Uniqueness Means Comparing Candidates
A uniqueness proof does not usually need to construct a second object. Instead, assume that two candidates both satisfy the defining condition, and compare them. When that condition is linear, subtraction often turns the comparison into a homogeneous equation. If the only solution to that homogeneous equation is zero, the candidates must be equal.
This reasoning has a useful logical distinction: showing that an object exists and showing that it is unique are separate tasks. A uniqueness argument begins by assuming two objects with the required property; it does not, by itself, produce even one such object. In applications, it is helpful to keep those tasks separate and to identify precisely what makes the difference between two candidates vanish.
For example, if a linear map \(T\) satisfies \(T(\mathbf{x})=\mathbf{b}\) and \(T(\mathbf{x}')=\mathbf{b}\), then linearity gives \(T(\mathbf{x}-\mathbf{x}')=\mathbf{0}\). To conclude \(\mathbf{x}=\mathbf{x}'\), we need the additional condition that the only vector mapped to \(\mathbf{0}\) is \(\mathbf{0}\). Without that condition, subtraction gives useful information, but not necessarily uniqueness.
Uniqueness for a Consistent Linear System
For a real matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), its null space is \(\operatorname{Nul}\mathbf{A}=\{\mathbf{z}\in\mathbb{R}^n:\mathbf{A}\mathbf{z}=\mathbf{0}\}\). The system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent when it has at least one solution. The next result gives an exact condition for a consistent system to have only one solution.
Proof. Suppose first that \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\), and let \(\mathbf{x}\) and \(\mathbf{x}'\) be any two solutions. Then \(\mathbf{A}\mathbf{x}=\mathbf{b}=\mathbf{A}\mathbf{x}'\). Subtracting the equations gives \(\mathbf{A}(\mathbf{x}-\mathbf{x}')=\mathbf{0}\), so \(\mathbf{x}-\mathbf{x}'\in\operatorname{Nul}\mathbf{A}\). The null space contains only \(\mathbf{0}\), hence \(\mathbf{x}-\mathbf{x}'=\mathbf{0}\) and \(\mathbf{x}=\mathbf{x}'\). The system is consistent, so at least one solution exists; therefore it has exactly one. Conversely, suppose the consistent system has a unique solution \(\mathbf{x}\), and choose any \(\mathbf{z}\in\operatorname{Nul}\mathbf{A}\). Then \(\mathbf{A}(\mathbf{x}+\mathbf{z})=\mathbf{A}\mathbf{x}+\mathbf{A}\mathbf{z}=\mathbf{b}+\mathbf{0}=\mathbf{b}\). Both \(\mathbf{x}\) and \(\mathbf{x}+\mathbf{z}\) solve the system, so uniqueness gives \(\mathbf{x}+\mathbf{z}=\mathbf{x}\), and thus \(\mathbf{z}=\mathbf{0}\). This holds for every \(\mathbf{z}\) in the null space, so \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\). \(\square\)
Worked Example: Proving a System Has One Solution
Solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\) and \(\mathbf{b}=\begin{bmatrix}5\\11\end{bmatrix}\), then prove that the solution is unique by subtraction.
Row-reduce the augmented matrix. First subtract \(3R_1\) from \(R_2\), then subtract \(2R_2\) from \(R_1\) and scale \(R_2\) by \(-\tfrac12\):
The reduced system gives \(x_1=1\) and \(x_2=2\). Check by multiplication: \(\mathbf{A}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}1+4\\3+8\end{bmatrix}=\begin{bmatrix}5\\11\end{bmatrix}=\mathbf{b}\), so the system is consistent.
For uniqueness, suppose \(\mathbf{x}\) and \(\mathbf{x}'\) are both solutions. Subtracting their equations gives \(\mathbf{A}(\mathbf{x}-\mathbf{x}')=\mathbf{0}\). Write \(\mathbf{z}=\mathbf{x}-\mathbf{x}'=\begin{bmatrix}z_1\\z_2\end{bmatrix}\). Then \(z_1+2z_2=0\) and \(3z_1+4z_2=0\). Subtracting three times the first equation from the second yields \(-2z_2=0\), so \(z_2=0\), and then \(z_1=0\). Thus \(\mathbf{x}-\mathbf{x}'=\mathbf{0}\), so the solution is unique.
Agreement on a Basis Determines a Linear Map
A basis is an ordered list \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) that is linearly independent and spans the vector space: every vector in the space can be written as a linear combination of the list, and the only combination equal to \(\mathbf{0}\) has all coefficients zero. In particular, every vector has a representation using the basis. For this result, the crucial fact is that linear maps preserve linear combinations.
Proof. If \(T=S\), then by equality of the maps, \(T(\mathbf{b}_j)=S(\mathbf{b}_j)\) for each basis vector. Conversely, assume that these basis values agree. Let \(\mathbf{v}\in V\). Since \(\mathcal{B}\) spans \(V\), there are scalars \(c_1,\dots,c_n\in\mathbb{F}\) such that \(\mathbf{v}=\sum_{j=1}^{n}c_j\mathbf{b}_j\). By linearity of both maps and the assumed agreement on each basis vector, \(T(\mathbf{v})=\sum_{j=1}^{n}c_jT(\mathbf{b}_j)=\sum_{j=1}^{n}c_jS(\mathbf{b}_j)=S(\mathbf{v})\). Because this holds for every \(\mathbf{v}\in V\), the maps are equal. \(\square\)
Only the spanning property is needed in the direction from basis values to equality: expressing an arbitrary vector in terms of the tested vectors lets linearity determine its image. The independence property ensures that a basis also gives unique coordinates, which makes those values a reliable way to specify a map.
