Four Checks Before Calling a Map an Isomorphism
An isomorphism is a particularly structured correspondence between vector spaces: it must respect the vector-space operations and pair every vector in one space with exactly one vector in the other. Common errors arise when one of these requirements is quietly dropped. A bijection of sets need not be linear; a linear map need not be one-to-one or onto; and matching dimensions do not guarantee that a particular map is an isomorphism.
Here, spaces and maps are over \(\mathbb{R}\), and all spaces discussed in dimension statements are finite-dimensional. The dimension of a vector space is the number of vectors in any basis. The tests in this tutorial are useful in sequence: first check whether an isomorphism is even possible from the dimensions, then check the map itself for linearity, injectivity, and surjectivity.
The word “bijective” by itself describes a property of a map between sets. It says nothing about whether the map respects addition and scalar multiplication. Conversely, linearity alone does not ensure that distinct inputs have distinct outputs or that every target vector is reached. These are separate requirements, so a correct argument must address each one.
Dimension Mismatch Rules Out an Isomorphism
Proof. Suppose \(T:V\to W\) is an isomorphism, and let \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) be a basis of \(V\), where \(n=\dim V\). The list \((T(\mathbf{v}_1),\dots,T(\mathbf{v}_n))\) spans \(W\): for any \(\mathbf{w}\in W\), surjectivity gives \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\mathbf{w}\). Write \(\mathbf{v}=\sum_{i=1}^n c_i\mathbf{v}_i\). Linearity then gives \(\mathbf{w}=\sum_{i=1}^n c_iT(\mathbf{v}_i)\). The list is also linearly independent. If \(\sum_{i=1}^n c_iT(\mathbf{v}_i)=\mathbf{0}\), linearity gives \(T(\sum_{i=1}^n c_i\mathbf{v}_i)=\mathbf{0}=T(\mathbf{0})\). Injectivity implies \(\sum_{i=1}^n c_i\mathbf{v}_i=\mathbf{0}\), so basis independence gives \(c_i=0\) for every \(i\). Therefore the images form a basis of \(W\), and \(\dim W=n=\dim V\). This argument also covers \(n=0\): the empty list spans \(W\) only when \(W=\{\mathbf{0}\}\), so both dimensions are zero. The contrapositive gives the final claim. \(\square\)
This is a necessary condition, not a test that proves a proposed map is an isomorphism. Unequal dimensions rule out an isomorphism immediately. Equal dimensions merely remove that particular obstacle; linearity, injectivity, and surjectivity still need attention.
Worked Example: A Linear Map Between Spaces of Different Dimensions
Consider \(T:\mathbb{R}^2\to\mathbb{R}^3\) defined by \(T(x,y)=(x,y,0)\). The dimensions are \(2\) and \(3\), so the dimension theorem already rules out an isomorphism. We can also identify which properties the map does and does not have.
For \(\mathbf{u}=(x_1,y_1)\), \(\mathbf{v}=(x_2,y_2)\), and \(c\in\mathbb{R}\),
and \(T(c\mathbf{u})=(cx_1,cy_1,0)=cT(\mathbf{u})\). Hence \(T\) is linear. It is injective: if \(T(x_1,y_1)=T(x_2,y_2)\), then \((x_1,y_1,0)=(x_2,y_2,0)\), so \(x_1=x_2\) and \(y_1=y_2\). But it is not surjective, because \((0,0,1)\in\mathbb{R}^3\) cannot equal \((x,y,0)\) for any \(x,y\). It is linear and injective, but not an isomorphism.
Check: \(T(2,-1)=(2,-1,0)\), and \(T(2,-1)\) has third coordinate \(0\), confirming directly that this output lies in the plane missed by the map.
Dimension mismatch does not mean there cannot be any bijection between the underlying sets. It means there cannot be a linear bijection, which is what an isomorphism requires. When dimensions differ, searching for a more clever formula cannot repair this obstruction.
A Bijection May Fail to Be Linear
Worked Example: A Nonlinear Bijection of \(\mathbb{R}\)
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^3\). This map is bijective: every real \(y\) has a real cube root \(x=\sqrt[3]{y}\), so \(f(x)=y\); and if \(f(x)=f(z)\), then \(x^3=z^3\), which implies \(x=z\) because the real cube function is strictly increasing.
However, \(f\) is not linear. Linearity would require \(f(1+1)=f(1)+f(1)\). The two sides are
Since \(8\ne2\), additivity fails. Therefore \(f\) is a bijection of sets but not an isomorphism of real vector spaces. No computation of its inverse can compensate for this failure: linearity is required of the map itself.
Proof. By definition, an isomorphism must be linear. A linear map must satisfy both \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) for every pair of inputs and \(T(c\mathbf{u})=cT(\mathbf{u})\) for every scalar and input. If either required equality fails for even one choice, the map is not linear. It therefore cannot meet the definition of an isomorphism, regardless of whether it is bijective. \(\square\)
Linear Does Not Mean Injective or Surjective
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 \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). These sets give direct tests for two separate properties.
