What Uniqueness Means
A matrix representation depends on choices: the linear map, an ordered basis for its domain, and an ordered basis for its codomain. Once all three are fixed, however, there is no freedom left in the matrix. The reason is that the coordinate identity must hold for every input, including each vector in the domain basis. Those basis-vector tests determine every column.
Let \(V\) and \(W\) be finite-dimensional vector spaces over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Fix ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) for \(V\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\) for \(W\), and let \(T:V\to W\) be linear. The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) lists the coefficients of \(\mathbf{v}\) in the ordered basis \(\mathcal{B}\); in particular, \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\), where \(\mathbf{e}_j\) is the \(j\)th standard coordinate vector.
The definition is an equality of coordinate vectors, not an equality between \(T(\mathbf{v})\) and a matrix product in the original spaces. The matrix product gives the coordinates of the output in \(\mathcal{C}\). Also, uniqueness does not assert that a matrix exists for an arbitrary collection of data: it says that if a representing matrix exists, no second, different matrix can represent the same map relative to the same bases.
The Basis-Vector Argument
For each \(j\), substitute \(\mathbf{v}=\mathbf{b}_j\) into the coordinate identity. Since \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\), the equation becomes \([T(\mathbf{b}_j)]_{\mathcal{C}}=\mathbf{A}\mathbf{e}_j\). Multiplication by \(\mathbf{e}_j\) selects column \(j\), so the \(j\)th column of \(\mathbf{A}\) must be \([T(\mathbf{b}_j)]_{\mathcal{C}}\). Thus the map and the two bases specify the required value of every column.
Proof. Assume that both matrices satisfy the coordinate identity for the same \(T,\mathcal{B},\mathcal{C}\). For each \(j\in\{1,\dots,n\}\), apply both identities to \(\mathbf{b}_j\). Because \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\), this gives $$ \mathbf{A}\mathbf{e}_j=[T(\mathbf{b}_j)]_{\mathcal{C}}=\mathbf{D}\mathbf{e}_j. $$ The vectors \(\mathbf{A}\mathbf{e}_j\) and \(\mathbf{D}\mathbf{e}_j\) are respectively the \(j\)th columns of \(\mathbf{A}\) and \(\mathbf{D}\). Therefore their \(j\)th columns agree for every \(j=1,\dots,n\). Matrices with the same entries in every column are equal, so \(\mathbf{A}=\mathbf{D}\). \(\square\)
This argument relies on having tested every domain basis vector. Testing just one of them determines only one column. It also relies on fixing the ordered bases: changing either basis changes the coordinate identity being tested, so uniqueness for one pair of bases does not say that representations using other bases must be the same matrix.
Worked Example: Testing Two Candidate Matrices in Standard Coordinates
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T(x,y)=(x+2y,3x-y)\), and use the standard basis in both spaces. Check the candidate matrices $$ \mathbf{A}=\begin{bmatrix}1&2\\3&-1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}1&2\\3&0\end{bmatrix}. $$
For an arbitrary input column \(\mathbf{x}=\begin{bmatrix}x\\y\end{bmatrix}\), the output coordinates are $$ [T(\mathbf{x})]=\begin{bmatrix}x+2y\\3x-y\end{bmatrix}. $$ For \(\mathbf{A}\), the matrix product is $$ \mathbf{A}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}1x+2y\\3x-1y\end{bmatrix} =\begin{bmatrix}x+2y\\3x-y\end{bmatrix}. $$ Thus \(\mathbf{A}\) satisfies the coordinate identity for every input.
For \(\mathbf{D}\), test the second basis vector \(\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}\). The map gives \(T(0,1)=(2,-1)\), while $$ \mathbf{D}\mathbf{e}_2 =\begin{bmatrix}1&2\\3&0\end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix} =\begin{bmatrix}2\\0\end{bmatrix}. $$ These are not the same coordinate vector. Hence \(\mathbf{D}\) does not represent \(T\). The basis-vector test isolates the error: its second column would need to be \(\begin{bmatrix}2\\-1\end{bmatrix}\).
