What Unequal Dimensions Can Tell Us
For a linear map between finite-dimensional vector spaces, the dimensions of the domain and codomain limit what the map can do. If the domain is larger, there are too many independent directions for the map to preserve all of them distinctly in the codomain. If the codomain is larger, the image cannot fill all of it. These are restrictions: they identify a property that is impossible in each dimension arrangement, but they do not say that every other property must hold.
Let \(V\) and \(W\) be finite-dimensional vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to W\) be linear. Write \(n=\dim V\) and \(m=\dim W\). 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 rank is \(\operatorname{rank}T=\dim(\operatorname{im}T)\), and the nullity is \(\operatorname{nullity}T=\dim(\ker T)\). A map is injective if equal outputs require equal inputs; it is surjective if every element of \(W\) is an output.
The key comparison is between the number of independent directions available in \(V\) and the number available in \(W\). Injectivity requires the kernel to contain only \(\mathbf{0}\), so it requires nullity zero. Surjectivity requires the image to be all of \(W\), so it requires rank \(m\). Combining those requirements with rank-nullity gives the dimension restrictions.
The Dimension Constraints
- If \(T\) is injective, then \(n\leq m\). Therefore, if \(n>m\), \(T\) is not injective.
- If \(T\) is surjective, then \(n\geq m\). Therefore, if \(n<m\), \(T\) is not surjective.
Proof. Suppose first that \(T\) is injective. The kernel then contains only \(\mathbf{0}\), so its dimension is zero. By rank-nullity, \(\operatorname{rank}T=n\). Because the image is a subspace of \(W\), its dimension is at most \(m\). Thus \(n=\operatorname{rank}T\leq m\). In particular, if \(n>m\), injectivity is impossible.
Now suppose that \(T\) is surjective. Then \(\operatorname{im}T=W\), so \(\operatorname{rank}T=m\). Rank-nullity gives \(n=\operatorname{nullity}T+m\), and nullity is a dimension, hence is nonnegative. Therefore \(n\geq m\). In particular, if \(n<m\), surjectivity is impossible. \(\square\)
The two restrictions point in opposite directions. A larger domain rules out injectivity but may still allow surjectivity. A larger codomain rules out surjectivity but may still allow injectivity. When dimensions are unequal, the theorem does not say that a map must fail at least one of these properties in a particular way beyond the stated restriction; it certainly does not force the remaining property to hold.
| Dimension comparison | Property ruled out | Property that may still occur |
|---|---|---|
| \(\dim V>\dim W\) | Injectivity | Surjectivity |
| \(\dim V<\dim W\) | Surjectivity | Injectivity |
| \(\dim V=\dim W\) | Neither by dimension alone | The equal-dimension result relates the two properties |
A Larger Domain: Surjective but Not Injective
Worked Example: A Map from \(\mathbb{R}^4\) to \(\mathbb{R}^2\)
Consider the linear map \(T:\mathbb{R}^4\to\mathbb{R}^2\) defined by \(T(a,b,c,d)=(a+c,b+d)\). Since \(4>2\), the theorem says that \(T\) cannot be injective. We can also determine directly whether it is surjective and check the kernel.
For any target vector \((x,y)\in\mathbb{R}^2\), choose the input \((x,y,0,0)\). Substitution gives $$ T(x,y,0,0)=(x+0,y+0)=(x,y). $$ Thus every target has a preimage, and \(T\) is surjective.
To find the kernel, set the output equal to \((0,0)\): $$ T(a,b,c,d)=(0,0) \quad\Longleftrightarrow\quad a+c=0,\quad b+d=0. $$ Hence \(c=-a\) and \(d=-b\), so $$ \ker T=\{(a,b,-a,-b):a,b\in\mathbb{R}\}. $$ For instance, \(T(1,0,-1,0)=(1-1,0+0)=(0,0)\), although \((1,0,-1,0)\ne(0,0,0,0)\). This explicitly confirms that \(T\) is not injective.
The image is all of \(\mathbb{R}^2\), so \(\operatorname{rank}T=2\). The kernel has two free parameters \(a,b\), so \(\operatorname{nullity}T=2\). The check \(4=2+2\) agrees with rank-nullity.
This example shows that a larger domain does not prevent a map from being surjective. It prevents the map from being injective: distinct inputs must share an output. Here, for example, \((0,0,0,0)\) and \((1,0,-1,0)\) both map to \((0,0)\).
A Larger Codomain: Injective but Not Surjective
Worked Example: A Map from \(\mathbb{R}^2\) to \(\mathbb{R}^4\)
Define \(S:\mathbb{R}^2\to\mathbb{R}^4\) by \(S(a,b)=(a,b,0,0)\). Since \(2<4\), the dimension theorem rules out surjectivity. We check that injectivity is possible for this dimension comparison.
If \(S(a,b)=S(c,d)\), then $$ (a,b,0,0)=(c,d,0,0), $$ so \(a=c\) and \(b=d\). Therefore the inputs are equal, and \(S\) is injective. Equivalently, its kernel is found by solving $$ S(a,b)=(0,0,0,0) \quad\Longleftrightarrow\quad a=0,\quad b=0, $$ so \(\ker S=\{(0,0)\}\).
The image consists exactly of vectors of the form \((a,b,0,0)\). The target vector \((0,0,1,0)\) is not of this form, so it has no preimage and \(S\) is not surjective. The image has dimension \(2\), and the kernel has dimension \(0\); rank-nullity checks as \(2=2+0\).
