Consider the mapping (transformation) from $\mathbb{R}^2 \mapsto \mathbb{R}^2$, given by:

$(x,y) \mapsto (-y,x)$

This mapping rotates vectors through an angle of $90^{\circ}$ (anticlockwise).

Definition: A mapping $T$ from a vector space $V$ into a vector space $W$ (written $T:V \rightarrow W$) is said to be a linear transformation, if for all $\mathbf{u}, \mathbf{v} \in V$ and $\alpha, \beta \in \mathbb{R}$:

or equivalently:

Suppose $T$ is an anti-clockwise rotation by $90^{\circ}$ in the $xy$-plane.

Geometrically, we can see that the diagonal of the sum vector parallelogram rotates along with its components. In other words,

Similarly, for scaling a vector by a scalar constant $\alpha$, under the same $90^{\circ}$ anti-clockwise rotation:

Since $T(\mathbf{u}) + T(\mathbf{v}) \neq T(\mathbf{u}+\mathbf{v})$, it is not a linear transformation.

Note: You can disprove linearity with a single counterexample. However, to prove a mapping is a linear transformation, you must show it holds in general.

Let's test with a counterexample. Say $ \mathbf{u}= \begin{pmatrix}1\\0\end{pmatrix}, \;\; \mathbf{v}= \begin{pmatrix}0\\1\end{pmatrix}. $

But computing them individually yields:

Since $ T(\mathbf{u}+\mathbf{v}) \neq T(\mathbf{u})+T(\mathbf{v}), $ the mapping is not a linear transformation.

Note: A single counterexample is enough to disprove linearity.

Let's test the additivity property. Take $ \mathbf{u}= \begin{pmatrix}1\\0\end{pmatrix}, \; \mathbf{v}= \begin{pmatrix}0\\1\end{pmatrix}. $

Individual computation gives: $ T(\mathbf{u}) = \begin{pmatrix}1\\0\\1\end{pmatrix}, \; T(\mathbf{v}) = \begin{pmatrix}0\\1\\1\end{pmatrix} $ $ \implies T(\mathbf{u})+T(\mathbf{v}) = \begin{pmatrix}1\\1\\2\end{pmatrix} $

So, for these vectors, $ T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u})+T(\mathbf{v}). $

Consider another transformation on $\mathbb{R}^2$ given by $T(\mathbf{v}) = 3\mathbf{v}$. We check the core properties to confirm if $T$ is linear.

Consider another transformation on $\mathbb{R}^2$ given by $T(\mathbf{v}) = 3\mathbf{v}$. We check the core properties to confirm if $T$ is linear.

(a) Check whether $T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$

(b) Check whether $T(\alpha \mathbf{u}) = \alpha T(\mathbf{u})$

Consider another transformation on $\mathbb{R}^2$ given by $T(\mathbf{v}) = 3\mathbf{v}$. We check the core properties to confirm if $T$ is linear.

(a) Check whether $T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$

(b) Check whether $T(\alpha \mathbf{u}) = \alpha T(\mathbf{u})$

Both rules are satisfied!

So $T(\mathbf{v}) = 3\mathbf{v}$ is rigorously proven to be a linear transformation!

Consider the mapping from $\mathbb{R}^2$ to $\mathbb{R}^3$ given by:

$ T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \left(x_2, x_1, x_1 + x_2\right)^T = \begin{pmatrix} x_2 \\ x_1 \\ x_1 + x_2 \end{pmatrix}$

We will show that this is linear using the single consolidated criteria: $$T\left(\alpha \mathbf{u} + \beta \mathbf{v}\right) = \alpha T(\mathbf{u}) + \beta T(\mathbf{v}).$$

Consider the mapping from $\mathbb{R}^2$ to $\mathbb{R}^3$ given by:

$ T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \left(x_2, x_1, x_1 + x_2\right)^T = \begin{pmatrix} x_2 \\ x_1 \\ x_1 + x_2 \end{pmatrix}$

