Comparing the Domain with the Image
For a linear map, injectivity asks whether distinct inputs can ever produce the same output. In finite-dimensional spaces, the Rank-Nullity Theorem gives a dimension-based way to answer that question: the dimension of the image reaches the dimension of the domain exactly when no nonzero input is sent to zero. The codomain need not be finite-dimensional for this test; what matters is that the domain has finite dimension.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to W\) be linear. 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 map is injective if \(T(\mathbf{v}_1)=T(\mathbf{v}_2)\) implies \(\mathbf{v}_1=\mathbf{v}_2\). The dimension of the image is also called the rank of the map.
The equation says that each dimension in the domain is accounted for either by directions that are lost in the kernel or by directions represented in the image. Consequently, the image dimension cannot exceed the domain dimension. To reach equality, the kernel must contribute no dimensions at all.
Proof. By the Rank-Nullity Theorem, $$ \dim V=\dim(\ker T)+\dim(\operatorname{im}T). $$ If \(\dim(\operatorname{im}T)=\dim V\), subtracting gives \(\dim(\ker T)=0\). A vector space has dimension zero exactly when it contains only its zero vector, so \(\ker T=\{\mathbf{0}\}\). Conversely, if \(\ker T=\{\mathbf{0}\}\), then \(\dim(\ker T)=0\), and rank-nullity gives \(\dim(\operatorname{im}T)=\dim V\). It remains to relate the zero kernel to injectivity. If \(T\) is injective and \(T(\mathbf{v})=\mathbf{0}\), then \(T(\mathbf{v})=T(\mathbf{0})\), so \(\mathbf{v}=\mathbf{0}\). Thus the kernel is trivial. If the kernel is trivial and \(T(\mathbf{v}_1)=T(\mathbf{v}_2)\), linearity gives \(T(\mathbf{v}_1-\mathbf{v}_2)=\mathbf{0}\). Hence \(\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}\), so \(\mathbf{v}_1=\mathbf{v}_2\) and \(T\) is injective. The two implications establish the equivalence. \(\square\)
Applying the Test in Coordinate Spaces
Worked Example: An Injective Map into a Larger Space
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(a,b)=(a,b,a+b)\). Determine the dimension of its image and decide whether it is injective.
The formula is linear because each output coordinate is a linear combination of the input coordinates. Its values can be written $$ T(a,b)=a(1,0,1)+b(0,1,1). $$ Thus \(\operatorname{im}T\) is spanned by \((1,0,1)\) and \((0,1,1)\). These two vectors are independent: if $$ \alpha(1,0,1)+\beta(0,1,1)=(0,0,0), $$ then the first coordinate gives \(\alpha=0\), and the second gives \(\beta=0\). They therefore form a basis for the image, so \(\dim(\operatorname{im}T)=2\). The domain also has dimension \(2\), with basis \((1,0),(0,1)\). By the dimension test, \(T\) is injective.
Check: Directly, \(T(a,b)=(0,0,0)\) requires \(a=0\) from the first coordinate and \(b=0\) from the second. Thus the kernel is trivial, in agreement with the dimension test. The third coordinate is then \(a+b=0\) as well.
Worked Example: A Noninjective Map with a Smaller Image
Define \(S:\mathbb{R}^3\to\mathbb{R}^2\) by \(S(x,y,z)=(x+y,y+z)\). Find the image dimension and use it to decide whether \(S\) is injective.
For every \((u,v)\in\mathbb{R}^2\), choose \((x,y,z)=(u,0,v)\). Then $$ S(u,0,v)=(u+0,0+v)=(u,v). $$ Every vector in \(\mathbb{R}^2\) is therefore in the image, so \(\operatorname{im}S=\mathbb{R}^2\) and \(\dim(\operatorname{im}S)=2\). The domain \(\mathbb{R}^3\) has dimension \(3\), so the image dimension is smaller than the domain dimension. The dimension test shows that \(S\) is not injective.
Check: Solving \(S(x,y,z)=(0,0)\) gives \(x+y=0\) and \(y+z=0\). Setting \(y=t\) gives \(x=-t\) and \(z=-t\), so the kernel consists of \((-t,t,-t)\). In particular, \(S(1,-1,1)=(0,0)\), although \((1,-1,1)\ne(0,0,0)\). This verifies noninjectivity directly. Rank-nullity also checks the dimensions: \(3=1+2\), since the kernel is spanned by \((-1,1,-1)\) and has dimension \(1\).
