1.8 Introduction to Linear Transformations
- A transformation \(T\) (not necessarily a linear transformation) that maps a vector \(\mathbf{x}\in\mathbb{R}^n\) to a vector \(T(\mathbf{x})\in\mathbb{R}^m\) is denoted as \(T: \mathbb{R}^n\to\mathbb{R}^m\). The \(\mathbb{R}^n\) is called the domain, and the \(\mathbb{R}^m\) is called the codomain.
Note that although the range of \(T\), \(\text{Range}(T)=\{T(\mathbf{x})\mid\mathbf{x}\in\mathbb{R}^n\}\), may not cover the entire codomain, the dimensionality is the same: \(\text{Range}(T)\subseteq \mathbb{R}^m\). - A matrix transformation is denoted as \(T(\mathbf{x})=A\mathbf{x}\) or \(\mathbf{x}\mapsto A\mathbf{x}\). Suppose \(A\) is an \(m\times n\) matrix, then the domain is \(\mathbb{R}^n\), and the codomain is \(\mathbb{R}^m\).
- A linear transformation, \(T\), satisfies the following conditions for all \(\mathbf{u}\), \(\mathbf{v}\), and \(c\):
- \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\);
- \(T(c\mathbf{u})=c T(\mathbf{u})\).
- The above two conditions are equivalent to \(T(c\mathbf{u}+d\mathbf{v})=c T(\mathbf{u})+d T(\mathbf{v})\).
- If \(T: \mathbb{R}^n \to \mathbb{R}^m\) is a linear transformation, \(T(\mathbf{0})=\mathbf{0}\), but the zero vector on the left hand side is in \(\mathbb{R}^n\), while the one on the right hand side is in \(\mathbb{R}^m\). This characteristic can be proved by substituting \(\mathbf{u}\) and \(\mathbf{v}\) with zero vectors and using either of the two conditions.
Solution (Solution to Exercise 1). \[\begin{bmatrix}1 & -3 & 3 \\ 3 & 5 & 2 \\ -1 & 7 & -5\end{bmatrix} \sim \begin{bmatrix}1 & -3 & 3 \\ 0 & 1 & -.5 \\ 0 & 0 & 0\end{bmatrix}\]
Since there are no free variables (the last column is augmented), there is exactly one \(x\) such that \(T(\mathbf{x})=\mathbf{b}\).
Remark. Having a row of zeros does not imply that the equation has multiple solutions. It all depends on free variables—namely, the coefficient columns.
Solution (Solution to Exercise 2). Though it’s possible to solve the transformation matrix because all are in \(\mathbb{R}^2\), the forms of \(\mathbf{e}_1\) and \(\mathbf{e}_2\) make us easy to represent any \(\mathbb{R}^2\) vectors using \(\mathbf{e}_1\) and \(\mathbf{e}_2\).
\[\begin{bmatrix}x_1\\ x_2\end{bmatrix}=x_1\mathbf{e}_1+x_2\mathbf{e}_2\]
Thus,
\[\begin{align*} T(\begin{bmatrix}x_1\\ x_2\end{bmatrix}) &= T(x_1\mathbf{e}_1+x_2\mathbf{e}_2) \\ &= x_1 T(\mathbf{e}_1) + x_2 T(\mathbf{e}_2) \\ &= x_1\mathbf{y}_1 + x_2\mathbf{y}_2 \end{align*}\]
A more general form will be discussed in the next chapter.
Solution (Solution to Exercise 3). Two points to notice at first:
- This is mentioned not because it could cause misunderstanding, but because it clarifies the meaning of vectors. The textbook is rigorously phrasing the problem as “the line through \(\mathbf{p}\) in the direction of \(\mathbf{v}\)”;
- The equation is a parametric equation, so all vectors and lines in this problem are in \(\mathbb{R}^n\). Do not confuse this with \(y=a+bx\), though the line equation can also be written as \(\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}0\\ a\end{bmatrix} + x\begin{bmatrix}1\\ b\end{bmatrix}\).
A paraphrase of the problem would be “does applying a linear transformation to this line always yield another line, or possibly a single point?”
\[T(\mathbf{x})=T(\mathbf{p}+t\mathbf{v})\] According to the definition of linear transformations, \[T(\mathbf{p}+t\mathbf{v})=T(\mathbf{p})+t T(\mathbf{v}).\]
Since \(T:\mathbb{R}^n\to\mathbb{R}^n\) maps a vector in \(\mathbb{R}^n\) to another vector in \(\mathbb{R}^n\), \(T(\mathbf{p})\) and \(T(\mathbf{v})\) are both vectors in \(\mathbb{R}^n\). Thus, the new line can be described as a line through \(T(\mathbf{p})\) in the direction of \(T(\mathbf{v})\).
When \(T(\mathbf{v})=\mathbf{0}\), the line will be mapped onto the single point at \(T(\mathbf{p})\).
Remark. We now know that applying a linear transformation on a line generates a new line or a point (a degenerate line). Quite linear, right?
We know that every matrix transformation is a linear transformation since \(A(\mathbf{u}+\mathbf{v})=A\mathbf{u}+A\mathbf{v}\) and \(A(c\mathbf{u})=c(A\mathbf{u})\).
If we are working in finite-dimensional vector spacese with chosen bases (e.g. \(\mathbb{R}^n\to\mathbb{R}^m\)), then yes — every linear transformation is a matrix transformation. The matrix depends on the choice of basis. (We will also learn this in the next chapter.)
If we are working in infinite-dimensional vector spaces (like function spaces), then not every linear transformation corresponds to a finite matrix. For example, the derivative operator \(D: C^\infty (\mathbb{R})\to C^\infty(\mathbb{R}),D(f)=f'\), is linear but cannot be represented by a finite matrix.
— ChatGPT