Math 511: Linear Algebra

6.1 Definition and Examples

---

6.1.1 Linear Transformations

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{0.0, 0.0, 1.0} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \definecolor{blueviolet}{rgb}{.537,.192,.937} \definecolor{jonquil}{rgb}{.949,.792,.098} \definecolor{shockingpink}{rgb}{1, 0, .741} \definecolor{royalblue}{rgb}{0, .341, .914} \definecolor{alien}{rgb}{.529,.914,.067} \definecolor{crimson}{rgb}{1, .094, .271} \def\ihat{\mathbf{\hat{\unicode{x0131}}}} \def\jhat{\mathbf{\hat{\unicode{x0237}}}} \def\khat{\mathrm{\hat{k}}} \def\tombstone{\unicode{x220E}} \def\contradiction{\unicode{x2A33}} $$

A linear transformation is a special kind of function. All linear transformations are functions, but not all functions are linear. Everyone should start with Grant Sanderson's video Linear Transformations

Definition of a Linear Transformation

A mapping $L$ from a vector space $V$ to a vector space $W$ is a linear transformation if

$$ \begin{align*} L(\mathbf{v} + \mathbf{w}) &= L(\mathbf{v}) + L(\mathbf{w}) \\ \\ L(\alpha\mathbf{v}) &= \alpha L(\mathbf{v}) \end{align*} $$

for all $\mathbf{v},\mathbf{w} \in V$ and for all scalars $\alpha$.

The properties are called the linearity properties and together they claim that for linear combinations of vectors

Lemma 6.1.1

If $\mathbf{v}_1$ and $\mathbf{v}_2$ are vectors in vector space $V$, and $\alpha$ and $\beta$ are scalars in $\mathbb{R}$, then $L:V\rightarrow W$ is a linear transformation from $V$ to $W$ if and only if

$$ L(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2) = \alpha L(\mathbf{v}_1) + \beta L(\mathbf{v}_2) $$

Proof:

1. $\implies$

If $L:V\rightarrow W$ is a linear transformation, then by property 1 of the definition

$$ \begin{align*} L(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2) &= L(\alpha\mathbf{v}_1) + L(\beta\mathbf{v}_2),\text{ and} \\ \\ &= \alpha\,L(\mathbf{v}_1) + \beta\,L(\mathbf{v}_2) \end{align*} $$
using the second property of a linear transformation to each term.

2. $\impliedby$

If transformation $L:V\rightarrow W$ has the property

$$ L(\alpha\mathbf{v}_1 + \beta\mathbf{v}_2) = \alpha L(\mathbf{v}_1) + \beta L(\mathbf{v}_2) $$
for any two vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ in $V$, and any two scalars $\alpha$ and $\beta$ in $\mathbb{R}$, then

$$ \begin{align*} L(\mathbf{v}_1 + \mathbf{v}_2) &= L((1)\mathbf{v}_1 + (1)\mathbf{v}_2) \\ &= (1)L(\mathbf{v}_1) + (1)L(\mathbf{v}_2) \\ &= L(\mathbf{v}_1) + L(\mathbf{v}_2) \\ \\ \text{and}&\ \\ \\ L(\alpha\mathbf{v}_1) &= L(\alpha\mathbf{v}_1 + \mathbf{0}) \\ &= L(\alpha\mathbf{v}_1 + 0\mathbf{v}_2) \\ &= \alpha\,L(\mathbf{v}_1) + 0\,L(\mathbf{v}_2) \\ &= \alpha\,L(\mathbf{v}_1) + \mathbf{0} \\ &= \alpha\,L(\mathbf{v}_1) \end{align*} $$ $\tombstone$

This means that the image of a linear combination of vectors in the domain of a linear transformation is the SAME linear combination of the images of each vector.

In Grant Sanderson's videos we see that the geometric interpretation of a linear transformation from vector space $\mathbb{R}^2$ to $\mathbb{R}^2$ is that

parallel lines stay parallel and evenly spaced

We can expand this geometric interpretation to $\mathbb{R}^3$.

parallel planes stay parallel and evenly spaced

In higher dimensional spaces $\mathbb{R}^n$ the best we can say is

parallel hyperplanes stay parallel and evenly spaced

In abstract vector spaces this can be difficult to put into words. We studied the geometric properties of abstract vector spaces in chapter 5.

We talked about linear combinations, however the definition and theorems discuss only linear combinations of two vectors. Will we get the same result for any linear combination of vectors in the domain of the linear transformation?

Theorem 6.1.2

If $L:V\rightarrow W$ is a linear transformation from vector space $V$ to vector space $W$, $\left\{\mathbf{v}_1, \mathbf{v}_2, \cdots, \mathbf{v}_k \right\}\subset V$ is a finite set of vectors in the domain $V$, and $\left\{\alpha_1, \alpha_2, \cdots, \alpha_k \right\}\in\mathbb{R}$ are scalars, then

$$ L\left(\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_k\mathbf{v}_k\right) = \alpha_1 L(\mathbf{v}_2) + \alpha_2 L(\mathbf{v}_2) + \cdots + \alpha_k L(\mathbf{v}_k) $$

Proof

There are three steps to proving a theorem using the principle of mathematical induction. You need a way to index the number of mathematical objects in a mathematical expression. In our proof we have

$$ \left\{\mathbf{v}_1, \mathbf{v}_2, \cdots, \mathbf{v}_k \right\}\subset V,\text{ and }\left\{\alpha_1, \alpha_2, \cdots, \alpha_k \right\}\in\mathbb{R} $$

Step 1. Prove that the assertion is true when $n=1$.