Thus $\alpha \mathbf{u} + \beta \mathbf{v} = \begin{pmatrix} \alpha u_1 + \beta v_1 \\ \alpha u_2 + \beta v_2 \end{pmatrix}$. Applying the rules of mapping $T$ gives:

1. Show that the following transformation $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is linear: $$T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2x_1 \\ x_2 \end{pmatrix}$$
2. Show that $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by the following mapping is a linear transformation: $$T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_2 \\ 3x_1 \end{pmatrix}$$

To prove that $T$ is linear, show that $ T(\alpha\mathbf{u}+\beta\mathbf{v}) = \alpha T(\mathbf{u}) + \beta T(\mathbf{v}) $ for all vectors $\mathbf{u},\mathbf{v}$ and scalars $\alpha,\beta$.

Since this holds for all $\mathbf{u},\mathbf{v}\in\mathbb{R}^2$ and all $\alpha,\beta\in\mathbb{R}$, $T$ is a linear transformation.

Consider the definite integral operator $I : C[a, b] \rightarrow \mathbb{R}$ defined by: $$I\left(f\right) = \ds \int_{a}^{b} f(t) \,dt$$

(where $C[a, b]$ represents the set of all continuous functions on the interval $[a, b]$).

Thus, this calculus integration operator is a linear transformation.

Consider the derivative operator $T = \dfrac{d}{dx}$ mapping $P_2 \rightarrow P_1$, where $P_n$ is the vector space of polynomials of degree $n$ or less.

Revisiting our prior example where $T\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_2 \\ x_1 \\ x_1 + x_2 \end{pmatrix}.$

Let's define matrix $A\,$ as $\,A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{pmatrix}.$ We can rewrite the transformation cleanly as matrix multiplication:

It is true in general that any operation defined by $T(\mathbf{x}) = A\mathbf{x}$ forms a linear transformation.

Theorem: If $T$ is a linear transformation mapping from $\mathbb{R}^n$ into $\mathbb{R}^m$, there exists a unique $m \times n$ matrix $A$ such that $T(\mathbf{x}) = A\mathbf{x}$ for all $\mathbf{x} \in \mathbb{R}^n$.

Crucially, the $j$-th column vector of $A$ is found by transforming the standard basis element $\mathbf{e}_j$. That is, $\,\mathbf{a}_j = T(\mathbf{e}_j)$ for $j = 1, 2, \dots, n$.

Theorem: If $T$ is a linear transformation mapping from $\mathbb{R}^n$ into $\mathbb{R}^m$, there exists a unique $m \times n$ matrix $A$ such that $T(\mathbf{x}) = A\mathbf{x}$ for all $\mathbf{x} \in \mathbb{R}^n$.

Proof Outline:

Expanding the linear combination $T(\mathbf{x}) = x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \dots + x_n\mathbf{a}_n$ into component forms:

Find the matrix representation for the linear mapping $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ given by

Geometric Interpretation

Suppose you are given the transformation matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$.

How would you describe the operation $T(\mathbf{v}) = A\mathbf{v}$ geometrically?

Conclusion: This transformation reflects points across the horizontal $x$-axis.

Geometric Interpretation

How would you describe the operation $T(\mathbf{v}) = A\mathbf{v}$ geometrically?

Reflection in the $x$-axis

When coordinates are reflected across the $x$-axis, the mappings translate as follows:

Reflection in the $y$-axis

When coordinates are instead reflected across the vertical $y$-axis, the mappings change to:

Reflection in the $y$-axis

Geometric interpretation

To rotate the standard 2D axes counterclockwise by a generic angle $\theta$:

Theorem: Let $T:\mathbb R^n \to \mathbb R^n$ be a linear transformation with matrix $A.$ Then the inverse transformation $T^{-1}$ exists if and only if $\det(A)\neq 0.$ Furthermore, $T^{-1}:\mathbb R^n \to \mathbb R^n$ is a linear transformation with matrix $A^{-1}.$

