Two Conditions That Test Linearity
A formula can look simple and still fail to define a linear map. For instance, a formula might square one coordinate, add a fixed vector, or combine entries in a way that does not respect scalar multiplication. The additivity-and-homogeneity test reduces the question to two precise requirements: the output must respect sums of inputs, and multiplying an input by a scalar must multiply its output by that same scalar.
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 when it preserves addition and scalar multiplication. The test is stated for every pair of vectors and every scalar, not just for selected examples.
The field matters: homogeneity must hold for every scalar in the field of the vector spaces. For a complex vector space, testing real scalars alone is insufficient. Additivity concerns arbitrary pairs of inputs, and homogeneity concerns arbitrary input-scalar pairs. Checking a few convenient values can reveal that a map is not linear, but it cannot prove linearity for all inputs.
Why the Test Is Equivalent to Linearity
The definition of a linear map is precisely that it preserves sums and scalar multiples. The theorem below makes explicit that testing these two properties separately is both necessary and sufficient.
Proof. Suppose first that \(T\) is linear. By the definition of linearity, it preserves addition, so \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) for every \(\mathbf{u},\mathbf{v}\in V\). It also preserves scalar multiplication, so \(T(c\mathbf{v})=cT(\mathbf{v})\) for every \(c\in\mathbb{F}\) and \(\mathbf{v}\in V\). Thus it is additive and homogeneous. Conversely, suppose \(T\) is additive and homogeneous. These are exactly the two preservation requirements in the definition of a linear map, so \(T\) is linear. Both implications hold for arbitrary inputs and scalars, as required. \(\square\)
Homogeneity includes the scalar \(c=0\). Thus a homogeneous map satisfies \(T(\mathbf{0})=\mathbf{0}\), because \(T(0\mathbf{v})=0T(\mathbf{v})=\mathbf{0}\). This is a useful quick check: a map that sends the zero vector somewhere other than the zero vector cannot be linear. The converse is not enough; having \(T(\mathbf{0})=\mathbf{0}\) by itself does not establish either additivity or homogeneity.
Proof. Since \(T\) is homogeneous, choosing the scalar \(0\) gives \(T(\mathbf{0})=T(0\mathbf{v})=0T(\mathbf{v})=\mathbf{0}\). Choosing the scalar \(-1\) gives \(T(-\mathbf{v})=T((-1)\mathbf{v})=(-1)T(\mathbf{v})=-T(\mathbf{v})\). \(\square\)
These consequences are convenient for disproving linearity. A failed zero-vector check or a failed negative-scalar check is decisive. Passing either check, however, is not a substitute for verifying both conditions for all inputs.
Applying the Test to Coordinate Formulas
Worked Example: A Map on \(\mathbb{R}^2\)
For \(\mathbf{x}=(x_1,x_2)\in\mathbb{R}^2\), define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T(x_1,x_2)=(2x_1-x_2,\;x_1+3x_2)\). Test whether \(T\) is linear.
To test additivity, take arbitrary \(\mathbf{u}=(u_1,u_2)\) and \(\mathbf{v}=(v_1,v_2)\). Their sum is \((u_1+v_1,u_2+v_2)\), so
For any real scalar \(c\),
Both conditions hold for arbitrary inputs and scalars, so the theorem shows that \(T\) is linear.
Check: Let \(\mathbf{u}=(1,-2)\), \(\mathbf{v}=(3,1)\), and \(c=-2\). Then \(T(\mathbf{u})=(4,-5)\), \(T(\mathbf{v})=(5,6)\), and \(T(\mathbf{u})+T(\mathbf{v})=(9,1)\). Since \(\mathbf{u}+\mathbf{v}=(4,-1)\), \(T(\mathbf{u}+\mathbf{v})=(9,1)\), as required. Also \(c\mathbf{u}=(-2,4)\), so \(T(c\mathbf{u})=(-8,10)=c(4,-5)=cT(\mathbf{u})\).
The key step is substituting arbitrary coordinates into both sides of each condition and simplifying. A formula written in coordinates does not need to be represented as a matrix to use the test. The same strategy works whenever the input objects have coefficients or entries that can be tracked under addition and scalar multiplication.
Applying the Test to Polynomial and Matrix Spaces
Worked Example: A Map on a Polynomial Space
Let \(P_2\) be the real vector space of polynomials of degree at most \(2\), and let \(P_1\) be the real vector space of polynomials of degree at most \(1\). Define \(S:P_2\to P_1\) by \(S(a+bx+cx^2)=(a-c)+(b+2c)x\). Test whether \(S\) is linear.
Write arbitrary inputs as \(p=a+bx+cx^2\) and \(q=d+ex+fx^2\). Their sum is \((a+d)+(b+e)x+(c+f)x^2\). Applying the formula gives
For a real scalar \(r\), \(rp=ra+rbx+rcx^2\), and therefore
Thus \(S\) is linear. The coefficients \(a,b,c\) are arbitrary, so these calculations establish the conditions for every polynomial in \(P_2\), not merely for a selected polynomial.
Check: Take \(p=1-2x+3x^2\) and \(q=2+x-x^2\). The formula gives \(S(p)=-2+4x\) and \(S(q)=3-x\), whose sum is \(1+3x\). Since \(p+q=3-x+2x^2\), \(S(p+q)=1+3x\), agreeing with the sum. Also \(2p=2-4x+6x^2\), so \(S(2p)=-4+8x=2S(p)\).
