Linear Transformations#
Table of Contents#
Definition of Transformation#
Example#
\(T : \mathbb{R}^2 \to \mathbb{R}^3\)
\( \begin{bmatrix} x \\ y \\ \end{bmatrix} \to \begin{bmatrix} x \\ y \\ xy \\ \end{bmatrix} \)
Example#
\(T : \mathbb{R}^2 \to \mathbb{R}^3\)
\( \begin{bmatrix} x \\ y \\ \end{bmatrix} \to \begin{bmatrix} \frac{2x }{x^2 + y^2 + 1} \\ \frac{2y }{x^2 + y^2 + 1} \\ \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \\ \end {bmatrix} \)
Definition of Linear Transformation#
If \(T\) is a linear transformation then
1. \(T(r \mathbf{u} + s \mathbf{v}) = \underbrace{T(r \mathbf{u}) + T(s \mathbf{v})}_{\textbf{by condition 1}} = \underbrace{rT(\mathbf{u}) + sT(\mathbf{v})}_{\textbf{by condition 2}}\)
If a transformation is linear then it respects linear combinations: transforming a linear combination of domain vectors is equivalent to taking the same linear combination of their images.
2. \(T(\mathbf{0}) = T(0 \cdot \mathbf{v}) = 0 \cdot T(\mathbf{v}) = \mathbf{0}\)
Transforming the zero domain vector yields the zero range vector.
Example#
PROBLEM STATEMENT
Is the transformation \(T : \textcolor{#FFC300}{\mathbb{R}^2} \to \textcolor{#C70039}{\mathbb{R}^3}\) defined by the following mapping linear?
\( \textcolor{#FFC300}{\begin{bmatrix} x \\ y \\ \end{bmatrix}} \mapsto \textcolor{#C70039}{\begin{bmatrix} x \\ y \\ x + y \\ \end{bmatrix}} \)
Let there be two vectors \(\mathbf{u, v} \in \mathbb{R}^2\) and two scalars \(r, s \in \mathbb{R}\).
\( T \left ( r \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} + s \begin{bmatrix} v_1 \\ v_2 \\ \end{bmatrix} \right) = T \left ( \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ \end{bmatrix} \right) = \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ ru_1 + ru_2 + sv_1 + sv_2 \\ \end{bmatrix} = \begin{bmatrix} ru_1 \\ ru_2 \\ ru_1 + ru_2 \\ \end{bmatrix} + \begin{bmatrix} sv_1 \\ sv_2 \\ sv_1 + sv_2 \\ \end{bmatrix} = r \begin{bmatrix} u_1 \\ u_2 \\ u_1 + u_2 \\ \end{bmatrix} + s \begin{bmatrix} v_1 \\ v_2 \\ v_1 + v_2 \\ \end{bmatrix} = sT \left ( \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} \right) + sT \left ( \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} \right) \)
It is linear.
\( \textcolor{#C70039}{ \begin{bmatrix} x \\ y \\ x + y \\ \end{bmatrix}} = \begin{bmatrix} x \\ 0 \\ x \\ \end{bmatrix} + \begin{bmatrix} 0 \\ y \\ y \\ \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} = \textcolor{#FF5733}{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ \end {bmatrix}} \textcolor{#FFC300}{ \begin{bmatrix} x \\ y \\ \end{bmatrix}} \)
Example#
PROBLEM STATEMENT
Is the transformation \(T : \mathbb{R}^2 \to \mathbb{R}^3\) defined by the following mapping linear?
\( \begin{bmatrix} x \\ y \\ \end{bmatrix} \to \begin{bmatrix} x \\ y \\ xy \\ \end{bmatrix} \)
Let there be two vectors \(\mathbf{u, v} \in \mathbb{R}^2\) and two scalars \(r, s \in \mathbb{R}\).