Example: If $T:\mathbb R^3 \to \mathbb R^3$ has a matrix \[A = \left( \begin{array}{ccc} 2 & 3 & 1\\ -1 & 2 & 3\\ 4 & 2 & 6\\ \end{array} \right), \]

then \(T^{-1}\) exists since \(\det(A) = 63\) \(\neq 0.\) This means that $A$ is invertible and $A^{-1}$ exists. Therefore $T^{-1}$ exists and is given by $T^{-1}(\mathbf x) = A^{-1}\mathbf x.$

Theorem: Let $T:\mathbb R^n \to \mathbb R^n$ be a linear transformation with matrix $A.$ Then the inverse transformation $T^{-1}$ exists if and only if $\det(A)\neq 0.$ Furthermore, $T^{-1}:\mathbb R^n \to \mathbb R^n$ is a linear transformation with matrix $A^{-1}.$

Example: To determine the inverse of $A$ we can apply the Gauss-Jordan elimination:

So \(\, T^{-1}\left(\mathbf x^T\right) \) \( =T^{-1}\left( \left(x_1, x_2, x_3\right)^T\right) \)

Example of a Product Transformation

Reflect each vector $\mathbf{x}\in\mathbb{R}^2$ across the $y$-axis and then rotate it $60^\circ$ anti-clockwise.

$$ R_y=\begin{pmatrix}-1&0\\0&1\end{pmatrix}, \qquad R_{60}= \begin{pmatrix} \frac12&-\frac{\sqrt3}{2}\\ \frac{\sqrt3}{2}&\frac12 \end{pmatrix}. $$

Since the reflection is applied first, $ T\rightarrow R_{60}R_y. $

Thus, the matrix representing the composition is $ \begin{pmatrix} -\frac12&-\frac{\sqrt3}{2}\\ -\frac{\sqrt3}{2}&\frac12 \end{pmatrix}. $

Example of a Product Transformation

Double the length of each vector $\mathbf{x}\in\mathbb{R}^2$ and then rotate it $30^\circ$ clockwise.

$$ S= \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}, \qquad R_{-30}= \begin{pmatrix} \frac{\sqrt3}{2}&\frac12\\ -\frac12&\frac{\sqrt3}{2} \end{pmatrix}. $$

Since the scaling is applied first, $ T\rightarrow R_{-30}S. $

Thus, the matrix representing the composition is $ \begin{pmatrix} \sqrt3&1\\ -1&\sqrt3 \end{pmatrix}. $

A More General Example of a Product Transformation

Let

be linear transformations defined as

where $M_n\big(\mathbb R\big)$ is the set of real $n\times n$ matrices and $\text{tr}(A)$ is the trace of $A$ (i.e. the sum of elements on the main diagonal).

Then $T_{2}\circ T_{1}:M_n\big(\mathbb R\big)\rightarrow\mathbb{R}$ is defined by:

A More General Example of a Product Transformation

Let

be linear transformations defined as

For example, if $A=\begin{pmatrix}2&-1\\ -3&6\end{pmatrix}$ then:

$T_{1}(A)=A+A^{T}$, $=\begin{pmatrix}4&-4\\ -4&12\end{pmatrix}$, so that $T_{2}\big(T_{1}(A)\big)$ $=\text{tr}\begin{pmatrix}4&-4\\ -4&12\end{pmatrix}$ $=16$.

• For any linear transformation $T$ from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$ we can find an $m\times n$ matrix $A$ so that $T(\mathbf{v})=A\mathbf{v}$.
• If the transformation is invertible, the inverse transformation has the matrix $A^{-1}$.
• The composition of two transformations $T_{1}(\mathbf{v})=A_{1}\mathbf{v}$ and $T_{2}(\mathbf{v})=A_{2}\mathbf{v}$ is the product matrix $A_{2}A_{1}$.