Three Tests for One Question
An isomorphism preserves the structure of a vector space without losing or missing any vectors. For a linear map \(T:V\to W\), the question is whether every vector of \(W\) is reached and whether distinct vectors of \(V\) remain distinct. The kernel and image translate these two questions into concrete tests. These tests apply whether or not \(V\) and \(W\) have the same dimension.
Throughout, take \(V\) and \(W\) to be vector spaces over \(\mathbb{R}\), and let \(T:V\to W\) be linear. The kernel collects the inputs sent to the zero vector, while the image collects all outputs the map can produce.
The kernel and image answer different questions. A trivial kernel means the only input mapped to zero is zero; it tests whether inputs can collapse together. The image tells us which outputs can be reached; it tests whether the map covers the codomain. For the kernel, “trivial” means \(\ker T=\{\mathbf{0}\}\), not merely that the kernel contains no vectors other than zero as a subset of the domain.
Why the Kernel Tests Injectivity
Proof. First suppose \(T\) is injective. If \(\mathbf{v}\in\ker T\), then \(T(\mathbf{v})=\mathbf{0}=T(\mathbf{0})\), since linearity gives \(T(\mathbf{0})=\mathbf{0}\). Injectivity implies \(\mathbf{v}=\mathbf{0}\). Also, \(\mathbf{0}\in\ker T\), so \(\ker T=\{\mathbf{0}\}\). Conversely, suppose \(\ker T=\{\mathbf{0}\}\), and suppose \(T(\mathbf{v}_1)=T(\mathbf{v}_2)\). Linearity gives \(T(\mathbf{v}_1-\mathbf{v}_2)=T(\mathbf{v}_1)-T(\mathbf{v}_2)=\mathbf{0}\). Thus \(\mathbf{v}_1-\mathbf{v}_2\in\ker T\), so \(\mathbf{v}_1-\mathbf{v}_2=\mathbf{0}\), which means \(\mathbf{v}_1=\mathbf{v}_2\). Hence \(T\) is injective. \(\square\)
The key step is to compare two inputs by subtracting them. If they produce the same output, their difference belongs to the kernel. A nonzero kernel vector therefore gives a direct witness that injectivity fails: if \(\mathbf{v}\ne\mathbf{0}\) and \(T(\mathbf{v})=\mathbf{0}\), then \(T(\mathbf{v})=T(\mathbf{0})\) even though \(\mathbf{v}\ne\mathbf{0}\).
Why the Image Tests Surjectivity
Proof. By definition, \(\operatorname{im}T\) consists exactly of the outputs \(T(\mathbf{v})\) as \(\mathbf{v}\) ranges over \(V\), and it is a subset of \(W\). If \(T\) is surjective, then for every \(\mathbf{w}\in W\) there is a \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\mathbf{w}\). Hence every \(\mathbf{w}\in W\) belongs to \(\operatorname{im}T\), so \(\operatorname{im}T=W\). Conversely, if \(\operatorname{im}T=W\), then every \(\mathbf{w}\in W\) belongs to the image and therefore equals \(T(\mathbf{v})\) for some \(\mathbf{v}\in V\). This is precisely surjectivity. \(\square\)
Combining the two tests gives the desired characterization, with no restriction that the spaces be finite-dimensional or have the same dimension.
Proof. If \(T\) is an isomorphism, it is injective and surjective. The first theorem gives \(\ker T=\{\mathbf{0}\}\), and the second gives \(\operatorname{im}T=W\). Conversely, if both equalities hold, the first theorem implies that \(T\) is injective and the second implies that it is surjective. Since \(T\) is already a linear map, it is an isomorphism by definition. \(\square\)
Matrix Tests for Maps Between Coordinate Spaces
For a linear map \(T:\mathbb{R}^n\to\mathbb{R}^m\) represented by an \(m\times n\) matrix \(\mathbf{A}\), the equation \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) makes both tests computational. The kernel consists of solutions to \(\mathbf{A}\mathbf{x}=\mathbf{0}\); the image is the set of all combinations of the columns of \(\mathbf{A}\). Thus, to test whether the map is an isomorphism, determine whether the homogeneous system has only the zero solution and whether \(\mathbf{A}\mathbf{x}=\mathbf{b}\) can be solved for every \(\mathbf{b}\in\mathbb{R}^m\).
Here is how row reduction expresses those conditions. Row operations are reversible, so they preserve the solutions of a homogeneous system. They also replace \(\mathbf{A}\) by \(\mathbf{R}=\mathbf{E}\mathbf{A}\) for an invertible matrix \(\mathbf{E}\): the new output is \(\mathbf{E}\) times the old output. Consequently, the kernel is unchanged, and the image is all of \(\mathbb{R}^m\) before the operations exactly when it is all of \(\mathbb{R}^m\) afterward.
