Two Requirements for an Isomorphism
A vector-space map can match every input with exactly one output and still fail to respect vector addition or scalar multiplication. Conversely, a map can preserve both operations and still identify distinct vectors or miss some outputs. An isomorphism requires both features: it is a linear map that is also a bijection. This definition captures a precise kind of equivalence between vector spaces without requiring them to be presented in the same way.
Throughout, let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A map \(T:V\to W\) is linear if, for all \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{F}\), it satisfies \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\). A map is injective if equal outputs imply equal inputs, and surjective if every element of its codomain is an output. It is bijective if it is both injective and surjective.
The two requirements must be checked separately. Linearity concerns how the map interacts with the vector-space operations; bijectivity concerns how inputs and outputs correspond. In particular, being one-to-one and onto does not by itself make a map an isomorphism. The field matters as well: the linearity condition uses scalars from the field over which both spaces are being considered.
What Linearity Preserves
The definition of linearity combines two operation rules. Setting \(a=b=1\) shows that a linear map preserves addition: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\). Setting one vector coefficient to zero and the other to \(a\) shows that it preserves scalar multiplication: \(T(a\mathbf{u})=aT(\mathbf{u})\). These rules also force a linear map to send the zero vector to the zero vector.
Proof. By additivity, \(T(\mathbf{0})=T(\mathbf{0}+\mathbf{0})=T(\mathbf{0})+T(\mathbf{0})\). Subtracting \(T(\mathbf{0})\) from both sides gives \(T(\mathbf{0})=\mathbf{0}\), where the zero vectors belong to the appropriate spaces. \(\square\)
More generally, linearity lets us take any finite linear combination through the map. For vectors \(\mathbf{v}_1,\dots,\mathbf{v}_p\in V\) and scalars \(c_1,\dots,c_p\in\mathbb{F}\), repeated use of additivity and scalar preservation gives \(T(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p)=c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)\). This applies to every positive integer \(p\). Bijectivity is not needed for this formula; linearity alone is enough.
Worked Examples: Checking the Two Requirements
Worked Example: A Linear Bijection from \(\mathbb{R}^2\) to \(\mathbb{R}^2\)
Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T(x,y)=(x+y,x-y)\). Verify that it is an isomorphism.
First check linearity. For \((x_1,y_1),(x_2,y_2)\in\mathbb{R}^2\) and \(a,b\in\mathbb{R}\),
Thus \(T\) is linear. To check bijectivity, let \((u,v)\) be an arbitrary output. If \(T(x,y)=(u,v)\), then \(x+y=u\) and \(x-y=v\). Adding these equations gives \(2x=u+v\), so \(x=(u+v)/2\). Subtracting the second equation from the first gives \(2y=u-v\), so \(y=(u-v)/2\). These values do produce the requested output:
Every \((u,v)\) therefore has a preimage. The equations also determine \(x\) and \(y\) uniquely, so that preimage is unique. Hence \(T\) is surjective and injective, and therefore bijective.
Check: \(T(2,-1)=(2-1,2-(-1))=(1,3)\). Using the formula for the preimage of \((1,3)\) gives \(((1+3)/2,(1-3)/2)=(2,-1)\), as required.
Worked Example: A Linear Map That Is Not an Isomorphism
Define \(S:\mathbb{R}^3\to\mathbb{R}^2\) by \(S(x,y,z)=(x,y)\). It preserves the operations: for vectors \(\mathbf{u}=(x_1,y_1,z_1)\), \(\mathbf{v}=(x_2,y_2,z_2)\), and \(a,b\in\mathbb{R}\),
So \(S\) is linear. It is surjective because any \((r,s)\in\mathbb{R}^2\) is \(S(r,s,0)\). But it is not injective: \(S(1,2,0)=(1,2)=S(1,2,5)\), although \((1,2,0)\ne(1,2,5)\). The map is therefore not bijective and is not an isomorphism.
Check: the two displayed inputs differ in their third coordinate, but that coordinate is discarded by \(S\). The failure of injectivity is exactly the failure of distinct inputs to remain distinct.
Worked Example: A Bijection That Is Not an Isomorphism
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x+1\). If \(f(x_1)=f(x_2)\), then \(x_1+1=x_2+1\), so \(x_1=x_2\); thus \(f\) is injective. For any \(y\in\mathbb{R}\), \(f(y-1)=(y-1)+1=y\), so \(f\) is surjective. It is bijective.
