A Practical Route Through Isomorphism Problems
An isomorphism is a linear map that is also bijective. In a problem, these conditions should be checked separately: a bijection need not be linear, and a linear map need not be bijective. When constructing an isomorphism, the most reliable method is to choose a basis in each space and specify where the basis vectors go. When deciding that no isomorphism exists, comparing dimensions can settle the question without searching through possible formulas.
All vector spaces and maps in this tutorial are over \(\mathbb{R}\), and the spaces considered are finite-dimensional. Recall that a map \(T:V\to W\) is linear when \(T(a\mathbf{v}+b\mathbf{w})=aT(\mathbf{v})+bT(\mathbf{w})\) for all \(\mathbf{v},\mathbf{w}\in V\) and \(a,b\in\mathbb{R}\). It is one-to-one if equal outputs imply equal inputs, and onto if every element of \(W\) is an output.
A useful first check is \(T(\mathbf{0})=\mathbf{0}\), which every linear map must satisfy: \(T(\mathbf{0})=T(0\mathbf{v})=0T(\mathbf{v})=\mathbf{0}\). Failure of this check proves that a candidate is not an isomorphism. Passing it does not prove linearity; the defining linearity equation still has to hold. For a linear map, one-to-one can be tested by its kernel: \(T\) is one-to-one exactly when the only vector mapped to \(\mathbf{0}\) is \(\mathbf{0}\). Onto asks a different question: whether every target vector occurs.
Worked Example: A Linear Map That Is Not an Isomorphism
Consider \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(x+2y,3x+6y)\). Its formula is linear because each output coordinate is a linear combination of \(x\) and \(y\). To test whether it is an isomorphism, solve \(T(x,y)=(0,0)\):
The second equation is three times the first. Taking \(y=1\) gives \(x=-2\), so \((-2,1)\ne(0,0)\) and
Thus \(T\) is not one-to-one. It is not onto either: every output has second coordinate three times its first, so \((0,1)\) is not an output. The kernel calculation and the missing target vector directly verify both failures.
Constructing a Map from Basis Data
Suppose \(V\) has an ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\). Every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). Therefore, if we prescribe vectors \(\mathbf{w}_1,\dots,\mathbf{w}_n\in W\), there is a natural rule: send \(\mathbf{v}\) to the same linear combination of the prescribed images. The uniqueness of the coordinates makes this a well-defined rule.
Proof. Each vector of \(V\) has a unique coordinate list relative to \(\mathcal{B}\), so the rule assigns exactly one output to each input. If \(\mathbf{v}=\sum_{i=1}^{n}c_i\mathbf{b}_i\) and \(\mathbf{u}=\sum_{i=1}^{n}d_i\mathbf{b}_i\), then for \(a,b\in\mathbb{R}\), the coordinates of \(a\mathbf{v}+b\mathbf{u}\) are \(ac_i+bd_i\). Consequently, \(T(a\mathbf{v}+b\mathbf{u})=\sum_{i=1}^{n}(ac_i+bd_i)\mathbf{w}_i=aT(\mathbf{v})+bT(\mathbf{u})\), so \(T\) is linear. Any linear map with \(T(\mathbf{b}_i)=\mathbf{w}_i\) must send \(\sum_i c_i\mathbf{b}_i\) to \(\sum_i c_i\mathbf{w}_i\), proving uniqueness.
If the image list is a basis of \(W\), every \(\mathbf{y}\in W\) has a unique expression \(\sum_i c_i\mathbf{w}_i\). The vector \(\sum_i c_i\mathbf{b}_i\) maps to \(\mathbf{y}\), proving onto. If \(T(\sum_i c_i\mathbf{b}_i)=\mathbf{0}\), then \(\sum_i c_i\mathbf{w}_i=\mathbf{0}\); independence of the image basis forces every \(c_i=0\), proving one-to-one. Hence \(T\) is an isomorphism.
Conversely, suppose \(T\) is an isomorphism. If \(\sum_i c_i\mathbf{w}_i=\mathbf{0}\), then \(T(\sum_i c_i\mathbf{b}_i)=\mathbf{0}\). Since \(T\) is one-to-one, its kernel contains only \(\mathbf{0}\), so \(\sum_i c_i\mathbf{b}_i=\mathbf{0}\), and independence of \(\mathcal{B}\) gives \(c_i=0\) for every \(i\). Thus the image list is independent. For any \(\mathbf{y}\in W\), surjectivity gives some \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{y}\). Writing \(\mathbf{v}=\sum_i c_i\mathbf{b}_i\) gives \(\mathbf{y}=\sum_i c_i\mathbf{w}_i\), so the image list spans \(W\). It is therefore a basis. \(\square\)
The theorem turns construction into a finite set of choices: select a basis in the domain, choose a basis in the target, and pair their vectors. The resulting map is determined on every vector by its coordinates. Its inverse is obtained by reversing the pairing.
Worked Example: Constructing an Isomorphism from a Nonstandard Basis
Construct an isomorphism from \(P_2\), the real polynomials of degree at most \(2\), to \(\mathbb{R}^3\) using the ordered list \(\mathcal{B}=(1+x,\;x+x^2,\;1+x^2)\) in \(P_2\), paired in order with the standard basis of \(\mathbb{R}^3\).