Applying the Test to Polynomial Maps
Worked Example: Differentiation on Polynomials
Let \(P_2\) be the real vector space of polynomials of degree at most \(2\), and \(P_1\) the real vector space of polynomials of degree at most \(1\). Define \(D:P_2\to P_1\) by \(D(p)=p'\). Determine whether \(D\) is injective by computing the image dimension.
Every \(p\in P_2\) has the form \(p(x)=a+bx+cx^2\), so $$ D(p)(x)=b+2cx. $$ The domain has basis \((1,x,x^2)\), hence dimension \(3\). The image consists of all polynomials \(b+2cx\). As \(b,c\) range over \(\mathbb{R}\), these outputs fill \(P_1\): a desired polynomial \(r+sx\) is obtained by choosing \(b=r\) and \(c=s/2\). Equivalently, the images of the domain basis vectors are \(D(1)=0\), \(D(x)=1\), and \(D(x^2)=2x\), which span \(P_1\). The polynomials \(1\) and \(2x\) are independent, since \(\alpha+2\beta x=0\) forces both coefficients \(\alpha\) and \(2\beta\) to be zero. Therefore \(\dim(\operatorname{im}D)=2\).
Since the image dimension \(2\) is smaller than the domain dimension \(3\), the dimension test says that \(D\) is not injective.
Check: The equation \(D(p)=0\) becomes \(b+2cx=0\). Equality of polynomials forces \(b=0\) and \(2c=0\), so \(c=0\), while \(a\) is unrestricted. Thus \(\ker D=\operatorname{span}\{1\}\), which has dimension \(1\). Rank-nullity gives \(3=1+2\), and the nonzero polynomial \(1\) is indeed sent to zero.
What the Dimension Comparison Does—and Does Not—Say
The test compares \(\dim V\) with \(\dim(\operatorname{im}T)\), not with \(\dim W\). The codomain may be larger than the domain, and the image may occupy only part of it. The first worked example illustrates this: its codomain has dimension \(3\), but its image has dimension \(2\), equal to the domain dimension. That equality is enough to establish injectivity.
Conversely, if the image dimension is less than the domain dimension, rank-nullity gives a positive kernel dimension. A positive-dimensional kernel contains a nonzero vector, so at least one nonzero input is sent to zero. Such a map cannot be injective. This is not merely a numerical pattern: rank-nullity explains precisely why the dimension comparison detects whether any information is lost.
The criterion requires a finite-dimensional domain. In that setting, rank-nullity applies and the image has a well-defined finite dimension. The codomain itself need not be finite-dimensional. In practical use, first determine the dimension of the domain, then find a basis or other justification for the image dimension, and finally compare the two. Finding a few outputs that are independent gives a lower bound on image dimension; to establish its exact dimension, also show that those outputs span the entire image.
Common Pitfall: Comparing with the Codomain
A tempting but incorrect rule is “\(T\) is injective exactly when the domain and codomain dimensions are equal.” This uses the wrong comparison. For a concrete correction, define \(U:\mathbb{R}^2\to\mathbb{R}^3\) by \(U(a,b)=(a,b,0)\). The domain has dimension \(2\), while the codomain has dimension \(3\), so those dimensions are not equal.
But the image is spanned by \((1,0,0)\) and \((0,1,0)\). These vectors are independent: \(\alpha(1,0,0)+\beta(0,1,0)=(0,0,0)\) gives \(\alpha=0\) and \(\beta=0\). Thus the image has dimension \(2\), equal to the domain dimension, and the correct test says \(U\) is injective. Directly, \(U(a,b)=(0,0,0)\) forces \(a=0\) and \(b=0\).
The incorrect calculation would compare \(\dim(\mathbb{R}^2)=2\) with \(\dim(\mathbb{R}^3)=3\) and conclude “not injective.” The correction is to compute \(\dim(\operatorname{im}U)=2\), not to substitute the codomain dimension. The codomain records where outputs are allowed to lie; the image records which outputs the map actually produces.
Check Your Understanding
Use the image dimension and the dimension test in each question. When deciding injectivity, explain which dimensions are being compared.
- A linear map has a finite-dimensional domain of dimension \(4\) and an image of dimension \(4\). What can you conclude about injectivity?
- A linear map \(T:\mathbb{R}^3\to\mathbb{R}^2\) has image dimension \(1\). What is the dimension of its kernel, and is \(T\) injective?
- For \(T:\mathbb{R}^2\to\mathbb{R}^3\), \(T(a,b)=(a,0,b)\), find a basis for the image and decide whether \(T\) is injective.
- For differentiation \(D:P_3\to P_2\), compute the dimensions of the domain and image, then apply the dimension test.
- State the finite-dimensional hypothesis in the dimension test and explain why comparing the codomain dimension is not the rule.
