Maps Have Inputs, Outputs, and a Codomain
A map between vector spaces assigns an output to each input. To discuss whether the assignment is linear, we need to keep track of where the inputs come from and where the outputs belong. These roles are not interchangeable: the domain is the space of allowable inputs, the codomain is the space in which outputs lie, and the output is the particular vector assigned to one chosen input.
Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A map \(T:V\to W\) assigns to each \(\mathbf{v}\in V\) exactly one vector \(T(\mathbf{v})\in W\). The space \(V\) is the domain, and \(W\) is the codomain. The notation \(T(\mathbf{v})\) means the output of \(T\) on the input \(\mathbf{v}\); it is an element of \(W\), not a new map. A map need not use every vector in its codomain as an output.
Both conditions are required. In the addition condition, \(\mathbf{u}+\mathbf{v}\) is formed in the domain \(V\), and \(T(\mathbf{u})+T(\mathbf{v})\) is formed in the codomain \(W\). Likewise, \(c\mathbf{v}\) is a domain vector, while \(cT(\mathbf{v})\) is a codomain vector. The vector-space operations make both sides of each equality meaningful.
The word “every” matters: a map is linear only if the two equalities hold for all choices of the indicated inputs and scalar. Checking a few inputs can reveal a failure, but matching outputs on a few inputs does not establish linearity. To prove that a formula defines a linear map, we take general inputs and a general scalar and verify both equalities.
A Coordinate Formula Between Different Spaces
Worked Example: A Map from R² to R³
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(x+2y,3x-y,0)\). Identify the domain, codomain, and output for an input, and verify that \(T\) is linear.
The domain is \(\mathbb{R}^2\), so an input is a column with two entries. The codomain is \(\mathbb{R}^3\), so each output has three entries. For example, with \(\mathbf{u}=(1,-2)\),
To check the two conditions for arbitrary inputs, let \(\mathbf{u}=(x_1,y_1)\), \(\mathbf{v}=(x_2,y_2)\), and \(c\in\mathbb{R}\). First, calculate the output on the sum:
Next, calculate the output on a scalar multiple:
Both defining conditions hold for arbitrary inputs and \(c\), so \(T\) is linear. As a numerical check of addition, take \(\mathbf{u}=(1,-2)\) and \(\mathbf{v}=(3,1)\). Then \(T(\mathbf{u})=(-3,5,0)\), \(T(\mathbf{v})=(5,8,0)\), and \(T(\mathbf{u})+T(\mathbf{v})=(2,13,0)\). Since \(\mathbf{u}+\mathbf{v}=(4,-1)\), we also get \(T(\mathbf{u}+\mathbf{v})=(4-2,12+1,0)=(2,13,0)\), as required.
The Same Definition Applies to Other Vector Spaces
The inputs and outputs need not be coordinate columns. For example, a polynomial space is a vector space: its addition is polynomial addition, and scalar multiplication multiplies each coefficient by the scalar. The same two linearity conditions apply, using the operations in the relevant domain and codomain.
Worked Example: A Map from Polynomials to R²
Let \(P_1(\mathbb{R})\) be the vector space of real polynomials of degree at most one, including the zero polynomial. Define \(S:P_1(\mathbb{R})\to\mathbb{R}^2\) by \(S(a+bx)=(a+2b,3a-b)\). Verify that this map is linear and find its output on \(p(x)=2-3x\).
Every input has the form \(p(x)=a+bx\), and every output is a two-entry vector. For arbitrary inputs \(p(x)=a+bx\) and \(q(x)=d+ex\), their sum is \((a+d)+(b+e)x\). Therefore,
For an arbitrary scalar \(c\in\mathbb{R}\), the polynomial \(cp\) has coefficients \(ca\) and \(cb\), so
Thus \(S\) is linear. For \(p(x)=2-3x\), the coefficients are \(a=2\) and \(b=-3\), giving
Substituting those coefficients into the defining formula gives the same output, so the calculation is consistent.
This example also illustrates why a map’s formula and its domain must be read together. The expression \(a+bx\) describes an input polynomial, while \((a+2b,3a-b)\) is an output in \(\mathbb{R}^2\). The output is not another polynomial, and the coefficients \(a,b\) are used to specify the input.
