What an Isomorphism Does to a Basis
A basis is a list that is both linearly independent and spanning: its vectors give unique coordinates for every vector in the space. An isomorphism is a linear map that is also a bijection. These two features have complementary roles. Injectivity prevents distinct combinations of basis vectors from becoming indistinguishable, while surjectivity ensures that every vector in the target is reached. Together, they show that an isomorphism carries a basis to a basis. Comparing the lengths of those bases then proves that the dimensions are equal.
We work over \(\mathbb{R}\). A list \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) in a vector space \(V\) is a basis if it is linearly independent and spans \(V\). A vector space is finite-dimensional if it has a finite basis. Its dimension \(\dim V\) is the number of vectors in a basis; the basis-cardinality theorem from the earlier material on bases and dimension ensures that this number is independent of the chosen basis.
The image list keeps the same number and order of entries as the original list. The claim is about this list being a basis of \(W\), not merely about its entries being distinct. Linear independence and spanning must both be established. We first prove each part directly from the definitions.
Why the Image List Is a Basis
Proof. First suppose \(n>0\). To prove independence, assume that scalars \(c_1,\dots,c_n\) satisfy \(c_1T(\mathbf{b}_1)+\cdots+c_nT(\mathbf{b}_n)=\mathbf{0}\). Linearity gives \(T(c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n)=\mathbf{0}=T(\mathbf{0})\). Since \(T\) is injective, equal images have equal inputs, so \(c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n=\mathbf{0}\). The original list is linearly independent, hence \(c_1=\cdots=c_n=0\). Thus its image list is linearly independent.
To prove spanning, take any \(\mathbf{w}\in W\). Since \(T\) is surjective, there is a \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{w}\). Since \(\mathcal{B}\) spans \(V\), there are scalars \(d_1,\dots,d_n\) such that \(\mathbf{v}=d_1\mathbf{b}_1+\cdots+d_n\mathbf{b}_n\). Applying linearity gives \(\mathbf{w}=T(\mathbf{v})=d_1T(\mathbf{b}_1)+\cdots+d_nT(\mathbf{b}_n)\). Every vector of \(W\) is therefore a linear combination of the image list. It spans \(W\), so it is a basis. If \(n=0\), then \(V=\{\mathbf{0}\}\). Surjectivity implies \(W=\{\mathbf{0}\}\), whose empty list is a basis. The claim holds in this case as well. \(\square\)
The proof separates the two hypotheses on purpose. Injectivity is used to transfer a relation among image vectors back to a relation among the original basis vectors. Surjectivity is used to start with an arbitrary target vector and express it using image vectors. Neither part of the argument establishes the other property by itself.
Proof. Choose a basis \(\mathcal{B}\) of \(V\). By the theorem, the image list \(T(\mathcal{B})\) is a basis of \(W\). The two lists have the same number of entries, since the image list contains one image for each vector in \(\mathcal{B}\). By the definition of dimension, the length of \(\mathcal{B}\) is \(\dim V\) and the length of \(T(\mathcal{B})\) is \(\dim W\). Therefore \(\dim V=\dim W\). In particular, \(W\) is finite-dimensional even if that was not assumed separately: the finite image list is a basis of \(W\). \(\square\)
Worked Examples
Worked Example: Coefficients Give a Basis of a Polynomial Space
Let \(P_1\) be the real vector space of polynomials of degree at most \(1\), and define \(T:\mathbb{R}^2\to P_1\) by \(T((a,b))=a+bx\). Track the standard basis of \(\mathbb{R}^2\).
The map is linear: for \((a,b),(c,d)\in\mathbb{R}^2\) and \(r,s\in\mathbb{R}\), \(T(r(a,b)+s(c,d))=T((ra+sc,rb+sd))=(ra+sc)+(rb+sd)x=rT((a,b))+sT((c,d))\). Every polynomial in \(P_1\) has the form \(a+bx\), so it is \(T((a,b))\); thus \(T\) is onto. If \(T((a,b))\) is the zero polynomial, its constant and \(x\)-coefficients are both zero, so \(a=b=0\); thus \(T\) is one-to-one.
The standard basis is \((\mathbf{e}_1,\mathbf{e}_2)=((1,0),(0,1))\). Its image list is \(T(\mathbf{e}_1)=1\) and \(T(\mathbf{e}_2)=x\), giving \((1,x)\). To check directly, if \(c_1\cdot 1+c_2x\) is the zero polynomial, its coefficients give \(c_1=c_2=0\). Also, an arbitrary \(a+bx\in P_1\) equals \(a\cdot 1+b x\). So \((1,x)\) is independent and spans \(P_1\), as the theorem predicts.
Check: \(T(a,b)=a+bx\) recovers the coefficients \(a,b\) of the input pair, so the displayed basis represents every target polynomial and only the zero pair maps to the zero polynomial.
Worked Example: A Basis of a Plane in \(\mathbb{R}^3\)
Define the subspace \(W=\{(a,b,a+2b):a,b\in\mathbb{R}\}\) of \(\mathbb{R}^3\), and set \(T:\mathbb{R}^2\to W\) by \(T((u,v))=(u,v,u+2v)\). Show how the standard basis maps to a basis of \(W\).
