From a Linear Map to a Coordinate Calculation
A linear map acts on vectors, while a matrix acts on coordinate columns. To use a matrix to calculate the value of a map, we must specify how vectors in both the domain and the codomain are written as coordinates. Those coordinate choices matter: the same linear map can be represented by different matrices when different bases are chosen.
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 \(W\). For \(\mathbf{v}\in V\), the coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in the unique expression \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\). Coordinates in \(W\) are written using \(\mathcal{C}\) in the same way.
\([\ T(\mathbf{v})\ ]_{\mathcal{C}}=\mathbf{M}[\mathbf{v}]_{\mathcal{B}}\).
When this equation holds, the matrix is denoted \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\). It converts the input’s \(\mathcal{B}\)-coordinates into the output’s \(\mathcal{C}\)-coordinates.
The arrow in the subscript records the direction of this conversion: start with coordinates in \(\mathcal{B}\), apply the matrix, and obtain coordinates in \(\mathcal{C}\). In particular, the matrix product is not usually the vector \(T(\mathbf{v})\) itself. It is the coordinate vector of \(T(\mathbf{v})\) in the specified codomain basis.
Using the Representation Equation
Once a representing matrix is known, evaluating \(T\) has two stages. First find the input coordinate vector in the domain basis. Then multiply by the matrix to obtain output coordinates. If the codomain basis is not the standard basis, a final conversion from coordinates to the actual vector in \(W\) is required.
Write \(\mathbf{v}\) as a linear combination of the ordered domain basis and form the coefficient column \([\mathbf{v}]_{\mathcal{B}}\).
Compute \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\).
Use the entries of \([T(\mathbf{v})]_{\mathcal{C}}\) as coefficients of the codomain basis vectors. If \(\mathcal{C}\) is the standard basis, the coordinate column already gives the usual vector entries.
Worked Example: A Nonstandard Domain Basis
Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \(T(x,y)=(2x+y,\;x+3y)\). Use the domain basis \(\mathcal{B}=((1,1),(1,-1))\) and the standard codomain basis to find a matrix representation and calculate \(T(3,1)\).
Write an arbitrary input in \(\mathcal{B}\)-coordinates as \(\mathbf{v}=a(1,1)+b(1,-1)\). Its usual entries are \(x=a+b\) and \(y=a-b\). Substituting these into the definition of \(T\) gives
The codomain basis is standard, so these two entries are also the output’s coordinates. Therefore, for every \(\mathbf{v}\),
To find the input coordinates of \((3,1)\), solve \(a+b=3\) and \(a-b=1\). Adding the equations gives \(2a=4\), so \(a=2\); then \(b=1\). Thus \([(3,1)]_{\mathcal{B}}=\begin{bmatrix}2\\1\end{bmatrix}\), and
Check: Direct substitution gives \(T(3,1)=(2\cdot3+1,\;3+3\cdot1)=(7,6)\), agreeing with the matrix calculation.
This example illustrates why the domain basis must be kept visible. The entries \((3,1)\) describe the vector in standard coordinates, but they are not its coordinates in \(\mathcal{B}\). The matrix multiplication uses \(\begin{bmatrix}2\\1\end{bmatrix}\), not \(\begin{bmatrix}3\\1\end{bmatrix}\).
Coordinate Equations Preserve the Map’s Values
Proof. By the definition of a representing matrix, \(\mathbf{M}[\mathbf{v}]_{\mathcal{B}}=[T(\mathbf{v})]_{\mathcal{C}}\). If \(T(\mathbf{v})=\mathbf{w}\), equal vectors have equal coordinates in the same basis, so \([T(\mathbf{v})]_{\mathcal{C}}=[\mathbf{w}]_{\mathcal{C}}\), giving the stated matrix equation. Conversely, suppose \(\mathbf{M}[\mathbf{v}]_{\mathcal{B}}=[\mathbf{w}]_{\mathcal{C}}\). The representation equation then gives \([T(\mathbf{v})]_{\mathcal{C}}=[\mathbf{w}]_{\mathcal{C}}\). A basis gives unique coordinates, so vectors with the same coordinate vector in \(\mathcal{C}\) are equal. Hence \(T(\mathbf{v})=\mathbf{w}\). \(\square\)
This equivalence lets us translate a question about a map into a question about coordinates. A solution of \(\mathbf{M}\mathbf{x}=\mathbf{y}\) corresponds to an input whose \(\mathcal{B}\)-coordinate vector is \(\mathbf{x}\) and whose image has \(\mathcal{C}\)-coordinate vector \(\mathbf{y}\). The coordinate bases must match the subscripts of the matrix; otherwise the equation does not express the desired map.
