From a Linear Map to a Matrix
A matrix can record how a linear map acts, but the entries depend on the coordinates used to describe inputs and outputs. To build that matrix, choose an ordered basis for the domain and an ordered basis for the codomain. The key observation is that a linear map is determined by its values on a basis: once those values are known, linearity determines its value on every linear combination of the basis vectors.
Throughout, the vector spaces are finite-dimensional over \(\mathbb{R}\). Let \(V\) have ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\), and let \(W\) have ordered basis \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\). The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) lists the coefficients in the unique expression of \(\mathbf{v}\) as a linear combination of the vectors in \(\mathcal{B}\). A map \(T:V\to W\) is linear when it preserves linear combinations: \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\in V\) and \(a,b\in\mathbb{R}\).
Each basis vector of the domain contributes one column, so there are \(n\) columns. Each image is described using \(m\) coordinates in the codomain basis, so each column has \(m\) entries. Consequently, the matrix has size \(m\times n\), even when the dimensions of \(V\) and \(W\) differ. Its entries are not necessarily the coordinates of \(T(\mathbf{b}_j)\) in the standard basis: they are coordinates in the chosen basis \(\mathcal{C}\).
The Coordinate-Matrix Property
The definition is designed so that multiplying the coordinate matrix by the input coordinate vector produces the output coordinate vector. This is its central use: it lets us calculate with a linear map using coordinates alone.
Proof. Write \([\mathbf{v}]_{\mathcal{B}}=(a_1,\dots,a_n)^{\top}\), so that \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\). Since \(T\) is linear, \(T(\mathbf{v})=a_1T(\mathbf{b}_1)+\cdots+a_nT(\mathbf{b}_n)\). Taking coordinates in \(\mathcal{C}\) preserves this linear combination, giving $$ [T(\mathbf{v})]_{\mathcal{C}} = a_1[T(\mathbf{b}_1)]_{\mathcal{C}}+\cdots+a_n[T(\mathbf{b}_n)]_{\mathcal{C}} = [T]_{\mathcal{C}\leftarrow\mathcal{B}} \begin{bmatrix}a_1\\ \vdots\\ a_n\end{bmatrix}. $$ The last column vector is \([\mathbf{v}]_{\mathcal{B}}\), which proves the equation. For uniqueness, suppose an \(m\times n\) matrix \(\mathbf{M}\) satisfies the equation for every \(\mathbf{v}\in V\). Substitute \(\mathbf{v}=\mathbf{b}_j\). Its coordinate vector is the \(j\)th standard basis vector in \(\mathbb{R}^n\), so \(\mathbf{M}[\mathbf{b}_j]_{\mathcal{B}}\) is the \(j\)th column of \(\mathbf{M}\). The equation therefore forces that column to be \([T(\mathbf{b}_j)]_{\mathcal{C}}\). This holds for every \(j\), so every column of \(\mathbf{M}\) equals the corresponding column in the defined coordinate matrix. Hence \(\mathbf{M}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}\). \(\square\)
The theorem also gives a reliable construction procedure: evaluate \(T\) on each domain basis vector, find the coordinates of each resulting vector in the codomain basis, and place those coordinate vectors side by side as columns. Once this is done, an arbitrary input is handled by first expressing it in the domain basis and then multiplying.
Computing Columns in Different Bases
Worked Example: A Map from \(\mathbb{R}^2\) to \(\mathbb{R}^3\)
Let \(T:\mathbb{R}^2\to\mathbb{R}^3\) be given by \(T(x,y)=(x+2y,3x-y,2x+y)\). Use the domain basis \(\mathcal{B}=((1,1),(1,-1))\) and the standard basis \(\mathcal{C}=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) of \(\mathbb{R}^3\) to find \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\).