Check: Multiplying \(\mathbf{A}\) by \(\begin{bmatrix}x\\y\end{bmatrix}\) reproduces both coordinates in the definition of \(T\), whereas evaluating \(T\) at \(\mathbf{e}_2\) gives \(\begin{bmatrix}2\\-1\end{bmatrix}\ne\begin{bmatrix}2\\0\end{bmatrix}=\mathbf{D}\mathbf{e}_2\). Only \(\mathbf{A}\) passes the required test.
Why Agreement on a Basis Controls a Linear Map
The same principle can be stated without matrices. If two linear maps agree on every vector in a basis, linearity forces them to agree on every vector in the space. This explains why testing all basis vectors is sufficient: each input is a linear combination of those vectors, and the map preserves that combination.
Proof. If \(S=T\), then the maps have equal values at every vector, in particular at every \(\mathbf{b}_j\). Conversely, assume \(S(\mathbf{b}_j)=T(\mathbf{b}_j)\) for all \(j\). Every \(\mathbf{v}\in V\) has a basis expansion \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\) for scalars \(a_1,\dots,a_n\). By linearity and the assumed equality on the basis, $$ S(\mathbf{v})=\sum_{j=1}^n a_jS(\mathbf{b}_j) =\sum_{j=1}^n a_jT(\mathbf{b}_j) =T(\mathbf{v}). $$ Thus \(S(\mathbf{v})=T(\mathbf{v})\) for every \(\mathbf{v}\in V\), which means \(S=T\). \(\square\)
For matrix representations, the theorem’s proof uses the coordinate identity directly, while this proposition gives its map-level interpretation. Each basis input is a complete test of one column. Once all those columns agree, the matrices agree; because the coordinate identity applies to every input, the matrices then give the same coordinate output for every input.
Worked Example: One Basis Test Is Not Enough
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be the identity map, with standard bases. Compare $$ \mathbf{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}1&0\\0&2\end{bmatrix}. $$ Both matrices give the same result on the first basis vector: $$ \mathbf{A}\mathbf{e}_1=\begin{bmatrix}1\\0\end{bmatrix} =\mathbf{D}\mathbf{e}_1=T(\mathbf{e}_1). $$ This single check does not establish that both matrices represent \(T\).
Now test the second basis vector. Since \(T(\mathbf{e}_2)=\begin{bmatrix}0\\1\end{bmatrix}\), $$ \mathbf{A}\mathbf{e}_2=\begin{bmatrix}0\\1\end{bmatrix}, \qquad \mathbf{D}\mathbf{e}_2=\begin{bmatrix}0\\2\end{bmatrix}. $$ Thus \(\mathbf{D}\) fails the second basis-vector test. For a direct check at a general input, $$ \mathbf{A}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}x\\y\end{bmatrix}=T(x,y), \qquad \mathbf{D}\begin{bmatrix}x\\y\end{bmatrix} =\begin{bmatrix}x\\2y\end{bmatrix}. $$ These outputs differ whenever \(y\ne0\). The first test fixed the first column but left the second column untested.
Check: \(\mathbf{A}\) returns the identity-map output for every \((x,y)\). The candidate \(\mathbf{D}\) changes the second coordinate from \(y\) to \(2y\), so it does not represent the identity map.
Coordinates Depend on the Chosen Basis
A basis-vector test must use coordinates in the specified codomain basis. If that basis is not the standard basis, the entries of the output vector in the standard coordinate system are not automatically its coordinates in the chosen basis. Confusing these two kinds of entries can produce a wrong candidate matrix even when the map itself has been evaluated correctly.
Worked Example: Checking a Representation with Nonstandard Bases
Let \(V=W=\mathbb{R}^2\), let \(T(x,y)=(2x+y,x-y)\), and use the ordered bases $$ \mathcal{B}=\big((1,1),(1,-1)\big), \qquad \mathcal{C}=\big((1,1),(1,-1)\big). $$ The following is a proposed representing matrix: $$ \mathbf{A}=\begin{bmatrix}\tfrac32&\tfrac32\\[2pt]\tfrac32&-\tfrac12\end{bmatrix}. $$ We check the coordinate identity, using inputs written in \(\mathcal{B}\)-coordinates.