When $k=1$ we have only one vector in a vector space, $\mathbf{v}_1\in V$ and one scalar $c_1\in\mathbb{R}$.

$$ L(c_1\mathbf{v}_1) = c_1L(\mathbf{v}_1) $$

because $V$ is a vector space, and $L$ is a linear transformation.

Step 2. Given that the Theorem is true for $k\le n$ vectors and scalars, show that it must also be true for $k=n+1$.

If our theorem is true for $k=n$, then we have by assumption

$$ L\left(\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n\right) = \alpha_1 L(\mathbf{v}_2) + \alpha_2 L(\mathbf{v}_2) + \cdots + \alpha_n L(\mathbf{v}_n) $$

Since $V$ and $W$ are vector spaces we have that $\mathbf{w} = \alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n$ and

$$ \begin{align*} L(\mathbf{w}) &= L\left(\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n\right) \\ \\ &= \alpha_1 L(\mathbf{v}_2) + \alpha_2 L(\mathbf{v}_2) + \cdots + \alpha_n L(\mathbf{v}_n) \end{align*} $$

Now we have from our definition of a linear transformation that

$$ \begin{align*} L\left(\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n + \alpha_{n+1}\mathbf{v}_{n+1}\right) &= L(\mathbf{w} + \alpha_{n+1}\mathbf{v}_{n+1}) \\ \\ &= L(\mathbf{w}) + \alpha_{n+1}L(\mathbf{v}_{n+1}) \\ \\ &= L\left(\alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2 + \cdots + \alpha_n\mathbf{v}_n\right) + \alpha_{n+1}L(\mathbf{v}_{n+1}) \\ \\ &= \alpha_1 L(\mathbf{v}_2) + \alpha_2 L(\mathbf{v}_2) + \cdots + \alpha_n L(\mathbf{v}_n) + \alpha_{n+1}L(\mathbf{v}_{n+1}) \end{align*} $$

We just proved that if the theorem is true for $k=n$, then it must also be true for $k=n+1$.

Step 3. Invoke the Principle of Mathematical Induction

By the Principle of Mathematical Induction, the theorem must be true for any positive integer $k$.$\tombstone$

This kind of proof works because we prove the theorem is true for $k=1$ in Step 1.

Because it is true for $k=1$, then by Step 2 the theorem is true for $k=2$.

Because it is true for $k=2$, then by Step 2 the theorem is true for $k=3$.

Because it is true for $k=3$, then by Step 2 the theorem is true for $k=4$.

$\cdots$

The Principle of Mathematical Induction basically states that this is sufficient to have shown that the theorem is true for all finite positive integers.

---

6.1.2 Linear Operators

If a linear transformation maps a vector space $V$, called the domain, into a vector space $W$, called the codomain, we will denote this as

$$ L: V\rightarrow W $$
For brevity in this course, when this arrow notation is used we will assume that both $V$ and $W$ are vector spaces.

Linear transformations are functions from one vector space to another, that respect (interesting choice of verb yes?) the algebraic structure of the vector spaces, their linear combinations.

People have a difficult time with using the term function for everything. It became common to refer to a function whose domain and codomain where vectors an operator.

Definition

An operator is a function whose inputs or outputs are a vector or some other mathematical object.

We are already familiar with operators.

• The differential operator maps a function to another function, its derivative

Recall that vector spaces of functions turn out to be important vector spaces in modern mathematical models. The linear operators acting upon vectors spaces of functions are similarly important.

If the domain and codomain of a linear operator (linear transformation, function) is the same vector space $V$ so that

$$ L:V\rightarrow V, $$

we say that $L$ is a linear operator on $V$. This way one doesn't need to say from $V$ to $V$.

Linear Operators on $\mathbb{R}^2$

Example 1

Let $L$ be a linear operator such that

$$ L(\mathbf{x}) = -2\mathbf{x} $$

for each $\mathbf{x}\in\mathbb{R}^2$. We have that

$$ \begin{align*} L(\alpha\mathbf{x} + \beta\mathbf{y}) &= -2(\alpha\mathbf{x} + \beta\mathbf{y}) \\ \\ &= -2(\alpha\mathbf{x}) + (-2)(\beta\mathbf{y}) \\ \\ &= \alpha(-2\mathbf{x}) + \beta(-2\mathbf{y}) \\ \\ & = \alpha L(\mathbf{x}) + \beta L(\mathbf{y}) \end{align*} $$

so this is a linear operator. (Alternatively, we could have shown that vector addition and scalar multiplication are linear separately.) Geometrically, we can think of this operator as stretching a vector by $2$ and reversing its direction.