First compute the images of the two domain basis vectors: $$ T(1,1)=(1+2,\;3-1,\;2+1)=(3,2,3), $$ $$ T(1,-1)=(1-2,\;3+1,\;2-1)=(-1,4,1). $$ Because \(\mathcal{C}\) is the standard basis, these vectors are already written as \(\mathcal{C}\)-coordinate vectors. They become the first and second columns, respectively: $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix} 3&-1\\ 2&4\\ 3&1 \end{bmatrix}. $$
Check the matrix on \(\mathbf{v}=(5,1)\). Its \(\mathcal{B}\)-coordinates satisfy \(a+b=5\) and \(a-b=1\), so \(2a=6\), \(a=3\), and \(b=2\). Thus \([\mathbf{v}]_{\mathcal{B}}=(3,2)^{\top}\). Multiplication gives $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix}3&-1\\2&4\\3&1\end{bmatrix} \begin{bmatrix}3\\2\end{bmatrix} = \begin{bmatrix}7\\14\\11\end{bmatrix}. $$ Direct substitution gives \(T(5,1)=(5+2,\;15-1,\;10+1)=(7,14,11)\), agreeing with the matrix calculation.
When the codomain basis is not standard, each image must first be expressed in that basis. The image itself is not automatically a column of the coordinate matrix; its coordinates are.
Worked Example: Using a Nonstandard Codomain Basis
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(x+y,2x-y)\). Take \(\mathcal{B}=((1,1),(1,-1))\) in the domain and \(\mathcal{C}=((1,1),(1,0))\) in the codomain. Find \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\).
A vector \((x,y)\) has \(\mathcal{C}\)-coordinates \((r,s)^{\top}\) when \(r(1,1)+s(1,0)=(r+s,r)=(x,y)\). Therefore \(r=y\) and \(s=x-y\). Apply \(T\) to the first domain basis vector: $$ T(1,1)=(2,1),\qquad [T(1,1)]_{\mathcal{C}}=(1,2-1)^{\top}=(1,1)^{\top}. $$ For the second basis vector, $$ T(1,-1)=(0,3),\qquad [T(1,-1)]_{\mathcal{C}}=(3,0-3)^{\top}=(3,-3)^{\top}. $$ Placing these coordinate vectors in columns gives $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix}1&3\\1&-3\end{bmatrix}. $$
For a check, take \(\mathbf{v}=(4,2)\). Its \(\mathcal{B}\)-coordinates solve \(a+b=4\), \(a-b=2\), hence \(a=3\), \(b=1\). Then $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}} = \begin{bmatrix}1&3\\1&-3\end{bmatrix} \begin{bmatrix}3\\1\end{bmatrix} = \begin{bmatrix}6\\0\end{bmatrix}. $$ This output coordinate vector reconstructs \(6(1,1)+0(1,0)=(6,6)\). Directly, \(T(4,2)=(4+2,8-2)=(6,6)\), so the result checks.
Maps Between Different Vector Spaces
The domain and codomain need not be coordinate spaces of the same dimension. The column method works in exactly the same way for abstract finite-dimensional spaces, provided that the basis images can be expressed in the chosen codomain basis. For example, a polynomial may be the input while an ordered pair of real numbers is the output. The matrix still has one column for each domain basis vector and one row for each codomain coordinate.
Worked Example: Evaluating a Polynomial at Two Points
Let \(P_1\) be the space of real polynomials of degree at most \(1\), and define \(T:P_1\to\mathbb{R}^2\) by \(T(p)=(p(0),p(1))\). Use \(\mathcal{B}=(1+x,1-x)\) for \(P_1\) and the standard basis for \(\mathbb{R}^2\). Find the coordinate matrix and check it on \(p(x)=2+3x\).