First verify that \(\mathcal{B}\) is a basis. Relative to the standard polynomial basis \((1,x,x^2)\), its coefficient columns form the matrix
The row operations are \(R_2\to R_2-R_1\), followed by \(R_3\to R_3-R_2\). There is a pivot in every column, so the three coefficient columns are independent; as \(P_2\) has dimension \(3\), they form a basis. If \(p(x)=a+bx+cx^2\), write \(p=\alpha(1+x)+\beta(x+x^2)+\gamma(1+x^2)\). Comparing coefficients gives \(\alpha+\gamma=a\), \(\alpha+\beta=b\), and \(\beta+\gamma=c\). Solving these equations yields
Define \(T(p)=(\alpha,\beta,\gamma)\). This sends the chosen basis to the standard basis, so the basis-image criterion proves that \(T\) is an isomorphism. For example, for \(p(x)=2-x+3x^2\), the coordinates are \(\alpha=(2-1-3)/2=-1\), \(\beta=(-2-1+3)/2=0\), and \(\gamma=(2+1+3)/2=3\). Thus \(T(p)=(-1,0,3)\). Check the result by reconstructing the polynomial: \(-1(1+x)+0(x+x^2)+3(1+x^2)=2-x+3x^2\).
The inverse map is also explicit: \(T^{-1}(u,v,w)=u(1+x)+v(x+x^2)+w(1+x^2)\). Indeed, its coordinates in \(\mathcal{B}\) are \((u,v,w)\), so \(T(T^{-1}(u,v,w))=(u,v,w)\). Conversely, the unique \(\mathcal{B}\)-coordinates of \(p\) reconstruct \(p\), so \(T^{-1}(T(p))=p\). These checks confirm both inverse identities.
Dimensions as a Classification Test
For finite-dimensional spaces, dimension gives a complete answer to the existence question. The Dimension Preservation Theorem, established in the earlier material on isomorphisms, states that if finite-dimensional real vector spaces \(V\) and \(W\) are isomorphic, then \(\dim V=\dim W\). This is a necessary condition. The basis-image criterion proves that it is also sufficient.
Proof. If \(V\) and \(W\) are isomorphic, the Dimension Preservation Theorem gives \(\dim V=\dim W\). Conversely, suppose their common dimension is \(n\). Choose bases \((\mathbf{b}_1,\dots,\mathbf{b}_n)\) of \(V\) and \((\mathbf{w}_1,\dots,\mathbf{w}_n)\) of \(W\). Assign \(T(\mathbf{b}_i)=\mathbf{w}_i\) for every \(i\), and extend the assignment linearly. By the basis-image criterion this produces an isomorphism \(T:V\to W\). Both directions hold, proving the equivalence. \(\square\)
This classification answers whether an isomorphism exists; it does not choose a unique one. Even when dimensions match, different choices of target basis generally give different maps. Conversely, when dimensions differ, no formula or clever choice of coordinates can produce an isomorphism.
Worked Example: Ruling Out an Isomorphism by Dimension
Can \(P_2\) be isomorphic to the space \(M_{2\times2}(\mathbb{R})\) of real \(2\times2\) matrices? The list \((1,x,x^2)\) is a basis for \(P_2\): every polynomial \(a+bx+cx^2\) has those coefficients uniquely, so \(\dim P_2=3\). The four matrices
form a basis of \(M_{2\times2}(\mathbb{R})\), since every matrix \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\) is their linear combination with coefficients \(a,b,c,d\), uniquely. Hence this matrix space has dimension \(4\). Because \(3\ne4\), the classification theorem shows that the spaces are not isomorphic. No candidate formula needs to be tested.
Common Pitfall: A Bijection Is Not Enough
Consider \(f:\mathbb{R}\to\mathbb{R}\) given by \(f(x)=x+1\). This map is bijective: for any \(y\in\mathbb{R}\), the unique input that gives \(y\) is \(x=y-1\). A wrong conclusion is that \(f\) is therefore an isomorphism. The correction is to check linearity as well: \(f(0)=1\ne0\), so \(f\) cannot be linear and is not an isomorphism.
There is a related trap in the basis construction. Matching independent image vectors is not enough if they fail to span the target. For example, three independent vectors in \(\mathbb{R}^4\) cannot be the image of a basis under an isomorphism onto \(\mathbb{R}^4\), because they do not form a basis of that target. The basis-image criterion requires both independence and spanning.
Check linearity using its definition; \(T(\mathbf{0})=\mathbf{0}\) is a quick necessary check.
For a linear map, check that the kernel is \(\{\mathbf{0}\}\) and that every target vector is attained.
Specify the images of a domain basis as a target basis and extend by linearity.
Equal finite dimensions permit a basis-matching construction; unequal dimensions rule out an isomorphism.
Check Your Understanding
For each problem, identify which condition or construction settles the question.
- Show that \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x-y,x+y)\), is an isomorphism, or give a specific reason it is not.
- Can \(\mathbb{R}^3\) be isomorphic to \(P_4\), the polynomials of degree at most \(4\)? State the dimensions used.
- Let \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\) be a basis of \(V\), and let \((\mathbf{w}_1,\mathbf{w}_2)\) be a basis of \(W\). Write the image of \(3\mathbf{b}_1-2\mathbf{b}_2\) under the map matching these bases.
- Give a bijection from \(\mathbb{R}\) to \(\mathbb{R}\) that is not an isomorphism, and use a linearity test to justify your answer.
- Explain why an independent image list of the right length is sufficient for the basis-image criterion only when it also spans the target.