For \(\mathbf{v}=s(1,1)+t(1,-1)=(s+t,s-t)\), apply \(T\): $$ T(\mathbf{v})=\big(2(s+t)+(s-t),(s+t)-(s-t)\big)=(3s+t,2t). $$ A vector \((X,Y)\) has \(\mathcal{C}\)-coordinates \((\alpha,\beta)\) when \(\alpha+\beta=X\) and \(\alpha-\beta=Y\). Solving gives \(\alpha=(X+Y)/2\) and \(\beta=(X-Y)/2\). Therefore $$ [T(\mathbf{v})]_{\mathcal{C}} =\begin{bmatrix}(3s+t+2t)/2\\(3s+t-2t)/2\end{bmatrix} =\begin{bmatrix}\tfrac32s+\tfrac32t\\[2pt]\tfrac32s-\tfrac12t\end{bmatrix}. $$ The proposed matrix gives the same result: $$ \mathbf{A}\begin{bmatrix}s\\t\end{bmatrix} =\begin{bmatrix}\tfrac32s+\tfrac32t\\[2pt]\tfrac32s-\tfrac12t\end{bmatrix}. $$ So it represents \(T\) for every input.
As a basis-vector check, \(T(1,1)=(3,0)\), whose \(\mathcal{C}\)-coordinates are \((3/2,3/2)\). Also, \(T(1,-1)=(1,2)\), whose \(\mathcal{C}\)-coordinates are \((3/2,-1/2)\). These are exactly the first and second columns of \(\mathbf{A}\), respectively.
Check: The coordinate formula above matches the matrix product for arbitrary \(s,t\). In particular, the two basis-vector outputs reproduce both columns, as required by the coordinate identity.
Common Mistake: Using the Wrong Coordinates
In the last example, a tempting but incorrect calculation is to use \(T(1,1)=(3,0)\) as the first column of the representation matrix. That pair lists the output in standard coordinates, not in the specified basis \(\mathcal{C}\). To find its \(\mathcal{C}\)-coordinates, solve $$ \alpha(1,1)+\beta(1,-1)=(3,0). $$ Comparing components gives \(\alpha+\beta=3\) and \(\alpha-\beta=0\). Adding the equations yields \(2\alpha=3\), so \(\alpha=3/2\); then \(\beta=3/2\). The correct first column is therefore \(\begin{bmatrix}3/2\\3/2\end{bmatrix}\), not \(\begin{bmatrix}3\\0\end{bmatrix}\).
This error is not a failure of uniqueness. It is a failure to check the right coordinate identity. Before comparing columns or accepting a candidate, identify the domain coordinates on the right side and the codomain coordinates on the left side. Only entries expressed in the fixed bases belong in the representation matrix.
Using Uniqueness as a Check
The basis-vector argument gives a concise way to audit a claimed representation. For each domain basis vector, evaluate the map, express the result in the fixed codomain basis, and compare that coordinate vector with the corresponding column. A mismatch at any stage rules out the candidate. If all columns match, linearity and the basis expansion ensure that the coordinate identity holds for every input.
Keep the quantifiers in view: the uniqueness theorem concerns one fixed map and one fixed pair of ordered bases. It does not say that every matrix is a representation, nor does it say that matrices for different basis choices must coincide. It says that two matrices satisfying the same coordinate identity for the same data must have equal columns, hence must be equal.
Check Your Understanding
Use the coordinate identity and the fixed-basis conditions in each question.
- In the uniqueness proof, why does applying the identity to \(\mathbf{b}_j\) select column \(j\) of a representing matrix?
- Suppose two \(3\times 2\) matrices both represent the same map relative to the same ordered bases. What must be true about their two columns, and why?
- For the identity map on \(\mathbb{R}^2\), explain why agreement on \(\mathbf{e}_1\) alone does not establish that a candidate matrix is the identity matrix.
- In the nonstandard-basis example, find the \(\mathcal{C}\)-coordinates of \((1,2)\) by solving for its basis coefficients.
- State both directions of the proposition relating equality of two linear maps to agreement on a domain basis.