The missing target direction matters: an injective map preserves distinctness of inputs, but that does not make every possible target an output. For a map into a larger-dimensional space, its image can occupy only a subspace of that space.
When an Allowed Property Still Fails
The inequalities in the theorem are necessary conditions, not sufficient conditions for a particular map. For example, \(\dim V\leq\dim W\) permits injectivity but does not guarantee it. Similarly, \(\dim V\geq\dim W\) permits surjectivity but does not guarantee it. A map can send many inputs to zero, or have an image that occupies only a small part of its codomain.
Worked Example: Neither Property Holds
Let \(U:\mathbb{R}^3\to\mathbb{R}^2\) be defined by \(U(a,b,c)=(a+b,2a+2b)\). Its outputs have the form \((s,2s)\), where \(s=a+b\), so its image is the line $$ \operatorname{im}U=\{(s,2s):s\in\mathbb{R}\}. $$ This is a one-dimensional subspace of \(\mathbb{R}^2\), and hence \(U\) is not surjective: \((0,1)\) is not in the image, because the first coordinate \(s=0\) would force the second coordinate to be \(2s=0\), not \(1\).
For the kernel, the equation \(U(a,b,c)=(0,0)\) gives \(a+b=0\); the second coordinate equation \(2a+2b=0\) is the same condition. Thus \(b=-a\), while \(c\) is unrestricted: $$ \ker U=\{(a,-a,c):a,c\in\mathbb{R}\}. $$ In particular, \(U(1,-1,0)=(1-1,2-2)=(0,0)\), so \(U\) is not injective. The kernel has dimension \(2\), the image has dimension \(1\), and \(3=2+1\) verifies rank-nullity. Here the larger domain rules out injectivity, but surjectivity also fails because the map’s image is only a line.
Which Properties Are Possible for Some Map?
There is a useful distinction between asking what must be true of a particular map and asking whether some map with the given dimensions can have a property. The dimension inequalities give exact answers to the second question: there exists an injective linear map \(V\to W\) if and only if \(\dim V\leq\dim W\); there exists a surjective linear map \(V\to W\) if and only if \(\dim V\geq\dim W\). The necessity follows from the theorem. The constructions below prove sufficiency.
Proof. Necessity in each statement is the dimension theorem. For sufficiency, let \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) be a basis of \(V\) and \((\mathbf{w}_1,\dots,\mathbf{w}_m)\) a basis of \(W\).
If \(n\leq m\), define a map on the basis vectors by \(T(\mathbf{v}_i)=\mathbf{w}_i\) for \(1\leq i\leq n\), and extend linearly. If \(T(\sum_{i=1}^{n}c_i\mathbf{v}_i)=\mathbf{0}\), then \(\sum_{i=1}^{n}c_i\mathbf{w}_i=\mathbf{0}\). The vectors \(\mathbf{w}_1,\dots,\mathbf{w}_n\) are part of a basis, so they are linearly independent. Thus every \(c_i=0\), which proves injectivity.
If \(n\geq m\), define \(S(\mathbf{v}_i)=\mathbf{w}_i\) for \(1\leq i\leq m\), and \(S(\mathbf{v}_i)=\mathbf{0}\) for \(m<i\leq n\), again extending linearly. Any \(\mathbf{w}\in W\) has an expression \(\mathbf{w}=\sum_{i=1}^{m}d_i\mathbf{w}_i\). Then \(S(\sum_{i=1}^{m}d_i\mathbf{v}_i)=\mathbf{w}\), so \(S\) is surjective. These constructions also cover a zero-dimensional space, with the corresponding empty basis or zero codomain. \(\square\)
Common Mistake: Subtracting from the Wrong Dimension
When using rank-nullity, the nullity is the domain dimension minus the rank, not the codomain dimension minus the rank. For the map \(U:\mathbb{R}^3\to\mathbb{R}^2\) above, a tempting but incorrect calculation is \(\operatorname{nullity}U=2-1=1\), using the codomain dimension \(2\). Rank-nullity instead gives $$ \operatorname{nullity}U=\dim V-\operatorname{rank}U=3-1=2. $$ This matches the explicit kernel \(\{(a,-a,c):a,c\in\mathbb{R}\}\), which has two free parameters.
A related error is to treat a permitted property as automatic. The inequality \(\dim V\geq\dim W\) does not by itself prove that a particular map is surjective; one must show that its image is all of \(W\). Likewise, \(\dim V\leq\dim W\) does not prove that a particular map is injective; one must check that its kernel is trivial or otherwise establish injectivity.
Dimension comparisons are therefore a fast first test. They can rule out injectivity or surjectivity without computing a formula for every output. When the relevant inequality does allow a property, inspect the kernel or image to decide whether the particular map actually has it.
Check Your Understanding
For each question, identify the dimension comparison and explain what it does—and does not—establish.
- If \(\dim V=5\) and \(\dim W=3\), which property is impossible for a linear map \(T:V\to W\)?
- If \(\dim V=2\) and \(\dim W=6\), can \(T\) be surjective? Is injectivity ruled out by dimensions?
- A map \(T:\mathbb{R}^4\to\mathbb{R}^3\) has rank \(2\). Find its nullity and state whether it can be injective or surjective.
- Explain why the condition \(\dim V\geq\dim W\) permits a surjective map but does not prove that a given map is surjective.
- For what dimension comparison can a linear map be both injective and surjective? Give the rank-nullity reasoning behind your answer.