\( T \left ( r \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} + s \begin{bmatrix} v_1 \\ v_2 \\ \end{bmatrix} \right) = T \left ( \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ \end{bmatrix} \right) = \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ (ru_1 + sv_1)(ru_2 + sv_2) \\ \end{bmatrix} = \\ \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ r^2u_1u_2 + rs(u_1v_2 + u_2v_1) + s^2v_1v_2 \\ \end{bmatrix} \ne \begin{bmatrix} ru_1 + sv_1 \\ ru_2 + sv_2 \\ ru_1u_2 + sv_1v_2 \\ \end{bmatrix} \\ = \begin{bmatrix} ru_1 \\ ru_2 \\ ru_1u_2 \\ \end{bmatrix} + \begin{bmatrix} sv_1 \\ sv_2 \\ sv_1v_2 \\ \end{bmatrix} = r \begin{bmatrix} u_1 \\ u_2 \\ u_1u_2 \\ \end{bmatrix} + s \begin{bmatrix} v_1 \\ v_2 \\ v_1v_2 \\ \end{bmatrix} = sT \left ( \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} \right) + sT \left ( \begin{bmatrix} u_1 \\ u_2 \\ \end{bmatrix} \right) \)
For instance
\( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \overset{T}{\mapsto} \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \overset{T}{\mapsto} \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \overset{T}{\mapsto} \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \)
\( T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = T \left ( \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} \right) = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \ne \begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} = T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) \)
It is not linear.
Example#
PROBLEM STATEMENT
Is the transformation \(T : \mathbb{R}^2 \to \mathbb{R}^3\) defined by the following mapping linear?
\( \begin{bmatrix} x \\ y \\ \end{bmatrix} \to \begin{bmatrix} \frac{2x }{x^2 + y^2 + 1} \\ \frac{2y }{x^2 + y^2 + 1} \\ \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \\ \end {bmatrix} \)
For instance
\( \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \overset{T}{\mapsto} \begin{bmatrix} 0 \\ 0 \\ -1 \\ \end{bmatrix} \)
It is not linear.
Linear transformations and matrices#
Let \(T : \textcolor{#FFC300}{\mathbb{R}^n} \to \textcolor{#C70039}{\mathbb{R}^m}\) be a linear transformation.
Let \(\mathbf{e_k} \in \mathbb{R}^n\) denote the vector whose only non zero entry is a \(1\) located in the \(k\)-th row.
The \(k\)-th column of the matrix associated with the linear transformation is the vector \(\mathbf{u_k} = T(\mathbf{e_k}) \in \mathbb{R}^m\)
\( \begin{aligned} T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{bmatrix} \right) &= T \left ( x_1 \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \\ \end{bmatrix} + \dots + x_n \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ \end{bmatrix} \right) \\ &= T (x_1 \mathbf{e_1} + x_2 \mathbf{e_2} + \dots + x_n \mathbf{e_n}) \\ &= x_1 T (\mathbf{e_1}) + x_2 T (\mathbf{e_2}) + \dots + x_n T (\mathbf{e_n}) \\ &= x_1 \mathbf{u_1} + x_2 \mathbf{u_2} + \dots + x_n \mathbf{u_n} \\ &= \underbrace{ \underbrace{ \begin{bmatrix} \vert & \vert & & \vert \\ \vert & \vert & & \vert \\ \mathbf{u_1} & \mathbf{u_2} & \dots & \mathbf{u_n} \\ \vert & \vert & & \vert \\ \vert & \vert & & \vert \\ \end {bmatrix} }_{\textcolor{#C70039}{m} \times \textcolor{#FFC300}{n}} \underbrace{ \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} }_{\textcolor{#FFC300}{n} \times 1} }_{\textcolor{#C70039}{m} \times 1} \end {aligned} \)
Each linear transformation from \(\textcolor{#FFC300}{\mathbb{R}^n}\) to \(\textcolor{#C70039}{\mathbb{R}^m}\) determines an \(\textcolor{#C70039}{m} \times \textcolor{#FFC300}{n}\) matrix.
Conversely, each \(\textcolor{#C70039}{m} \times \textcolor{#FFC300}{n}\) matrix \(\mathbf{A}\) determines a linear transformation \(\mathbf{x \mapsto Ax}\) from \(\textcolor{#FFC300}{\mathbb{R}^n}\) to \(\textcolor{#C70039}{\mathbb{R}^m}\).