Justification. In the homogeneous system given by the RREF, a column without a pivot corresponds to a free variable. If one exists, set that free variable to \(1\), set the other free variables to \(0\), and solve for the pivot variables; this produces a nonzero solution. If every column has a pivot, there are no free variables and the homogeneous system has only the zero solution. This proves the injectivity test by the kernel theorem. For surjectivity, if every row has a pivot, the reduced system can be solved for any right-hand side by assigning the free variables value \(0\) and solving for the pivot variables. If some row has no pivot, the RREF has a zero row, so every output of the reduced matrix has zero in that row and the image is not all of \(\mathbb{R}^m\). Reversibility of the row operations transfers this conclusion to the original matrix. The isomorphism test now follows from the kernel–image theorem. \(\square\)
Worked Tests
Worked Example: A Map That Is an Isomorphism
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\), where \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\). Test its kernel and image.
To test the image, solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for an arbitrary \(\mathbf{b}=(b_1,b_2)^\top\). Row-reduce the augmented matrix:
The operations are \(R_2\to R_2-3R_1\), \(R_2\to-\tfrac12R_2\), and \(R_1\to R_1-2R_2\). For every \(\mathbf{b}\), a solution is \(\mathbf{x}=(b_2-2b_1,\frac{3b_1-b_2}{2})^\top\). Therefore every vector in the codomain is reached. Setting \(b_1=b_2=0\) gives the only homogeneous solution \(\mathbf{x}=\mathbf{0}\), so the kernel is trivial as well.
Check: Multiplying the proposed solution back gives \(x_1+2x_2=b_2-2b_1+3b_1-b_2=b_1\) and \(3x_1+4x_2=3b_2-6b_1+6b_1-2b_2=b_2\). Thus \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for every \(\mathbf{b}\), and the map is an isomorphism.
Worked Example: Surjective but Not Injective
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x_1,x_2,x_3)=(x_1+2x_2,x_2+x_3)\). Its matrix is \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&1&1\end{bmatrix}\). Find its kernel and image.
For an arbitrary \((b_1,b_2)\in\mathbb{R}^2\), choose \((x_1,x_2,x_3)=(b_1,0,b_2)\). Then \(T(b_1,0,b_2)=(b_1,b_2)\), so every codomain vector is reached and \(\operatorname{im}T=\mathbb{R}^2\). To find the kernel, solve \(x_1+2x_2=0\) and \(x_2+x_3=0\). Taking \(x_3=t\) gives \(x_2=-t\) and \(x_1=2t\), so \(\ker T=\{t(2,-1,1):t\in\mathbb{R}\}\), which contains nonzero vectors.
Check: For any \(t\), \(T(2t,-t,t)=(2t-2t,-t+t)=(0,0)\). Also \(T(b_1,0,b_2)=(b_1,b_2)\) directly. Thus the map is surjective but not injective, and so is not an isomorphism.
Worked Example: Injective but Not Surjective
Let \(S:\mathbb{R}^2\to\mathbb{R}^3\) be \(S(x,y)=(x,y,0)\). Its matrix is \(\mathbf{B}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\).
If \(S(x,y)=(0,0,0)\), then \(x=0\) and \(y=0\), so \(\ker S=\{(0,0)\}\). The map is injective. Every output has third coordinate \(0\), however, so \(\operatorname{im}S=\{(a,b,0):a,b\in\mathbb{R}\}\), which is not all of \(\mathbb{R}^3\). For example, \((0,0,1)\) is not reached.
Check: Reaching \((0,0,1)\) would require \(x=0\), \(y=0\), and \(0=1\) simultaneously, which is impossible. The kernel is trivial, but the image is not the codomain; therefore \(S\) is not an isomorphism.
A Common Mistake: Checking Only the Kernel
A trivial kernel is enough to conclude that a map is injective, not that it is an isomorphism. For the map \(S(x,y)=(x,y,0)\), the calculation \(S(x,y)=(0,0,0)\Rightarrow x=0,\ y=0\) is correct. The incorrect next step would be to conclude that \(S\) is therefore an isomorphism. That ignores whether the map reaches every vector in its codomain.
The correction is to test the image separately. To reach \((0,0,1)\), one would need \(x=0\), \(y=0\), and \(0=1\); hence this vector is not in the image. The two conditions in the isomorphism test have separate jobs, and neither can be omitted. Similarly, a map with full image need not be injective, as the preceding example shows.
| Kernel | Image | Conclusion |
|---|---|---|
| \(\{\mathbf{0}\}\) | All of \(W\) | Isomorphism |
| \(\{\mathbf{0}\}\) | A proper subset of \(W\) | Injective, not surjective |
| More than \(\{\mathbf{0}\}\) | All of \(W\) | Surjective, not injective |
| More than \(\{\mathbf{0}\}\) | A proper subset of \(W\) | Neither injective nor surjective |
Check Your Understanding
For each map, determine its kernel and image before deciding whether it is an isomorphism.
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) given by \(T(x,y)=(x+y,2x+2y)\), is the map injective, surjective, both, or neither?
- Find the kernel and image of \(T:\mathbb{R}^2\to\mathbb{R}^3\), \(T(x,y)=(x,0,y)\). Is it an isomorphism?
- For a matrix representing a map \(\mathbb{R}^n\to\mathbb{R}^m\), what pivot conditions test injectivity and surjectivity?
- Prove directly from the definitions that if \(\ker T\) contains a nonzero vector, then \(T\) is not injective.
- Give an example of a surjective linear map with a nontrivial kernel, and verify both properties.