Worked Example: A Map on a Matrix Space
Define \(U:\mathbb{R}^{2\times2}\to\mathbb{R}^2\) by \(U\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=(a+d,\;b-c)\). Test whether \(U\) is linear.
Let \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}e&f\\g&h\end{bmatrix}\). Entrywise addition gives \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}a+e&b+f\\c+g&d+h\end{bmatrix}\), so
For any \(r\in\mathbb{R}\), every entry of \(r\mathbf{A}\) is \(r\) times the corresponding entry of \(\mathbf{A}\). Consequently,
Both conditions hold for arbitrary matrices and scalars, so \(U\) is linear.
Check: Take \(\mathbf{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}-2&1\\4&0\end{bmatrix}\). Then \(U(\mathbf{A})=(4,3)\) and \(U(\mathbf{B})=(-2,-3)\), whose sum is \((2,0)\). Also \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}-1&3\\3&3\end{bmatrix}\), and \(U(\mathbf{A}+\mathbf{B})=(2,0)\). For \(r=-2\), \(U(-2\mathbf{A})=(-8,-6)=-2(4,3)\).
What Counts as a Valid Test
For a proposed map, the proof should match the universal quantifiers in the test. To prove additivity, begin with arbitrary \(\mathbf{u},\mathbf{v}\) in the domain and show the two expressions are equal. To prove homogeneity, begin with an arbitrary scalar \(c\) from the correct field and an arbitrary vector \(\mathbf{v}\), then show the required equality. If a formula is defined in terms of coefficients or matrix entries, write those coefficients or entries explicitly before applying it.
Confirm the formula sends every allowed input to an element of the stated codomain, and identify whether scalars are real or complex.
Use arbitrary inputs, form their sum, and compare the formula applied to the sum with the sum of the separate outputs.
Use an arbitrary scalar from the field and an arbitrary input. Include negative and zero scalars in the argument, not just positive examples.
If both identities hold for all the required inputs, the map is linear. If either identity fails for one allowed choice, it is not linear.
A single counterexample is enough to disprove a universal condition. For example, if an additivity calculation fails at even one pair of inputs, the map is not additive. By contrast, several successful numerical checks do not establish the condition for every pair; the proof must use arbitrary inputs or another argument that covers all of them.
Common Pitfalls: Checking Only One Condition
Additivity alone is not sufficient. Over \(\mathbb{C}\), consider conjugation \(C:\mathbb{C}\to\mathbb{C}\), \(C(z)=\overline{z}\). It is additive because \(\overline{z+w}=\overline z+\overline w\). But for \(z=1\) and scalar \(c=i\), \(C(i\cdot1)=\overline i=-i\), while \(iC(1)=i\). Since \(-i\ne i\), conjugation is not homogeneous over \(\mathbb{C}\), and therefore is not a complex-linear map. It would be incorrect to conclude that conjugation is complex-linear just because it preserves sums.
Homogeneity alone is not sufficient. Define \(G:\mathbb{R}^2\to\mathbb{R}\) by \(G(0,0)=0\), and, for \((x,y)\ne(0,0)\), by \(G(x,y)=\frac{x^3}{x^2+y^2}\). For any nonzero real \(c\) and nonzero \((x,y)\), the denominator of \(G(cx,cy)\) is \(c^2(x^2+y^2)\), so \(G(cx,cy)=cG(x,y)\). If \(c=0\) or \((x,y)=(0,0)\), both sides of the homogeneity identity are \(0\). Thus \(G\) is homogeneous for every real scalar. However, \(G(1,0)=1\), \(G(0,1)=0\), and \(G(1,1)=\tfrac12\). Additivity would require \(G(1,1)=G(1,0)+G(0,1)=1\), which is false. So \(G\) is not linear.
A related error is testing a formula on only one input or one scalar and then treating that calculation as a proof. For instance, verifying \(T(2\mathbf{v})=2T(\mathbf{v})\) for one particular \(\mathbf{v}\) says nothing about all other vectors and scalars. The complete test requires arbitrary inputs in both identities. Conversely, when the goal is to disprove linearity, use a concrete failed identity and show both sides explicitly.
These checks are useful because they separate the work into two manageable questions. A formula may pass one and fail the other, and the test identifies exactly what still needs to be established. The same reasoning applies regardless of whether the inputs are coordinate vectors, polynomials, or matrices, provided the domain and codomain are vector spaces over the stated field.
Check Your Understanding
For each candidate, decide what the test requires and justify your conclusion with a calculation or a general argument.
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(x-y,2x)\), verify additivity and homogeneity using arbitrary inputs.
- Let \(S:P_2\to\mathbb{R}\) be given by \(S(a+bx+cx^2)=a+bc\). Which test condition fails? Give specific inputs or scalars that demonstrate the failure.
- For conjugation \(C:\mathbb{C}\to\mathbb{C}\), verify additivity and test homogeneity using scalar \(i\) and input \(1\). Is \(C\) linear over \(\mathbb{C}\)?
- Explain why checking \(T(\mathbf{0})=\mathbf{0}\) is necessary for linearity but is not enough to prove linearity.
- For \(U:\mathbb{R}^{2\times2}\to\mathbb{R}\) defined by \(U\left(\begin{bmatrix}a&b\\c&d\end{bmatrix}\right)=a-d\), verify both conditions for arbitrary matrices and a real scalar.