Example#
PROBLEM STATEMENT
Is the transformation linear?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] x_1 - x_2 \\ 3x_1 + x_2 \\ \end {bmatrix*} \)
In other words, is there a \(2 \times 2\) matrix \(\mathbf{A}\) s.t. \(\mathbf{x \mapsto Ax}\)?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] x_1 - x_2 \\ 3x_1 + x_2 \\ \end {bmatrix*} = \begin{bmatrix*}[r] x_1 \\ 3x_1 \\ \end{bmatrix*} + \begin{bmatrix*}[r] -x_2 \\ x_2 \\ \end{bmatrix*} = x_1 \begin{bmatrix*}[r] 1 \\ 3 \\ \end{bmatrix*} + x_2 \begin{bmatrix*}[r] -1 \\ 1 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 1 & -1 \\ 3 & 1 \\ \end {bmatrix*} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \)
Yes, the transformation is linear.
Example#
PROBLEM STATEMENT
Is the transformation linear?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] x_1 &-& 2x_2 \\ x_1 &-& 3 \\ 2x_1 &-& 5x_2 \\ \end {bmatrix*} \)
In other words, is there a \(3 \times 2\) matrix \(\mathbf{A}\) s.t. \(\mathbf{x \mapsto Ax}\)?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] x_1 &-& 2x_2 \\ x_1 &-& 3 \\ 2x_1 &-& 5x_2 \\ \end {bmatrix*} = \begin{bmatrix*}[r] 0 \\ -3 \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] x_1 \\ x_1 \\ 2x_1 \\ \end{bmatrix*} + \begin{bmatrix*}[r] -2x_2 \\ 0 \\ -5x_2 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ -3 \\ 0 \\ \end{bmatrix*} + x_1 \begin{bmatrix*}[r] 1 \\ 1 \\ 2 \\ \end{bmatrix*} + x_2 \begin{bmatrix*}[r] -2 \\ 0 \\ -5 \\ \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ -3 \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -2 \\ 1 & 0 \\ 2 & -5 \\ \end {bmatrix*} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \)
No, the transformation is not linear.
For example
\( T(\mathbf{0}) = \begin{bmatrix*}[r] 0 \\ -3 \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -2 \\ 1 & 0 \\ 2 & -5 \\ \end {bmatrix*} \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \ne \mathbf{0} \)
Example#
PROBLEM STATEMENT
Is the transformation linear?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] \sqrt{x_1} & + & x_2 \\ -x_1 & + & x_2 \\ \end {bmatrix*} \)
In other words, is there a \(2 \times 2\) matrix \(\mathbf{A}\) s.t. \(\mathbf{x \mapsto Ax}\)?
\( T \left ( \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] \sqrt{x_1} & + & x_2 \\ -x_1 & + & x_2 \\ \end {bmatrix*} = \begin{bmatrix*}[r] \sqrt{x_1} \\ -x_1 \\ \end{bmatrix*} + \begin{bmatrix*}[r] x_2 \\ x_2 \\ \end{bmatrix*} = x_1 \begin{bmatrix*}[r] \frac{1}{\sqrt{x_1}} \\ -1 \\ \end{bmatrix*} + x_2 \begin{bmatrix*}[r] 1 \\ 1 \\ \end{bmatrix*} = \begin{bmatrix*}[r] \frac{1}{\sqrt{x_1}} & 1 \\ -1 & 1 \\ \end {bmatrix*} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \)
No, the transformation is not linear.
For example
\( T \left ( \begin{bmatrix} 2 \\ 0 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] \sqrt{2} & + & 0 \\ -2 & + & 0 \\ \end {bmatrix*} = \begin{bmatrix*}[r] \sqrt{2} \\ -2 \\ \end {bmatrix*} \)
\( T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) = \begin{bmatrix*}[r] \sqrt{1} & + & 0 \\ -1 & + & 0 \\ \end {bmatrix*} = \begin{bmatrix*}[r] 1 \\ -1 \\ \end {bmatrix*} \)
\( \begin{bmatrix*}[r] \sqrt{2} \\ -2 \\ \end {bmatrix*} = T \left ( 2 \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) \ne 2T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) = 2 \begin{bmatrix*}[r] 1 \\ -1 \\ \end {bmatrix*} \)
Example#
Let \(\mathbb{R}^2 \to \mathbb{R}^3\) be a linear transformation which sends \(\mathbf{e_1}\) to \(\mathbf{u_1}\) and \(\mathbf{e_2}\) to \(\mathbf{u_2}\).