For the first domain basis vector, \((1+x)(0)=1\) and \((1+x)(1)=2\), so \(T(1+x)=(1,2)\). For the second, \((1-x)(0)=1\) and \((1-x)(1)=0\), so \(T(1-x)=(1,0)\). As the codomain basis is standard, these are the columns: $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix}1&1\\2&0\end{bmatrix}. $$ To find the input coordinates of \(p(x)=2+3x\), solve \(a(1+x)+b(1-x)=(a+b)+(a-b)x=2+3x\). Thus \(a+b=2\) and \(a-b=3\); adding gives \(2a=5\), so \(a=\tfrac52\), and then \(b=2-\tfrac52=-\tfrac12\). Therefore $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}}[p]_{\mathcal{B}} = \begin{bmatrix}1&1\\2&0\end{bmatrix} \begin{bmatrix}\tfrac52\\-\tfrac12\end{bmatrix} = \begin{bmatrix}2\\5\end{bmatrix}. $$ Direct evaluation gives \(p(0)=2\) and \(p(1)=2+3=5\), so \(T(p)=(2,5)\), as required.
A Common Coordinate-Matrix Mistake
Using standard input coordinates when the domain basis is different gives the wrong multiplication. In the first worked example, the matrix is \(\begin{bmatrix}3&-1\\2&4\\3&1\end{bmatrix}\), and the input \((5,1)\) has standard coordinates \((5,1)^{\top}\). Multiplying by those coordinates gives \((14,14,16)^{\top}\), which is not \(T(5,1)\). The matrix expects input coordinates in \(\mathcal{B}\), not standard coordinates. The correct input coordinates are \((3,2)^{\top}\), and multiplication gives \((7,14,11)^{\top}\), which agrees with direct evaluation.
A second way to detect this error is to check the definition column by column. The first column must be \([T(\mathbf{b}_1)]_{\mathcal{C}}\), not \([T(\mathbf{e}_1)]_{\mathcal{C}}\) unless the first domain basis vector actually is \(\mathbf{e}_1\). The basis order matters as well: switching the order of the domain basis vectors switches the corresponding columns. Switching the codomain basis changes the coordinates within every column.
Using the Matrix as a Coordinate Rule
The coordinate matrix translates one coordinate description into another through the map: domain coordinates go in, and codomain coordinates come out. The vector represented by those output coordinates is \(T(\mathbf{v})\), reconstructed using \(\mathcal{C}\). This distinction is useful whenever a linear map is applied repeatedly or when its input and output naturally use different bases. It also explains why a matrix is not the map itself: the map acts on vectors in \(V\), while its coordinate matrix acts on coordinate columns in \(\mathbb{R}^n\).
To construct and use a coordinate matrix consistently:
- Keep the domain basis order fixed and evaluate the map on each basis vector in that order.
- Express each image using the codomain basis, then place its coordinate column in the matching position.
- Multiply by \([\mathbf{v}]_{\mathcal{B}}\), not by an unrelated coordinate description of \(\mathbf{v}\).
- Interpret the resulting column as \([T(\mathbf{v})]_{\mathcal{C}}\), and reconstruct the vector with basis \(\mathcal{C}\) if needed.
These steps provide both a construction and a check: the matrix must send the coordinate vector of each domain basis vector to the coordinate vector of its image. Since every domain vector is a linear combination of those basis vectors, linearity then guarantees the same rule for every input.
Check Your Understanding
For each problem, keep track of which basis supplies the input coordinates and which basis supplies the output coordinates.
- Let \(T(x,y)=(2x-y,x+3y)\) and use the standard bases in both copies of \(\mathbb{R}^2\). Find the two columns of \([T]\).
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\), use domain basis \(((1,0),(1,1))\), codomain basis \(((1,1),(0,1))\), and \(T(x,y)=(x+y,y)\). Find \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\).
- Explain why the coordinate matrix has \(m\) rows and \(n\) columns when the codomain basis has \(m\) vectors and the domain basis has \(n\) vectors.
- Let \(\mathbf{v}=a_1\mathbf{b}_1+a_2\mathbf{b}_2\). Use linearity to explain why the columns \([T(\mathbf{b}_1)]_{\mathcal{C}}\) and \([T(\mathbf{b}_2)]_{\mathcal{C}}\) produce \([T(\mathbf{v})]_{\mathcal{C}}\) when multiplied by \((a_1,a_2)^{\top}\).
- Describe a concrete check that would reveal whether a proposed second column is not \([T(\mathbf{b}_2)]_{\mathcal{C}}\).
