Why the Columns Record Basis-Vector Images
A matrix representation encodes how a linear map acts, but its entries depend on the bases used to describe inputs and outputs. The columns have a concrete meaning: each one records the output of the map on a particular domain basis vector, expressed in the codomain basis. This connection explains how to construct a representation and how to check that the resulting matrix uses the intended coordinates.
Throughout, spaces and scalars are over \(\mathbb{R}\), and the vector spaces considered here are finite-dimensional. An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) of a vector space \(V\) is an ordered list that is linearly independent and spans \(V\). Consequently, every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\). Its coordinate vector is the column \([\mathbf{v}]_{\mathcal{B}}=(a_1,\dots,a_n)^\top\). The order matters: changing the order of a basis changes the order of the coordinates.
Let \(T:V\to W\) be a linear map, with domain basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) and codomain basis \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\). The matrix representation \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\) is the matrix that takes the coordinates of an input in \(\mathcal{B}\) to the coordinates of its image in \(\mathcal{C}\). In particular, for every \(\mathbf{v}\in V\), it satisfies \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\).
Proof. Write \(\mathbf{e}_j\) for the \(j\)th standard basis vector in \(\mathbb{R}^n\). The basis expansion of \(\mathbf{b}_j\) in \(\mathcal{B}\) has coefficient \(1\) on \(\mathbf{b}_j\) and \(0\) on every other basis vector. Thus \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\). By the defining property of the representation matrix, \([T(\mathbf{b}_j)]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{e}_j\). Multiplication by \(\mathbf{e}_j\) selects the \(j\)th column of a matrix, so this column is exactly \([T(\mathbf{b}_j)]_{\mathcal{C}}\). \(\square\)
The theorem gives a direct construction. Apply \(T\) to each domain basis vector, express each resulting image in the codomain basis, and place the coordinate vectors in order as the columns. The image \(T(\mathbf{b}_j)\) itself is a vector in \(W\); the column is its coordinate vector in \(\mathcal{C}\). These are the same information in different forms, not necessarily the same list of numbers as the image in some other coordinates.
Proof. The coordinates of \(\mathbf{v}\) in \(\mathcal{B}\) are \((x_1,\dots,x_n)^\top\). By the definition of the representation matrix, \([T(\mathbf{v})]_{\mathcal{C}}=\mathbf{A}(x_1,\dots,x_n)^\top\). The rule for multiplying a matrix by a column vector gives \(\mathbf{A}(x_1,\dots,x_n)^\top=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), which proves the claim. \(\square\)
Constructing a Representation in Chosen Bases
Worked Example: A Map from \(\mathbb{R}^2\) to \(\mathbb{R}^3\)
Define \(T:\mathbb{R}^2\to\mathbb{R}^3\) by \(T(x,y)=(2x-y,\ x+3y,\ 4x+y)^\top\). Use the domain basis \(\mathcal{B}=((1,1)^\top,(1,-1)^\top)\) and the standard basis \(\mathcal{C}\) of \(\mathbb{R}^3\) to construct the representation matrix.
The proposed domain list is a basis: for any \((x,y)^\top\), the equation \((x,y)^\top=a(1,1)^\top+b(1,-1)^\top\) gives \(a+b=x\) and \(a-b=y\), hence \(a=(x+y)/2\) and \(b=(x-y)/2\). These coefficients exist and are unique. Now apply \(T\) to each basis vector. Since \(\mathcal{C}\) is the standard basis, the output coordinates are its ordinary entries.
By the column theorem, these coordinate vectors form the first and second columns, in that order:
Check the representation on an input with \(\mathcal{B}\)-coordinates \((2,-1)^\top\). The corresponding vector is \(2(1,1)^\top-(1,-1)^\top=(1,3)^\top\). Directly, \(T(1,3)=(-1,10,7)^\top\). Using the representation gives
The result agrees with the direct computation, and the output coordinates are in the standard basis \(\mathcal{C}\).
When the codomain basis is not standard, the images must be converted to coordinates before they are put into the matrix. Solving for those coordinates is part of the construction, not an optional adjustment made after writing down the image vectors.
Worked Example: A Nonstandard Codomain Basis
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be \(T(x,y)=(2x+y,\ x-y)^\top\). Use the standard basis \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2)\) in the domain and \(\mathcal{C}=((1,1)^\top,(1,-1)^\top)\) in the codomain.