\( \mathbf{e_1} = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}, \mathbf{u_1} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}, \mathbf{e_2} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}, \mathbf{u_2} = \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} \)
What is the matrix representing \(T\)?
\( T(\mathbf{x}) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 1 \\ \end {bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix} \)
Find the images of the vectors \(\begin{bmatrix} 3 \\ 2 \\ \end{bmatrix}, \begin{bmatrix} a \\ b \\ \end{bmatrix}\) under \(T\).
\( T \left ( \begin{bmatrix} 3 \\ 2 \\ \end{bmatrix} \right) = T \left ( 3 \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = T \left ( 3 \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + T \left ( 2 \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = 3T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + 2T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = 3 \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix} + 2 \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 3 + 0 \\ 0 + 2 \\ 6 + 2 \\ \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 8 \\ \end{bmatrix} \)
\( T \left ( \begin{bmatrix} a \\ b \\ \end{bmatrix} \right) = T \left ( a \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = T \left ( a \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + T \left ( b \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = aT \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + bT \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = a \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix} + b \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} a + 0 \\ 0 + b \\ 2a + b \\ \end{bmatrix} = \begin{bmatrix} a \\ b \\ 2a + b \\ \end{bmatrix} \)
Example#
Let \(\mathbb{R}^3 \to \mathbb{R}^2\) be a linear transformation which sends \(\mathbf{e_1}\) to \(\mathbf{u_1}\), \(\mathbf{e_2}\) to \(\mathbf{u_2}\), and \(\mathbf{e_3}\) to \(\mathbf{u_3}\).
\( \mathbf{e_1} = \begin{bmatrix*}[r] 1 \\ 0 \\ 0 \\ \end{bmatrix*}, \mathbf{u_1} = \begin{bmatrix*}[r] 1 \\ 4 \\ \end{bmatrix*}, \mathbf{e_2} = \begin{bmatrix*}[r] 0 \\ 1 \\ 0 \\ \end{bmatrix*}, \mathbf{u_2} = \begin{bmatrix*}[r] -2 \\ 9 \\ \end{bmatrix*}, \mathbf{e_3} = \begin{bmatrix*}[r] 0 \\ 0 \\ 1 \\ \end{bmatrix*}, \mathbf{u_3} = \begin{bmatrix*}[r] 3 \\ -8 \\ \end{bmatrix*} \)
What is the matrix representing \(T\)?
\( T(\mathbf{x}) = \begin{bmatrix*}[r] 1 & -2 & 3 \\ 4 & 9 & -8 \\ \end {bmatrix*} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} \)
Example#
PROBLEM STATEMENT
Let there be a linear transformation \(T(\mathbf{x}) = \mathbf{Ax}\) associated with the matrix
\( \mathbf{A} = \begin{bmatrix*}[r] 1 & -3 & 2 \\ 3 & -8 & 8 \\ 0 & 1 & 2 \\ 1 & 0 & 8 \\ \end {bmatrix*} \)
Find all vectors \(\mathbf{x}\) having vector \(\mathbf{b} = \begin{bmatrix*}[r] 1 \\ 6 \\ 3 \\ 10 \\ \end{bmatrix*}\) as image under \(T\).
In other words, solve the matrix equation.