Worked Example: A Map from R³ to R²
Let \(T:\mathbb{R}^3\to\mathbb{R}^2\) be given by \(T(x,y,z)=(x+2y-z,\;3x-y+2z)\). Use the domain basis \(\mathcal{B}=((1,0,1),(0,1,1),(1,1,0))\) and the standard basis in the codomain. Find a representing matrix and use it to evaluate \(T(3,1,0)\).
First verify that the listed domain vectors form a basis. Placing them as columns and row reducing gives
The row reduction shows that every coordinate column has a unique solution when expressed using these three vectors, so they form a basis of \(\mathbb{R}^3\). Write an arbitrary input as \(\mathbf{v}=a(1,0,1)+b(0,1,1)+c(1,1,0)\). Then \(x=a+c,\;y=b+c,\;z=a+b\). Substitution into \(T\) yields
Thus the coordinate equation for every input is
For \((3,1,0)\), the equations \(a+c=3,\;b+c=1,\;a+b=0\) give \(a=1,\;b=-1,\;c=2\). Therefore
Check: Directly, \(T(3,1,0)=(3+2-0,\;9-1+0)=(5,8)\). This agrees with the matrix product because the codomain basis is standard.
Linearity and Coordinate Calculations
The representation equation also respects addition and scalar multiplication. This is not an extra property imposed on the matrix: it follows from the linearity of \(T\), the coordinate representation, and the rules of matrix multiplication.
\([T(r\mathbf{u}+s\mathbf{v})]_{\mathcal{C}}=\mathbf{M}\bigl(r[\mathbf{u}]_{\mathcal{B}}+s[\mathbf{v}]_{\mathcal{B}}\bigr)\).
Proof. Since \(T\) is linear, \(T(r\mathbf{u}+s\mathbf{v})=rT(\mathbf{u})+sT(\mathbf{v})\). Coordinates in a fixed basis preserve linear combinations, so \([T(r\mathbf{u}+s\mathbf{v})]_{\mathcal{C}}=r[T(\mathbf{u})]_{\mathcal{C}}+s[T(\mathbf{v})]_{\mathcal{C}}\). Apply the representation equation to each term: this equals \(r\mathbf{M}[\mathbf{u}]_{\mathcal{B}}+s\mathbf{M}[\mathbf{v}]_{\mathcal{B}}\). Distributing matrix multiplication over addition and scalar multiplication gives the claimed expression. \(\square\)
Worked Example: Standard Bases and a Rectangular Matrix
Let \(T:\mathbb{R}^2\to\mathbb{R}^3\) be defined by \(T(x,y)=(x+2y,\;2x-y,\;x+y)\), using standard bases in both spaces. Calculate \(T(-1,3)\) by matrix multiplication.
For an arbitrary input, the output entries are linear expressions in \(x\) and \(y\), so the coordinate equation is
Substitute the input coordinate column:
Check: The defining formula gives \(T(-1,3)=(-1+2\cdot3,\;2(-1)-3,\;-1+3)=(5,-5,2)\), as required.
Common Mistake: Multiplying by the Wrong Coordinates
In the first worked example, the matrix \(\begin{bmatrix}3&1\\4&-2\end{bmatrix}\) represents \(T\) relative to the domain basis \(((1,1),(1,-1))\), not the standard domain basis. A tempting but incorrect calculation is
This treats the usual entries of \((3,1)\) as though they were its coordinates in \(\mathcal{B}\). But \((3,1)=2(1,1)+1(1,-1)\), so its \(\mathcal{B}\)-coordinate column is \(\begin{bmatrix}2\\1\end{bmatrix}\), and the correct product is \(\begin{bmatrix}7\\6\end{bmatrix}\). The direct map calculation also gives \((7,6)\). Always check which basis the matrix expects before multiplying.
A matrix representation is therefore more than a list of coefficients: it is a coordinate-dependent rule for calculating a coordinate-dependent output. Keep the domain basis, codomain basis, and coordinate columns aligned, and matrix multiplication evaluates the original linear map exactly.
Check Your Understanding
For each question, identify the domain and codomain coordinates before carrying out any multiplication.
- If \(\mathcal{B}=((1,1),(1,-1))\), find \([(4,2)]_{\mathcal{B}}\).
- Using the matrix from the first worked example, calculate \(T(1,3)\).
- For the second worked example, use the coordinate vector \(\begin{bmatrix}0\\2\\-1\end{bmatrix}\) to find the output coordinates.
- Explain why a representing matrix’s product with \([\mathbf{v}]_{\mathcal{B}}\) gives coordinates of \(T(\mathbf{v})\), not automatically the usual entries of \(T(\mathbf{v})\).
- State the coordinate equation that is equivalent to \(T(\mathbf{v})=\mathbf{w}\).