However, \(f\) is not linear. A linear map must send zero to zero, while \(f(0)=0+1=1\ne0\). Equivalently, scalar multiplication already fails for the scalar \(0\) and input \(3\):
Thus this bijection is not an isomorphism: it matches inputs and outputs bijectively but does not preserve the vector-space operations.
Isomorphisms Preserve Linear Relations
A linear relation among a list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) is an equation \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). For a linear map, any such relation among the inputs gives a relation among their images. If the map is also injective, the reverse implication holds as well.
Proof. Write \(\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). If \(\mathbf{v}=\mathbf{0}\), linearity and preservation of zero give \(c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)=T(\mathbf{v})=T(\mathbf{0})=\mathbf{0}\). For the reverse direction, suppose \(c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)=\mathbf{0}\). Linearity says this left-hand side is \(T(\mathbf{v})\), so \(T(\mathbf{v})=\mathbf{0}=T(\mathbf{0})\). Since \(T\) is injective, \(\mathbf{v}=\mathbf{0}\). This proves both implications. \(\square\)
A list is linearly independent if its only linear relation has every coefficient equal to zero. The theorem therefore has an immediate consequence, in both directions.
Proof. By the theorem, a choice of coefficients gives a zero linear combination of the original list exactly when it gives a zero linear combination of the image list. Consequently, one list has only the all-zero relation exactly when the other does. That is precisely independence for each list. \(\square\)
This result explains why both parts of the isomorphism definition matter. Linearity carries relations forward; injectivity ensures that a nonzero vector cannot be sent to zero and gives the reverse implication. A merely linear map may lose relations by sending distinct vectors to the same output, while a merely bijective map need not preserve any linear relation.
Isomorphisms Carry Bases to Bases
Recall that a list spans a vector space if every vector in that space is a linear combination of the list. A basis is a list that is both linearly independent and spanning. An isomorphism preserves both conditions, so it carries a basis of its domain to a basis of its codomain.
Proof. First suppose \(\mathbf{v}_1,\dots,\mathbf{v}_p\) spans \(V\). For any \(\mathbf{w}\in W\), surjectivity gives a vector \(\mathbf{v}\in V\) such that \(T(\mathbf{v})=\mathbf{w}\). Since the list spans \(V\), there are scalars \(c_1,\dots,c_p\) with \(\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). Applying linearity gives \(\mathbf{w}=c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)\), so the image list spans \(W\).
Conversely, suppose the image list spans \(W\). For any \(\mathbf{v}\in V\), there are scalars \(c_1,\dots,c_p\) such that \(T(\mathbf{v})=c_1T(\mathbf{v}_1)+\cdots+c_pT(\mathbf{v}_p)\). By linearity, the right-hand side is \(T(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p)\). Injectivity now gives \(\mathbf{v}=c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p\). Since this holds for every \(\mathbf{v}\in V\), the original list spans \(V\). \(\square\)
Combining this theorem with the corollary on independence shows that a list is a basis of \(V\) exactly when its image list is a basis of \(W\). Indeed, spanning and independence each hold for one list exactly when they hold for the other. No assumption that the spaces are finite-dimensional is needed for these statements about a given finite list.
Common Mistake: Checking Only One Requirement
A common error is to call any bijection an isomorphism. For \(f(x)=x+1\), one might correctly show that every real number \(y\) has the unique preimage \(y-1\), then incorrectly conclude that \(f\) is an isomorphism. The calculation \(f(0)=1\ne0\) corrects the conclusion: every linear map must preserve zero, so this bijection is not linear.
The converse error is to call every linear map an isomorphism. The projection \(S(x,y,z)=(x,y)\) preserves addition and scalar multiplication and reaches every vector of \(\mathbb{R}^2\), but \(S(1,2,0)=S(1,2,5)\). It is not injective, so it is not bijective. In practice, write down the two questions separately: does the map preserve the vector-space operations, and is it both one-to-one and onto?
Check Your Understanding
For each question, distinguish carefully between linearity and bijectivity.
- Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T(x,y)=(x,0)\). Is it linear? Is it injective? Is it surjective?
- For \(f:\mathbb{R}\to\mathbb{R}\) given by \(f(x)=2x\), verify whether it is an isomorphism by checking linearity and bijectivity separately.
- Explain why a linear map \(T:V\to W\) must satisfy \(T(\mathbf{0})=\mathbf{0}\).
- If an isomorphism sends a list to a linearly dependent image list, what does the relation theorem imply about the original list?
- State why an isomorphism sends a spanning list in its domain to a spanning list in its codomain. Identify which part of the definition is used to find a preimage of an arbitrary codomain vector.