\( \underset{\mathbf{A}}{ \begin{bmatrix*}[r] 1 & -3 & 2 \\ 3 & -8 & 8 \\ 0 & 1 & 2 \\ 1 & 0 & 8 \\ \end {bmatrix*} } \underset{\mathbf{x}}{ \begin{bmatrix*}[r] x_1 \\ x_2 \\ x_3 \\ \end{bmatrix*} } = \underset{\mathbf{b}}{ \begin{bmatrix*}[r] 1 \\ 6 \\ 3 \\ 10 \\ \end{bmatrix*} } \)
ROW REDUCTION
\( \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 3 & -8 & 8 \\ 0 & 1 & 2 \\ 1 & 0 & 8 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 6 \\ 3 \\ 10 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-3)r_1}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \\ 1 & 0 & 8 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 3 \\ 10 \\ \end {matrix*} \right. \right] \underset{r_4 \leftarrow\,\,\, r_4 + (-1)r_1}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & \boxed{1} & 2 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 3 \\ 9 \\ \end {matrix*} \right. \right] \)
\( \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & \boxed{1} & 2 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 3 \\ 9 \\ \end {matrix*} \right. \right] \underset{r_3 \leftarrow\,\,\, r_3 + (-1)r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & \boxed{1} & 2 \\ 0 & 0 & 0 \\ 0 & 3 & 6 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 0 \\ 9 \\ \end {matrix*} \right. \right] \underset{r_4 \leftarrow\,\,\, r_4 + (-3)r_2}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & \boxed{1} & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \)
\( \left [ \begin{matrix*}[r] \boxed{1} & -3 & 2 \\ 0 & \boxed{1} & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 1 \\ 3 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + 3r_2}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & 0 & 8 \\ 0 & \boxed{1} & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 10 \\ 3 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{reduced}} \)
\( \begin{aligned} x_1 + 8x_3 &= 10 && \iff && x_1 = -8x_3 + 10 \\ x_2 + 2x_3 &= 3 && \iff && x_2 = -2x_3 + 3 \\ & && && x_3 \,\,\,\text{free} \\ \end {aligned} \)
\( \mathbf{x} = \begin{bmatrix*}[r] -8t & + & 10 \\ -2t & + & 3 \\ t & \end {bmatrix*} = \begin{bmatrix*}[r] -8t \\ -2t \\ t \\ \end{bmatrix*} + \begin{bmatrix*}[r] 10 \\ 3 \\ 0 \\ \end{bmatrix*} = t \begin{bmatrix*}[r] -8 \\ -2 \\ 1 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 10 \\ 3 \\ 0 \\ \end{bmatrix*} \)
All vectors \(\mathbf{x}\) having vector \(\mathbf{b}\) as image under \(T\) include the line through the point \((10, 3, 0)\) parallel to the vector \(\begin{bmatrix*}[r] -8 \\ -2 \\ 1 \\ \end{bmatrix*}\)
Example#
PROBLEM STATEMENT
Let there be a linear transformation \(T(\mathbf{x}) = \mathbf{Ax}\) associated with the matrix
\( \mathbf{A} = \begin{bmatrix*}[r] 3 & 2 & 10 & -6 \\ 1 & 0 & 2 & -4 \\ 0 & 1 & 2 & 3 \\ 1 & 4 & 10 & 8 \\ \end {bmatrix*} \)
Find all vectors \(\mathbf{x}\) having the zero vector as image under \(T\).
In other words, solve the matrix equation.