Two Consequences of the Definition
Proof. Choose any \(\mathbf{v}\in V\) and apply preservation of scalar multiplication with scalar \(0\). Since \(0\mathbf{v}=\mathbf{0}\) in \(V\), linearity gives \(T(\mathbf{0})=T(0\mathbf{v})=0T(\mathbf{v})=\mathbf{0}\) in \(W\). \(\square\)
This gives a quick way to disprove linearity: if a proposed map sends the domain’s zero vector to a nonzero vector in the codomain, it cannot satisfy the definition.
Proof. The domain vectors \(\mathbf{v}\) and \(-\mathbf{v}\) add to \(\mathbf{0}\). By preservation of addition and the result just proved, \(T(\mathbf{v})+T(-\mathbf{v})=T(\mathbf{v}+(-\mathbf{v}))=T(\mathbf{0})=\mathbf{0}\) in \(W\). The unique additive inverse of \(T(\mathbf{v})\) in \(W\) is \(-T(\mathbf{v})\), so \(T(-\mathbf{v})=-T(\mathbf{v})\). \(\square\)
When One Linearity Condition Fails
Worked Example: A Map That Is Not Linear
Define \(F:\mathbb{R}^2\to\mathbb{R}^2\) by \(F(x,y)=(x^2,y)\). The formula assigns exactly one vector in \(\mathbb{R}^2\) to every input in \(\mathbb{R}^2\), so it defines a map. Determine whether it is linear.
It is enough to find one input and one scalar for which the scalar-multiplication condition fails. Take \(\mathbf{v}=(1,0)\) and \(c=2\). The output on the scalar-multiplied input is
But scaling the original output gives
Since \((4,0)\ne(2,0)\), the scalar-multiplication condition fails, so \(F\) is not linear. As a check on the output calculation, substituting \((2,0)\) into the formula gives first coordinate \(2^2=4\) and second coordinate \(0\), exactly as above.
A single failed equality settles the question, because linearity requires both conditions to hold for every permitted choice. Conversely, a few successful numerical checks are not a proof: they only verify the chosen inputs. For a proof, the calculations must use arbitrary vectors and an arbitrary scalar, as in the first two examples.
Common Mistake: Confusing a Formula with Linearity
A frequent error is to call a map linear merely because its formula looks simple or acts coordinate by coordinate. For example, consider \(G:\mathbb{R}^2\to\mathbb{R}^2\), \(G(x,y)=(x+1,y)\). It is a well-defined map, but the added constant changes its behavior. A wrong calculation might ignore the \(1\) and claim \(G(\mathbf{0})=\mathbf{0}\). Substitution corrects this immediately:
The zero-vector proposition therefore rules out linearity. The addition condition also fails concretely if \(\mathbf{u}=\mathbf{v}=(0,0)\): \(G(\mathbf{u}+\mathbf{v})=(1,0)\), while \(G(\mathbf{u})+G(\mathbf{v})=(1,0)+(1,0)=(2,0)\). The correction is to check the actual formula against the definition, rather than infer linearity from how familiar the expression appears.
Keep the three roles separate when reading notation: \(T:V\to W\) names the map and its domain and codomain; \(\mathbf{v}\in V\) is an input; \(T(\mathbf{v})\in W\) is the output for that input. Linearity is a further property of the map, established only by verifying preservation of addition and scalar multiplication across the whole domain.
Check Your Understanding
For each question, distinguish the map’s spaces from a particular input and output, and use the definition where appropriate.
- For \(T:\mathbb{R}^2\to\mathbb{R}^3\) defined by \(T(x,y)=(x-y,2x,3y)\), state the domain and codomain, and calculate \(T(2,-1)\).
- What two equalities must hold for a map \(T:V\to W\) to be linear?
- For \(S(a+bx)=(a-b,2a+b)\), with domain \(P_1(\mathbb{R})\) and codomain \(\mathbb{R}^2\), calculate \(S(1+4x)\).
- Show that \(H:\mathbb{R}\to\mathbb{R}\), \(H(t)=t+3\), is not linear by evaluating it at the zero input.
- State what the zero-vector proposition implies about \(T(\mathbf{0})\) when \(T\) is linear, and identify the space containing that output.
