Recovering a Matrix from Transformation Constraints

What Transformation Constraints Determine

A linear transformation is determined by how it acts on enough inputs to describe every vector in its domain. In this tutorial, the domain is \(\mathbb{R}^2\), and the output space is either \(\mathbb{R}^2\) or \(\mathbb{R}^3\). The unknown standard matrix therefore has two columns: each column records the image of one standard basis vector. If images are prescribed for other independent inputs, we can first express the standard basis vectors using those inputs and then compute their images.

Throughout, vectors are columns and scalars are real. A map \(T:\mathbb{R}^2\to\mathbb{R}^m\), where \(m=2\) or \(m=3\), is linear if \(T(a\mathbf{x}+b\mathbf{y})=aT(\mathbf{x})+bT(\mathbf{y})\) for all \(\mathbf{x},\mathbf{y}\in\mathbb{R}^2\) and all \(a,b\in\mathbb{R}\). The standard basis vectors are \(\mathbf{e}_1=(1,0)^\top\) and \(\mathbf{e}_2=(0,1)^\top\).

The columns are outputs, not rows of input-output data. Indeed, if \(\mathbf{x}=(x_1,x_2)^\top=x_1\mathbf{e}_1+x_2\mathbf{e}_2\), linearity gives \(T(\mathbf{x})=x_1T(\mathbf{e}_1)+x_2T(\mathbf{e}_2)\). This is exactly the product of the matrix with columns \(T(\mathbf{e}_1)\) and \(T(\mathbf{e}_2)\) by \(\mathbf{x}\).

Proof. Let \(T\) be linear and set \(\mathbf{A}=\begin{bmatrix}T(\mathbf{e}_1)&T(\mathbf{e}_2)\end{bmatrix}\). For any \(\mathbf{x}=(x_1,x_2)^\top\), the equality \(\mathbf{x}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\) gives \(T(\mathbf{x})=x_1T(\mathbf{e}_1)+x_2T(\mathbf{e}_2)=\mathbf{A}\mathbf{x}\). Thus \(\mathbf{A}\) represents \(T\). If another matrix \(\mathbf{B}\) also represents \(T\), then \(\mathbf{B}\mathbf{e}_1=T(\mathbf{e}_1)\) and \(\mathbf{B}\mathbf{e}_2=T(\mathbf{e}_2)\); these products are the first and second columns of \(\mathbf{B}\), so \(\mathbf{B}=\mathbf{A}\). Conversely, given any \(m\times2\) matrix \(\mathbf{A}\), define \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). Matrix multiplication gives \(T(a\mathbf{x}+b\mathbf{y})=aT(\mathbf{x})+bT(\mathbf{y})\), so \(T\) is linear, and its values on \(\mathbf{e}_1,\mathbf{e}_2\) are the columns of \(\mathbf{A}\). The formula also shows that no other map with those columns is possible. \(\square\)

Starting with the Standard Basis

Recovering a Matrix from Other Independent Inputs

Now suppose the prescribed inputs are \(\mathbf{u}\) and \(\mathbf{v}\), not necessarily \(\mathbf{e}_1,\mathbf{e}_2\). They must be linearly independent: neither may be obtained as a scalar multiple of the other, and the only equation \(s\mathbf{u}+t\mathbf{v}=\mathbf{0}\) must have \(s=t=0\). In \(\mathbb{R}^2\), two independent vectors provide unique coordinates for every input. One way to see this computationally is to place them as columns in \(\mathbf{U}=\begin{bmatrix}\mathbf{u}&\mathbf{v}\end{bmatrix}\). The earlier pivot-column criterion says independent columns have a pivot in each column. Here \(\mathbf{U}\) has two rows and two columns, so its reduced row echelon form is \(\mathbf{I}_2\); consequently, \(\mathbf{U}\mathbf{c}=\mathbf{x}\) has exactly one solution \(\mathbf{c}\) for each \(\mathbf{x}\in\mathbb{R}^2\).

Proof. Because \(\mathbf{u},\mathbf{v}\) are independent, the pivot-column criterion applied to \(\mathbf{U}=\begin{bmatrix}\mathbf{u}&\mathbf{v}\end{bmatrix}\) gives a pivot in each of its two columns. Its reduced row echelon form is therefore \(\mathbf{I}_2\), so for every \(\mathbf{x}\in\mathbb{R}^2\) there is a unique pair \(s,t\in\mathbb{R}\) with \(\mathbf{x}=s\mathbf{u}+t\mathbf{v}\). Define \(T(\mathbf{x})=s\mathbf{p}+t\mathbf{q}\) using this unique representation. If \(\mathbf{x}=s\mathbf{u}+t\mathbf{v}\) and \(\mathbf{y}=r\mathbf{u}+w\mathbf{v}\), then \(\mathbf{x}+\mathbf{y}=(s+r)\mathbf{u}+(t+w)\mathbf{v}\), and for any scalar \(c\), \(c\mathbf{x}=(cs)\mathbf{u}+(ct)\mathbf{v}\). The definition of \(T\) now gives \(T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})\) and \(T(c\mathbf{x})=cT(\mathbf{x})\), so \(T\) is linear. Its values on \(\mathbf{u}\) and \(\mathbf{v}\) are \(\mathbf{p}\) and \(\mathbf{q}\), respectively. Finally, any linear map with those prescribed values must satisfy \(T(s\mathbf{u}+t\mathbf{v})=s\mathbf{p}+t\mathbf{q}\) for every \(\mathbf{x}=s\mathbf{u}+t\mathbf{v}\); thus it must be the map just defined. \(\square\)

This proof also suggests a practical method. Find the coefficients representing \(\mathbf{e}_1\) and \(\mathbf{e}_2\) in terms of \(\mathbf{u},\mathbf{v}\). Apply the same coefficients to \(\mathbf{p},\mathbf{q}\); the resulting output vectors are the columns of the standard matrix.

Extra Constraints and Consistency

Sometimes more than two input-output pairs are given. Once two independent inputs determine the map, every additional condition must agree with the same linear combination of outputs. This is not an extra choice: linearity forces it.

Proof. If a linear map satisfies \(T(\mathbf{u})=\mathbf{p}\) and \(T(\mathbf{v})=\mathbf{q}\), then \(T(\mathbf{w})=T(a\mathbf{u}+b\mathbf{v})=a\mathbf{p}+b\mathbf{q}\). Therefore consistency requires \(\mathbf{r}=a\mathbf{p}+b\mathbf{q}\). Conversely, if that equality holds, the unique map from the recovery theorem has \(T(\mathbf{w})=a\mathbf{p}+b\mathbf{q}=\mathbf{r}\), so it satisfies the additional condition as well. \(\square\)

A Common Mistake: Placing Images in Rows

Suppose \(T(\mathbf{e}_1)=(2,-1)^\top\) and \(T(\mathbf{e}_2)=(3,4)^\top\). A common incorrect arrangement is \(\begin{bmatrix}2&-1\\3&4\end{bmatrix}\), formed by writing the image coordinates as rows. But multiplying this matrix by \(\mathbf{e}_1\) gives \((2,3)^\top\), not the required \((2,-1)^\top\). The correction is to place each image in a column:

A reliable check is to multiply the proposed matrix by each standard basis vector: the products select its columns, so they must reproduce the prescribed images in order. For nonstandard inputs, check each given input directly instead. If there are extra constraints, first express each extra input as a combination of the independent inputs; its output must use the same coefficients.