)=(a+0,b+0)=(a,b)\). Every target has a preimage, confirming the dimension test.
Common Pitfall: Confusing Rank with Domain Dimension
For the differentiation map \(D:P_2\to P_1\), it is tempting to say that the rank is \(3\) because the domain has dimension \(3\), with basis \(1,x,x^2\). That calculation is wrong: rank is the dimension of the image, not the number of basis vectors in the domain. In fact, \(D(1)=0\), \(D(x)=1\), and \(D(x^2)=2x\); the outputs span only the two-dimensional space \(P_1\). The corrected calculation is \(\operatorname{rank}D=2\), which equals \(\dim P_1\), so the map is surjective.
A related error is to treat a list of outputs as if its length automatically gave the image dimension. The outputs \(D(1),D(x),D(x^2)\) make a list of three vectors, but one of them is the zero polynomial and the others span a space of dimension \(2\). To find rank, determine the dimension of the span of the outputs, accounting for zero vectors and linear dependence.
What the Test Tells You
The dimension test compresses a question about every possible target into a comparison of two numbers. It does not require finding a preimage for each target separately. First determine the image, or a basis for it; then compare its dimension with the codomain dimension. If the dimensions match, the image cannot be a proper subspace, by the lemma. If the rank is smaller, the image is proper and some codomain vectors are necessarily missing.
Keep the roles of the spaces distinct: \(\operatorname{rank}T\) measures the image, while \(\dim W\) measures the codomain. The theorem says nothing about whether different inputs can have the same output; its conclusion concerns whether all outputs in \(W\) are reached. The domain can be larger, smaller, or infinite-dimensional without changing the exact form of this surjectivity test, provided \(W\) is finite-dimensional.
Check Your Understanding
For each question, identify the image dimension and the codomain dimension before drawing a conclusion.
- A linear map has image \(\operatorname{span}\{(1,0,0),(0,1,0)\}\subseteq\mathbb{R}^3\). What is its rank, and can it be surjective onto \(\mathbb{R}^3\)?
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(x+y,0)\), find a basis for the image and decide whether \(T\) is surjective.
- Explain why \(\dim(\operatorname{im}T)=\dim W\) implies \(\operatorname{im}T=W\) when \(W\) is finite-dimensional.
- For differentiation \(D:P_3\to P_2\), describe the image and use dimensions to decide whether \(D\) is surjective.
- Does the dimension test require \(V\) to be finite-dimensional, or \(W\)? State the condition and explain its role.