\( \underset{\mathbf{A}}{ \begin{bmatrix*}[r] 3 & 2 & 10 & -6 \\ 1 & 0 & 2 & -4 \\ 0 & 1 & 2 & 3 \\ 1 & 4 & 10 & 8 \\ \end {bmatrix*} } \underset{\mathbf{x}}{ \begin{bmatrix*}[r] x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix*} } = \underset{\mathbf{b}}{ \begin{bmatrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix*} } \)
ROW REDUCTION
\( \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 1 & 0 & 2 & -4 \\ 0 & 1 & 2 & 3 \\ 1 & 4 & 10 & 8 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, r_2 + (-\frac{1}{3})r_1}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & -\frac{2}{3} & -\frac{4}{3} & -2 \\ 0 & 1 & 2 & 3 \\ 1 & 4 & 10 & 8 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_4 \leftarrow\,\,\, r_4 + (-\frac{1}{3})r_1}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{-\frac{2}{3}} & -\frac{4}{3} & -2 \\ 0 & 1 & 2 & 3 \\ 0 & \frac{10}{3} & \frac{20}{3} & 10 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \)
\( \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{-\frac{2}{3}} & -\frac{4}{3} & -2 \\ 0 & 1 & 2 & 3 \\ 0 & \frac{10}{3} & \frac{20}{3} & 10 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_3 \leftarrow\,\,\, r_3 + \frac{3}{2} r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{-\frac{2}{3}} & -\frac{4}{3} & -2 \\ 0 & 0 & 0 & 0 \\ 0 & \frac{10}{3} & \frac{20}{3} & 10 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_4 \leftarrow\,\,\, r_4 + 5r_2}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{-\frac{2}{3}} & -\frac{4}{3} & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{row echelon}} \)
\( \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{-\frac{2}{3}} & -\frac{4}{3} & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_2 \leftarrow\,\,\, (-\frac{3}{2}) r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{3} & 2 & 10 & -6 \\ 0 & \boxed{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, r_1 + (-2)r_2}{\rightarrow} \left [ \begin{matrix*}[r] \boxed{3} & 0 & 6 & -12 \\ 0 & \boxed{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \)
\( \left [ \begin{matrix*}[r] \boxed{3} & 0 & 6 & -12 \\ 0 & \boxed{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] \underset{r_1 \leftarrow\,\,\, \frac{1}{3} r_1}{\rightarrow} \underbrace{ \left [ \begin{matrix*}[r] \boxed{1} & 0 & 2 & -4 \\ 0 & \boxed{1} & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end {matrix*} \left | \begin{matrix*}[r] 0 \\ 0 \\ 0 \\ 0 \\ \end {matrix*} \right. \right] }_{\textbf{reduced}} \)
\( \begin{aligned} x_1 + 2x_3 - 4x_4 &= 0 && \iff && x_1 = -2x_3 + 4x_4 \\ x_2 + 2x_3 + 3x_4 &= 0 && \iff && x_2 = -2x_3 - 3x_4 \\ & && && x_3 \,\,\,\text{free} \\ & && && x_4 \,\,\,\text{free} \\ \end {aligned} \)
\( \mathbf{x} = \begin{bmatrix*}[r] -2s & + & 4t \\ -2s & - & 3t \\ s & & \\ & & t \\ \end {bmatrix*} = \begin{bmatrix*}[r] -2s \\ -2s \\ s \\ 0 \\ \end{bmatrix*} + \begin{bmatrix*}[r] 4t \\ -3t \\ 0 \\ t \\ \end{bmatrix*} = s \begin{bmatrix*}[r] -2 \\ -2 \\ 1 \\ 0 \\ \end{bmatrix*} + t \begin{bmatrix*}[r] 4 \\ -3 \\ 0 \\ 1 \\ \end{bmatrix*} \)
All vectors \(\mathbf{x}\) having the zero vector as image under \(T\) include
\( \text{span} \left \{ \begin{bmatrix*}[r] -2 \\ -2 \\ 1 \\ 0 \\ \end{bmatrix*}, \begin{bmatrix*}[r] 4 \\ -3 \\ 0 \\ 1 \\ \end{bmatrix*} \right\} \)
through the origin.
Onto#
One-to-one#
Example#
The transformation
\( \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \mapsto \begin{bmatrix} y \\ z \\ \end{bmatrix} \)
is linear and onto.
LINEAR
\( T \left ( \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \right) = T \left ( x \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \right) = T \left ( x \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \right) + T \left ( y \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} \right) + T \left ( z \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \right) = xT \left ( \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \right) + yT \left ( \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} \right) + zT \left ( \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \right) = x \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + z \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} y \\ z \\ \end{bmatrix} \)
\( T(\mathbf{x}) = \underset{\mathbf{A}}{ \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix} } \underset{\mathbf{x}}{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} } \)
ONTO
The transformation is onto because \(\mathbf{A}\) is a (reduced) row echelon matrix with a pivot in each row.
ONE-TO-ONE
The transformation is not one-to-one because \(\mathbf{A}\) does not contain a pivot in each column.