Proof. First suppose \(T\) is injective. If \(\mathbf{v}\in\ker T\), then \(T(\mathbf{v})=\mathbf{0}=T(\mathbf{0})\), so injectivity gives \(\mathbf{v}=\mathbf{0}\). Since \(T(\mathbf{0})=\mathbf{0}\) for a linear map, \(\ker T=\{\mathbf{0}\}\). Conversely, suppose \(\ker T=\{\mathbf{0}\}\) and \(T(\mathbf{u})=T(\mathbf{v})\). Linearity gives \(T(\mathbf{u}-\mathbf{v})=\mathbf{0}\), so \(\mathbf{u}-\mathbf{v}\in\ker T\). Thus \(\mathbf{u}-\mathbf{v}=\mathbf{0}\), and \(\mathbf{u}=\mathbf{v}\); \(T\) is injective. For surjectivity, its definition says precisely that every \(\mathbf{w}\in W\) equals \(T(\mathbf{v})\) for some \(\mathbf{v}\in V\). This is exactly the statement that every \(\mathbf{w}\in W\) belongs to \(\operatorname{im}T\), or \(\operatorname{im}T=W\). \(\square\)
Worked Example: Onto but Not One-to-One
Define \(S:\mathbb{R}^3\to\mathbb{R}^2\) by \(S(x,y,z)=(x,y)\). For \(\mathbf{u}=(x_1,y_1,z_1)\), \(\mathbf{v}=(x_2,y_2,z_2)\), and \(c\in\mathbb{R}\),
and \(S(c\mathbf{u})=(cx_1,cy_1)=cS(\mathbf{u})\), so \(S\) is linear. For any target \((a,b)\in\mathbb{R}^2\), \(S(a,b,0)=(a,b)\), so \(S\) is surjective. But \(S(0,0,0)=S(0,0,1)=(0,0)\), although \((0,0,0)\ne(0,0,1)\). Thus \(S\) is not injective and cannot be an isomorphism.
Check: The nonzero input \((0,0,1)\) is in the kernel, since \(S(0,0,1)=(0,0)\). The kernel test gives the same conclusion.
Worked Example: Equal Dimensions Do Not Guarantee an Isomorphism
Define \(U:\mathbb{R}^2\to\mathbb{R}^2\) by \(U(x,y)=(x,0)\). The domain and codomain both have dimension \(2\), and \(U\) is linear: \(U((x_1,y_1)+(x_2,y_2))=(x_1+x_2,0)=U(x_1,y_1)+U(x_2,y_2)\), while \(U(c(x,y))=(cx,0)=cU(x,y)\).
Nevertheless, \(U\) is not injective because \(U(0,0)=(0,0)=U(0,1)\) with different inputs. It is not surjective either: every output has second coordinate \(0\), so \((0,1)\) is not in the image. Equal dimensions make an isomorphism possible, but do not make this particular map one.
Check: The kernel contains \((0,1)\ne(0,0)\), and the image omits \((0,1)\). These verify failure of injectivity and surjectivity separately.
Using the Checks Together
For finite-dimensional spaces of equal dimension, a useful additional result is that a linear map is injective if and only if it is surjective. This fact does not mean the properties can be ignored: it applies only after linearity and equality of finite dimensions have been established. The basis-count property behind it is that in an \(n\)-dimensional space, any \(n\) linearly independent vectors form a basis, and any \(n\) spanning vectors form a basis.
Proof. Let \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) be a basis of \(V\). If \(T\) is injective, the list \((T(\mathbf{v}_1),\dots,T(\mathbf{v}_n))\) is linearly independent: a relation among its vectors gives \(T(\sum_i c_i\mathbf{v}_i)=\mathbf{0}=T(\mathbf{0})\), so injectivity gives \(\sum_i c_i\mathbf{v}_i=\mathbf{0}\), and hence every \(c_i=0\). The list has \(n\) independent vectors in the \(n\)-dimensional space \(W\), so it is a basis of \(W\). Thus the image of \(T\) contains a basis of \(W\), and \(T\) is surjective. Conversely, if \(T\) is surjective, the list \((T(\mathbf{v}_1),\dots,T(\mathbf{v}_n))\) spans \(W\): any \(\mathbf{w}\in W\) equals \(T(\mathbf{v})\) for some \(\mathbf{v}\in V\), and expressing \(\mathbf{v}\) in the chosen basis expresses \(\mathbf{w}\) as a combination of the image list. A spanning list of \(n\) vectors in an \(n\)-dimensional space is a basis, so the image list is independent. If \(T(\mathbf{v})=\mathbf{0}\), express \(\mathbf{v}=\sum_i c_i\mathbf{v}_i\). Then \(\sum_i c_iT(\mathbf{v}_i)=\mathbf{0}\); independence gives \(c_i=0\) for every \(i\), so \(\mathbf{v}=\mathbf{0}\). The kernel test, or the definition directly, now gives injectivity. When \(n=0\), both spaces are \(\{\mathbf{0}\}\), and the unique map between them is both injective and surjective, so the conclusion still holds. \(\square\)
A common mistake is to reason, “The dimensions match, so the map is an isomorphism.” That conclusion does not follow. In the last example the dimensions match and the map is linear, but both injectivity and surjectivity fail. Another mistake is to show only that every target vector is reached: the map \(S\) above is onto, but it identifies distinct inputs. The reverse mistake is to check only that distinct inputs have distinct outputs: \(T\) above is injective, but misses a target vector.
For a proposed isomorphism, keep the requirements distinct. A dimension mismatch rules out an isomorphism between finite-dimensional spaces. With matching dimensions, verify that the map is linear, then establish injectivity and surjectivity (or use their equivalence for a linear map between equal finite dimensions). A single failed condition is enough to reject the claim; satisfying one condition does not silently establish the others.
Check Your Understanding
For each claim, identify which isomorphism requirement must be checked or which one fails.
- Why can no finite-dimensional real vector space of dimension \(3\) be isomorphic to one of dimension \(2\)?
- For \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=x^3\), show a specific addition test that proves \(f\) is not linear.
- For \(T:\mathbb{R}^2\to\mathbb{R}^3\), \(T(x,y)=(x,y,0)\), give a target vector not in the image.
- For \(S:\mathbb{R}^3\to\mathbb{R}^2\), \(S(x,y,z)=(x,y)\), find a nonzero vector in the kernel.
- What additional hypotheses are needed to infer surjectivity from injectivity using the equal-dimension theorem?
