Why Every Linear Transformation Has One Standard Matrix

Why a Few Values Determine the Whole Transformation

A linear transformation acts on every vector in its domain, so it may seem that describing it requires infinitely many pieces of information. In \(\mathbb{R}^n\), however, every input is a combination of just \(n\) standard coordinate vectors. Linearity means that once the transformation is known on those vectors, its value on every other input is forced. The corresponding matrix records those \(n\) images as its columns.

For \(j=1,\dots,n\), the standard basis vector \(\mathbf{e}_j\in\mathbb{R}^n\) has a \(1\) in position \(j\) and \(0\) in every other position. An input \(\mathbf{x}=(x_1,\dots,x_n)^\top\) can therefore be written as \(\mathbf{x}=\sum_{j=1}^n x_j\mathbf{e}_j\): its \(j\)th coordinate on the right is \(x_j\). A map \(T:\mathbb{R}^n\to\mathbb{R}^m\) is linear if it preserves addition and scalar multiplication: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) and \(T(c\mathbf{u})=cT(\mathbf{u})\) for all \(\mathbf{u},\mathbf{v}\in\mathbb{R}^n\) and \(c\in\mathbb{R}\).

Each column has \(m\) entries because each image \(T(\mathbf{e}_j)\) belongs to \(\mathbb{R}^m\). The matrix has \(n\) columns because the domain has \(n\) standard basis vectors. The key point is that this is not merely a convenient way to record some values: the matrix reproduces the transformation on every input.

Existence: The Columns Give the Transformation

Proof. Let \(\mathbf{x}=(x_1,\dots,x_n)^\top\). The coordinate expansion gives \(\mathbf{x}=\sum_{j=1}^n x_j\mathbf{e}_j\). By additivity, applying \(T\) to this finite sum gives the sum of the images: this follows by repeatedly applying \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\), one term at a time. By homogeneity, \(T(x_j\mathbf{e}_j)=x_jT(\mathbf{e}_j)\) for each \(j\). Consequently,

The last equality is the definition of matrix-vector multiplication: it forms the linear combination of the columns of \(\mathbf{A}\) with coefficients \(x_1,\dots,x_n\). This proves the formula for every input \(\mathbf{x}\). \(\square\)

The proof uses only the coordinate expansion and the two defining properties of a linear map. In particular, it applies whether \(m=n\) or the matrix is rectangular. When \(m\ne n\), the product still has the right shape: an \(m\times n\) matrix times an \(n\times1\) vector produces an \(m\times1\) vector, as required for an output in \(\mathbb{R}^m\).

Uniqueness: No Other Matrix Can Do the Job

Existence says that the matrix made from the basis-vector images works. Uniqueness says that a different matrix cannot produce the same transformation on all inputs. To prove uniqueness, evaluate any proposed representation at each standard basis vector. Multiplication by \(\mathbf{e}_j\) selects column \(j\), so these evaluations reveal every column of the proposed matrix.

Proof. Fix any \(j\in\{1,\dots,n\}\). The two assumed formulas hold at \(\mathbf{x}=\mathbf{e}_j\), so \(\mathbf{A}\mathbf{e}_j=T(\mathbf{e}_j)=\mathbf{B}\mathbf{e}_j\). The product of a matrix with \(\mathbf{e}_j\) is its \(j\)th column: in the matrix-vector product, the coefficient of column \(j\) is \(1\), and the coefficients of all other columns are \(0\). Hence the \(j\)th columns of \(\mathbf{A}\) and \(\mathbf{B}\) are equal. This holds for every \(j=1,\dots,n\), so all corresponding columns agree and \(\mathbf{A}=\mathbf{B}\). Existence was proved in the preceding theorem, so together the results give exactly one standard matrix. \(\square\)

Equality of Linear Maps Can Be Checked on the Standard Basis

The existence and uniqueness results also give a useful test for equality. Two linear maps with the same domain and codomain are equal exactly when they agree on every standard basis vector. Agreement on a few other inputs might happen by coincidence, but agreement on all standard basis vectors fixes every column and therefore fixes the value at every input.

Proof. If \(T=S\), then \(T(\mathbf{e}_j)=S(\mathbf{e}_j)\) for every \(j\). These images are the corresponding columns of \(\mathbf{A}\) and \(\mathbf{B}\), so the matrices have equal columns and \(\mathbf{A}=\mathbf{B}\). Conversely, suppose \(\mathbf{A}=\mathbf{B}\). For any \(\mathbf{x}\in\mathbb{R}^n\), the representation theorem gives \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) and \(S(\mathbf{x})=\mathbf{B}\mathbf{x}\). Therefore \(T(\mathbf{x})=\mathbf{A}\mathbf{x}=\mathbf{B}\mathbf{x}=S(\mathbf{x})\), so \(T=S\). Finally, equality of all basis-vector images is exactly equality of the columns, which is equivalent to equality of the matrices. \(\square\)

This test is useful when formulas for two transformations look different. Instead of comparing their outputs for every possible input, compute the \(n\) basis-vector images for each map and compare them. The reasoning depends on both maps being linear: without linearity, values on the standard basis need not determine values elsewhere.

Common Pitfall: Putting the Images in the Wrong Orientation

The images \(T(\mathbf{e}_1),\dots,T(\mathbf{e}_n)\) are columns, in the same order as the basis vectors. For the first worked example, a mistaken arrangement would be to use the images as rows, producing the \(2\times3\) matrix \(\begin{bmatrix}1&3&0\\2&-1&4\end{bmatrix}\). This cannot be the standard matrix of a map from \(\mathbb{R}^2\) to \(\mathbb{R}^3\): its product with a \(2\times1\) input is not defined, and its dimensions do not match the required \(3\times2\) shape.

A different error is to keep the correct dimensions but swap the columns, using \(\widetilde{\mathbf{A}}=\begin{bmatrix}2&1\\-1&3\\4&0\end{bmatrix}\). Then \(\widetilde{\mathbf{A}}\mathbf{e}_1=(2,-1,4)^\top=T(\mathbf{e}_2)\), not \(T(\mathbf{e}_1)=(1,3,0)^\top\). The correction is to put \(T(\mathbf{e}_1)\) in column 1 and \(T(\mathbf{e}_2)\) in column 2. Testing the proposed matrix on each \(\mathbf{e}_j\) catches both orientation and ordering errors.

The standard matrix is a complete record of a linear map in standard coordinates: its shape records the domain and codomain sizes, and its columns record the images that determine the map. This is why matrix multiplication is more than a convenient calculation rule: it is precisely the coordinate form of applying a linear transformation.