Example#
The transformation
\( \begin{bmatrix} y \\ z \\ \end{bmatrix} \mapsto \begin{bmatrix} 0 \\ x \\ y \\ \end{bmatrix} \)
is linear and one-to-one.
LINEAR
\( T \left ( \begin{bmatrix} x \\ y \\ \end{bmatrix} \right) = T \left ( x \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = T \left ( x \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + T \left ( y \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = xT \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right) + yT \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right) = x \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ x \\ y \\ \end{bmatrix} \)
\( T(\mathbf{x}) = \underset{\mathbf{A}}{ \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 1 \\ \end {bmatrix} } \underset{\mathbf{x}}{ \begin{bmatrix} x \\ y \\ \end{bmatrix} } \)
ONE-TO-ONE
The transformation is one-to-one because \(\mathbf{A}\) is row-equivalent to a (reduced) row echelon matrix with a pivot in each column.
ONTO
The transformation is not onto because \(\mathbf{A}\) does not contain a pivot in each row.
Example#
PROBLEM STATEMENT
Determine whether the transformation
\( \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \mapsto \begin{bmatrix} \frac{x + y + z}{3} \\ \frac{x + y + z}{3} \\ \frac{x + y + z}{3} \\ \end{bmatrix} \)
is linear, onto, and one-to-one.
LINEAR
\( T \left ( \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \right) = T \left ( x \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix} \right) = xT \left( \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \right) + yT \left( \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \right) + zT \left( \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} \right) = x \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \\ \end{bmatrix} + y \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \\ \end{bmatrix} + z \begin{bmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \\ \end{bmatrix} = \begin{bmatrix} \frac{x + y + z}{3} \\ \frac{x + y + z}{3} \\ \frac{x + y + z}{3} \\ \end{bmatrix} \)
\( T(\mathbf{x}) = \underset{\mathbf{A}}{ \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end {bmatrix} } \underset{\mathbf{x}}{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} } \)
The transformation is linear.
ONTO and ONE-TO-ONE
\( \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \end {bmatrix} \underset{\text{row reduce}}{\rightarrow} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end {bmatrix} \)
The transformation is neither onto (because not all rows contain a pivot) nor one-to-one (because not all columns contain a pivot).
Example#
PROBLEM STATEMENT
Determine whether the linear transformation \(T : \mathbb{R}^4 \to \mathbb{R}^3\) defined by \(T(\mathbf{x}) = \mathbf{Ax}\) with
\( \mathbf{A} = \begin{bmatrix*}[r] 1 & -4 & 8 & 1 \\ 0 & 2 & -1 & 3 \\ 0 & 0 & 0 & 5 \\ \end {bmatrix*} \)
is onto and one-to-one.
ONTO and ONE-TO-ONE
\( \mathbf{A} = \begin{bmatrix*}[r] \boxed{1} & -4 & 8 & 1 \\ 0 & \boxed{2} & -1 & 3 \\ 0 & 0 & 0 & \boxed{5} \\ \end {bmatrix*} \)
is already in row echelon form and so we can determine that the linear transformation is onto because there is a pivot in each row but not one-to-one because there is not a pivot in each column.
Example#
PROBLEM STATEMENT
Determine whether the linear transformation \(S : \mathbb{R}^3 \to \mathbb{R}^3\) defined by \(S(\mathbf{x}) = \mathbf{Bx}\) with
\( \mathbf{B} = \begin{bmatrix*}[r] 2 & 1 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \\ \end {bmatrix*} \)
is onto and one-to-one.