The formula is linear, since for scalars \(r,s\) and pairs \((u,v),(p,q)\), \(T(r(u,v)+s(p,q))=T((ru+sp,rv+sq))\) \(=(ru+sp,rv+sq,r(u+2v)+s(p+2q))=rT((u,v))+sT((p,q))\). The map is one-to-one: if \(T((u,v))=(0,0,0)\), its first two coordinates give \(u=0\) and \(v=0\). It is onto \(W\): every \((a,b,a+2b)\in W\) equals \(T((a,b))\). Hence \(T\) is an isomorphism.
The standard basis vectors map to \(T(\mathbf{e}_1)=(1,0,1)\) and \(T(\mathbf{e}_2)=(0,1,2)\). Let \(\mathbf{w}=(a,b,a+2b)\in W\). Then \(\mathbf{w}=a(1,0,1)+b(0,1,2)\), so these two vectors span \(W\). For independence, if \(c_1(1,0,1)+c_2(0,1,2)=(0,0,0)\), the first coordinate gives \(c_1=0\) and the second gives \(c_2=0\). Thus \(((1,0,1),(0,1,2))\) is a basis of \(W\).
Check: Substituting the coefficients back gives \(a(1,0,1)+b(0,1,2)=(a,b,a+2b)\), exactly the general vector in \(W\). The image list is a basis and has two vectors, matching \(\dim\mathbb{R}^2=2\).
Worked Example: Shifting Polynomials Preserves a Basis
On \(P_2\), the real polynomials of degree at most \(2\), define \(T(p)(x)=p(x+1)\). Apply \(T\) to the standard basis \((1,x,x^2)\), and verify the resulting list is a basis.
For polynomials \(p,q\) and scalars \(r,s\), \(T(rp+sq)(x)=(rp+sq)(x+1)=rp(x+1)+sq(x+1)\). Thus \(T\) is linear. It is bijective: the map \(S(q)(x)=q(x-1)\) is its inverse, because \(S(T(p))(x)=T(p)(x-1)=p((x-1)+1)=p(x)\) and \(T(S(q))(x)=S(q)(x+1)=q((x+1)-1)=q(x)\).
The basis vectors have images \(T(1)=1\), \(T(x)=x+1\), and \(T(x^2)=(x+1)^2=x^2+2x+1\). To check independence and spanning by coefficients, use the standard coordinates relative to \((1,x,x^2)\). The columns of the image vectors are \(\begin{bmatrix}1&1&1\\0&1&2\\0&0&1\end{bmatrix}\). Row operations give
There is a pivot in every column, so the coefficient columns are independent and span \(\mathbb{R}^3\). Equivalently, the three image polynomials are independent and span \(P_2\).
Check: Applying the inverse shift \(S(q)(x)=q(x-1)\) to the three images returns \(1,x,x^2\), respectively. This confirms that the map used is bijective and that its displayed basis images are the correct ones.
Common Pitfall: Injectivity Alone Is Not Enough
A frequent error is to prove that the images of a basis are independent and then conclude they are a basis of the target. Independence alone does not prove spanning. For example, consider \(J:\mathbb{R}^2\to\mathbb{R}^3\) given by \(J((a,b))=(a,b,0)\). The standard basis images are \((1,0,0)\) and \((0,1,0)\). They are independent: a combination equal to \((0,0,0)\) has first coordinate \(c_1=0\) and second coordinate \(c_2=0\). But they do not span \(\mathbb{R}^3\), because \((0,0,1)\) cannot be written as \((a,b,0)\). This map is one-to-one but not onto, so it is not an isomorphism. In the theorem, surjectivity is the step that ensures spanning in the codomain.
Conversely, checking only that a map is onto would not justify the independence step: that step uses injectivity to infer that equal images have equal inputs. The proof of basis preservation needs both hypotheses for different reasons. A useful way to organize a solution is to write “independent” and “spanning” as separate goals, then identify which property of the isomorphism proves each one.
Why Dimension Is an Isomorphism Invariant
The dimension equality is a consequence of a structural fact, not an accidental agreement between coordinate counts. An isomorphism preserves the linear-combination relationships that define a basis: independence is preserved because no nonzero relation can collapse to zero, and spanning is preserved because every target vector has a preimage. Since the image basis has exactly one entry for each vector in the original basis, the number of basis vectors is unchanged.
This conclusion gives a necessary condition for isomorphism: two finite-dimensional real vector spaces with different dimensions cannot be isomorphic. If their dimensions differ, no linear bijection can exist, because any such map would have to send a basis to a basis of the same length. The converse—that equal dimensions suffice for an isomorphism—is a separate construction result. Here the proved direction is that an existing isomorphism forces equality of dimensions.
Check Your Understanding
For each question, distinguish the role of injectivity from the role of surjectivity.
- Suppose \(T:V\to W\) is an isomorphism and \((\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)\) is a basis of \(V\). What are the two properties you must establish to show its image list is a basis of \(W\)?
- In the independence part of the theorem’s proof, where is injectivity used?
- In the spanning part of the theorem’s proof, where is surjectivity used?
- For \(J((a,b))=(a,b,0)\), identify a vector in \(\mathbb{R}^3\) that is not in the span of the standard basis images.
- If \(V\) and \(W\) are finite-dimensional and isomorphic, explain why a basis of \(V\) and its image basis in \(W\) have the same length.