Proof. Because the basis spans \(V\), at least one such list of scalars exists. Suppose two lists \(c_1,\dots,c_n\) and \(d_1,\dots,d_n\) both represent \(\mathbf{v}\). Subtract their representations to obtain \(\sum_{j=1}^{n}(c_j-d_j)\mathbf{b}_j=\mathbf{0}\). Linear independence of the basis means that every coefficient in this zero combination is zero. Thus \(c_j-d_j=0\), or \(c_j=d_j\), for every \(j\). The representation is therefore unique. \(\square\)
Worked Example: Two Maps with the Same Basis Values
Define linear maps \(T,S:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(x,y,x+y)\) and \(S(x,y)=(x,y,x+y)\). Verify that their values agree on the standard basis, then use those values to establish agreement everywhere.
The standard basis of \(\mathbb{R}^2\) is \(\mathbf{e}_1=(1,0)\), \(\mathbf{e}_2=(0,1)\). Evaluating each map gives \(T(\mathbf{e}_1)=(1,0,1)=S(\mathbf{e}_1)\) and \(T(\mathbf{e}_2)=(0,1,1)=S(\mathbf{e}_2)\). Now take any \((x,y)\in\mathbb{R}^2\). Its basis representation is \((x,y)=x\mathbf{e}_1+y\mathbf{e}_2\). Linearity gives \(T(x,y)=x(1,0,1)+y(0,1,1)=(x,y,x+y)\), and the same computation gives \(S(x,y)=(x,y,x+y)\). Thus the maps agree for every input. For a numerical check, at \((2,-1)\) both maps give \((2,-1,1)\), since \(2+(-1)=1\).
Matrix Columns Record Basis Values
For a linear map \(T:\mathbb{R}^n\to\mathbb{R}^m\), its matrix in standard coordinates is the \(m\times n\) matrix whose \(j\)th column is \(T(\mathbf{e}_j)\). This column rule follows from the standard basis representation \(\mathbf{x}=\sum_{j=1}^{n}x_j\mathbf{e}_j\): linearity gives \(T(\mathbf{x})=\sum_{j=1}^{n}x_jT(\mathbf{e}_j)\), which is exactly the product of the matrix with columns \(T(\mathbf{e}_j)\) and the column vector \(\mathbf{x}\).
Worked Example: Recovering a Matrix from Its Basis Values
Find the matrix of the linear map \(T:\mathbb{R}^2\to\mathbb{R}^2\) satisfying \(T(\mathbf{e}_1)=(2,1)\) and \(T(\mathbf{e}_2)=(-1,3)\), and verify its action.
Place the two image vectors in the columns, in the same order as the basis vectors:
For an arbitrary column vector \(\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\), linearity requires \(T(\mathbf{x})=x_1(2,1)+x_2(-1,3)=(2x_1-x_2,\;x_1+3x_2)\). Multiplication by the proposed matrix gives \(\mathbf{A}\mathbf{x}=\begin{bmatrix}2x_1-x_2\\x_1+3x_2\end{bmatrix}\), so it has the required action on every input. In particular, \(\mathbf{A}\mathbf{e}_1=\begin{bmatrix}2\\1\end{bmatrix}\) and \(\mathbf{A}\mathbf{e}_2=\begin{bmatrix}-1\\3\end{bmatrix}\), exactly the prescribed images.
These values also show that no other matrix can represent the same map: any representing matrix must send \(\mathbf{e}_1\) to its first column and \(\mathbf{e}_2\) to its second. Those columns are already fixed, so the matrix must be \(\mathbf{A}\).
Common Pitfall: Subtraction Does Not Always Prove Equality
A common error is to stop after deriving \(T(\mathbf{x}-\mathbf{x}')=\mathbf{0}\) and claim that \(\mathbf{x}=\mathbf{x}'\) without checking whether \(T(\mathbf{z})=\mathbf{0}\) forces \(\mathbf{z}=\mathbf{0}\). Consider the projection \(P:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(P(x,y)=(x,0)\). The distinct inputs \((1,0)\) and \((1,2)\) have the same image \((1,0)\). Subtracting the inputs gives \((0,-2)\), and \(P(0,-2)=(0,0)\), but the difference is not zero. Thus a zero image alone does not imply a zero input.
The correction is to state and use the required condition: the only vector sent to zero must be zero. Equivalently, \(P\) fails the condition because \(P(0,1)=(0,0)\) while \((0,1)\ne\mathbf{0}\). In a proof of uniqueness, always identify the difference equation and justify why it forces the difference itself to vanish.
The same care applies to basis values. Agreement on a few vectors does not generally imply agreement everywhere; the tested vectors must span the domain, and the maps must be linear. A basis supplies a spanning list, while linearity transports agreement on that list to every linear combination.
Check Your Understanding
For each question, identify which condition makes the uniqueness argument work.
- A consistent system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has two solutions \(\mathbf{x}\) and \(\mathbf{x}'\). What equation does their difference satisfy?
- Why is consistency needed in the theorem about a unique solution and \(\operatorname{Nul}\mathbf{A}=\{\mathbf{0}\}\)?
- If two linear maps agree on every vector in a basis, explain how to compare their values on a general vector.
- Suppose \(\mathbf{v}=c_1\mathbf{b}_1+c_2\mathbf{b}_2=d_1\mathbf{b}_1+d_2\mathbf{b}_2\), where \((\mathbf{b}_1,\mathbf{b}_2)\) is a basis. Which basis property proves \(c_1=d_1\) and \(c_2=d_2\)?
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\), suppose \(T(\mathbf{e}_1)=(1,2)\) and \(T(\mathbf{e}_2)=(3,0)\). Write the matrix of \(T\) in standard coordinates.