The list \(\mathcal{C}\) is a basis: the equations \((u,v)^\top=a(1,1)^\top+b(1,-1)^\top\) give \(a+b=u\), \(a-b=v\), and therefore \(a=(u+v)/2\), \(b=(u-v)/2\). In particular, for any output \((u,v)^\top\), its \(\mathcal{C}\)-coordinates are \(((u+v)/2,(u-v)/2)^\top\).
The domain basis vectors have images \(T(\mathbf{e}_1)=(2,1)^\top\) and \(T(\mathbf{e}_2)=(1,-1)^\top\). Convert the first image to \(\mathcal{C}\)-coordinates: its coefficients are \((2+1)/2=3/2\) and \((2-1)/2=1/2\). The second image is the second vector of \(\mathcal{C}\), so its coordinates are \((0,1)^\top\). The columns therefore give
For an arbitrary input \((x,y)^\top\), its domain coordinates are \((x,y)^\top\), so the matrix produces \((3x/2,\ x/2+y)^\top\). Reconstruct the output from these \(\mathcal{C}\)-coordinates:
This verifies the matrix for every input, not just the two basis vectors.
More Domain Basis Vectors Than Output Coordinates
The column rule also applies when the number of domain basis vectors and the number of codomain basis vectors differ. In that case, each image is still expressed in the codomain basis, and each resulting coordinate vector is still one column. The following example makes this distinction visible: the original map outputs ordinary vectors with two entries, while the matrix columns record those outputs in a chosen basis.
Worked Example: A Map from \(\mathbb{R}^3\) to \(\mathbb{R}^2\)
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x,y,z)=(x+y,\ y+z)^\top\). Use the standard basis in the domain and \(\mathcal{C}=((1,1)^\top,(1,-1)^\top)\) in the codomain.
As above, a vector \((u,v)^\top\) has \(\mathcal{C}\)-coordinates \(((u+v)/2,(u-v)/2)^\top\). The three standard domain basis vectors have images
Convert each image to \(\mathcal{C}\)-coordinates. For \((1,0)^\top\), the coordinates are \((1/2,1/2)^\top\); for \((1,1)^\top\), they are \((1,0)^\top\); and for \((0,1)^\top\), they are \((1/2,-1/2)^\top\). Thus
For an arbitrary input, matrix multiplication gives the output coordinates \(((x+2y+z)/2,\ (x-z)/2)^\top\). Reconstructing the vector represented by these coordinates yields
The reconstruction checks both rows of the matrix against the map for every \(x,y,z\).
A Common Coordinate Error
A frequent mistake is to put the entries of \(T(\mathbf{b}_j)\) directly into a column even when those entries are written in coordinates other than the chosen codomain basis. In the second example, the image \(T(\mathbf{e}_1)=(2,1)^\top\) is written in standard coordinates. It is not the coordinate vector of that image in \(\mathcal{C}\).
The incorrect matrix \(\begin{bmatrix}2&1\\1&-1\end{bmatrix}\) would treat \((2,1)^\top\) as the \(\mathcal{C}\)-coordinates of \(T(\mathbf{e}_1)\). But those coordinates would represent \(2(1,1)^\top+1(1,-1)^\top=(3,1)^\top\), which is not \(T(\mathbf{e}_1)=(2,1)^\top\). The correction is to solve for coordinates in the specified basis: \(T(\mathbf{e}_1)=\frac32(1,1)^\top+\frac12(1,-1)^\top\), so the first column is \((3/2,1/2)^\top\). Always label the basis used for each coordinate vector before assembling the columns.
To construct and verify a representation, keep the two bases in view: the domain basis determines which inputs to apply \(T\) to and the order of the columns; the codomain basis determines how each image is converted into column entries. Once the columns have been assembled, a general input is handled by weighting those columns with its domain coordinates. This is precisely why matrix multiplication reproduces the coordinates of the mapped vector.
Check Your Understanding
For each question, keep track of which basis is used for the input coordinates and which for the output coordinates.
- Explain why multiplying a matrix by \(\mathbf{e}_j\) selects its \(j\)th column, and use this to prove the column rule for a map representation.
- Let \(T(x,y)=(x+2y,3x-y)^\top\), with domain basis \(((1,0)^\top,(1,1)^\top)\) and standard codomain basis. Find the representation matrix by computing both basis-vector images.
- For the codomain basis \(((1,1)^\top,(1,-1)^\top)\), find the coordinates of \((3,1)^\top\).
- In the third worked example, multiply the representation matrix by \((1,-2,3)^\top\), then reconstruct the output in the codomain basis and compare with direct evaluation of \(T(1,-2,3)\).
- Describe the error in using the ordinary entries of an image as a matrix column when the codomain basis is nonstandard.