ROW REDUCE
\( \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + (-\frac{3}{2})r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \frac{1}{2} & -\frac{7}{2} \\ 1 & 3 & 2 \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + (-\frac{1}{2})r_1}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & -\frac{7}{2} \\ 0 & \frac{5}{2} & \frac{1}{2} \\ \end {bmatrix*} \)
\( \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & -\frac{7}{2} \\ 0 & \frac{5}{2} & \frac{1}{2} \\ \end {bmatrix*} \underset{r_3 \leftarrow\,\,\, r_3 + (-5)r_2}{\rightarrow} \underbrace{ \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & -\frac{7}{2} \\ 0 & 0 & \boxed{18} \\ \end {bmatrix*} }_{\textbf{row echelon}} \underset{r_3 \leftarrow\,\,\, \frac{1}{18} r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & -\frac{7}{2} \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)
\( \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & -\frac{7}{2} \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, r_2 + \frac{7}{2} r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 3 \\ 0 & \boxed{\frac{1}{2}} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-3)r_3}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 0 \\ 0 & \boxed{\frac{1}{2}} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)
\( \begin{bmatrix*}[r] \boxed{2} & 1 & 0 \\ 0 & \boxed{\frac{1}{2}} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_2 \leftarrow\,\,\, 2r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 1 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, r_1 + (-1)r_2}{\rightarrow} \begin{bmatrix*}[r] \boxed{2} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \)
\( \begin{bmatrix*}[r] \boxed{2} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} \underset{r_1 \leftarrow\,\,\, \frac{1}{2} r_1}{\rightarrow} \underbrace{ \begin{bmatrix*}[r] \boxed{1} & 0 & 0 \\ 0 & \boxed{1} & 0 \\ 0 & 0 & \boxed{1} \\ \end {bmatrix*} }_{\textbf{reduced}} \)
ONTO and ONE-TO-ONE
The transformation is both onto and one-to-one because the matrix \(\mathbf{B}\) is row-equivalent to a row echelon matrix in which all rows contain a pivot and all columns contain a pivot.
Composition of Linear Transformations#
Theorem#
Let \(T : \mathbb{R}^p \to \mathbb{R}^n\) and \(L : \mathbb{R}^n \to \mathbb{R}^m\) be two linear transformations.
\(\mathbb{R}^m \overset{L}{\leftarrow} \mathbb{R}^n \overset{T}{\leftarrow} \mathbb{R}^p\)
The composition \(L \circ T : \mathbb{R}^p \to \mathbb{R}^m\) is a linear transformation.
Proof#
Since \(T\) is linear it is the case that for some vectors \(\mathbf{u, v} \in \mathbb{R}^p\) and some scalars \(r, s\)
\(T(r \mathbf{u} + s \mathbf{v}) = rT(\mathbf{u}) + sT(\mathbf{v})\)
and since \(L\) is linear it is the case that for some vectors \(\mathbf{x, y} \in \mathbb{R}^n\) and some scalars \(h, k\)
\(L(h \mathbf{x} + k \mathbf{y}) = hL(\mathbf{x}) + kL(\mathbf{y})\)
Let \(\mathbf{x} = rT(\mathbf{u}), \mathbf{y} = sT(\mathbf{v})\).
\( L(h(rT(\mathbf{u})) + k(sT(\mathbf{v}))) = hrL(T(\mathbf{u})) + ksL(T(\mathbf{v})) \)
Theorem#
The composition \(L \circ T\) of the linear transformations \(L, T\) is the linear transformation represented by the product \(\mathbf{AB}\) of the matrices \(\mathbf{A, B}\).
Proof#
The linear transformations \(L, T\) are represented by the matrices \(\mathbf{A, B}\) respectively.
\( \begin{aligned} L(\mathbf{e_k}) &= \sum_{i = 1}^m A_{ik} \mathbf{e_i} \\ T(\mathbf{e_j}) &= \sum_{k = 1}^n B_{kj} \mathbf{e_k} \\ \end {aligned} \)
\( \begin{aligned} L \circ T (\mathbf{e_j}) = L \left( \sum_{k = 1}^n B_{kj} \mathbf{e_k} \right) = \sum_{k = 1}^n B_{kj} L (\mathbf{e_k}) = \sum_{k = 1}^n B_{kj} \left( \sum_{i = 1}^m A_{ik} \mathbf{e_i} \right) = \sum_{i = 1}^m \left( \sum_{k = 1}^n A_{ik} B_{kj} \right) \mathbf{e_i} = \sum_{i = 1}^m (\mathbf{AB})_{ij} \mathbf{e_i} \end {aligned} \)