Example 2

Another type of linear operator we can consider is a projection onto a vector. Considering the mapping $L$ given by

$$ L(\mathbf{x}) = x_1 \mathbf{e}_1 $$

for all $\mathbf{x}\in\mathbb{R}^2$. Then we have that for any scalars $\alpha,\beta\in\mathbb{R}$ and vectors $\mathbf{x},\mathbf{y}\in\mathbb{R}^2$

$$ L(\alpha\mathbf{x} + \beta\mathbf{y}) = (\alpha x_1 + \beta y_1)\mathbf{e}_1 = \alpha(x_1\mathbf{e}_1) + \beta(y_1\mathbf{e}_1) = \alpha L(\mathbf{x}) + \beta L(\mathbf{y}) $$\

$L$ is therefore a linear operator. Pictorially, the operator takes a vector and reduces it to only the part of it that points in the direction of the $x_1$-axis.

Note:

The vector in the picture is labeled $\mathbf{y}$ so as to not be ambiguous about the $x_1$ component of the vector and the $x_1$-axis.

Example 3

Another possible linear operator is a reflection. The mapping $L$ given by

$$ L(\mathbf{x}) = \begin{bmatrix} -y_1 \\ \ \ y_2 \end{bmatrix} $$

takes a vector $\mathbf{x}\in\mathbb{R}^2$ and reflects it about the $x_2$-axis. This operator is linear since

$$ \begin{align*} L(\alpha\mathbf{x} + \beta\mathbf{y}) &= \begin{bmatrix} -(\alpha x_1 + \beta y_1) \\ \ \ \alpha x_2 + \beta y_2 \end{bmatrix} \\ \\ &= \alpha \begin{bmatrix} -x_1 \\ \ \ x_2 \end{bmatrix} + \beta \begin{bmatrix} -y_1 \\ \ \ y_2 \end{bmatrix} \\ \\ &= \alpha L(\mathbf{x}) + \beta L(\mathbf{y}) \end{align*} $$

---

6.1.3 Linear Transformations from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$

Example 4

Let $L: \mathbb{R}^2 \rightarrow \mathbb{R}^1$ be the linear operator given by

$$ L(\mathbf{x}) = x_1 + 2x_2 $$

which is a linear transformation because

$$ \begin{align*} L(\alpha\mathbf{x} + \beta\mathbf{y}) &= L\left(\alpha\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} + \beta\begin{bmatrix} y_1 \\ y_2 \end{bmatrix}\right) \\ \\ &= L\left(\begin{bmatrix} \alpha x_1 + \beta y_1 \\ \alpha x_2 + \beta y_2 \end{bmatrix}\right)\\ \\ &= (\alpha x_1 + \beta y_1) + 2(\alpha x_2 + \beta y_2) \\ \\ &= \alpha(x_1 + 2x_2) + \beta(y_1 + 2y_2) \\ \\ &= \alpha L(\mathbf{x}) + \beta L(\mathbf{y}) \end{align*} $$

Visualizations for linear transformations such as these can be difficult, since we quickly run out of dimensions with which to represent what his happening. In the image here the two axes on the lower part of the image are the $x_1$- and $x_2$-axis, whereas the vertical axis represents height and the quantity $x_1 + 2 x_2$.

Definition

A linear operator $L:V\rightarrow\mathbb{R}$ or $L:V\rightarrow\mathbb{C}$ from a vector space $V$ to the field of real or complex scalars is called a

linear functional.

Example 5

Let a mapping $M$ be given by

$$ M(\mathbf{x}) = x_1^2 + x_2^2 $$

This mapping is not linear since

$$ M(\alpha\mathbf{x}) = (\alpha x_1)^2 + (\alpha x_2)^2 = \alpha^2 (x_1^2 + x_2^2) = \alpha^2 M(\mathbf{x}) \neq \alpha M(\mathbf{x}) $$

Note:

Here we use $M$ to be very explicit about the fact that this is not a linear transformation, but the naming does not matter. Whether or not a map is a linear transformation must be shown by its adherence to or violation of the linearity properties.

Example 6

We can also have linear transformations from $\mathbb{R}^2$ to $\mathbb{R}^3$, from a lower dimensional to higher dimensional vector space. Let our $L: \mathbb{R}^2\rightarrow\mathbb{R}^3$ be given by

$$ L(\mathbf{x}) = \begin{bmatrix} x_2 \\ x_1 + x_2 \\ x_1 \end{bmatrix} $$

This is a linear transformation because

$$ L(\alpha\mathbf{x}) = L\left( \begin{bmatrix} \alpha x_1 \\ \alpha x_2 \end{bmatrix} \right) = \begin{bmatrix} \alpha x_2 \\ \alpha x_1 + \alpha x_2 \\ \alpha x_1 \end{bmatrix} = \begin{bmatrix} \alpha x_2 \\ \alpha (x_1 + x_2) \\ \alpha x_1 \end{bmatrix} = \alpha\begin{bmatrix} x_2 \\ x_1 + x_2 \\ x_1 \end{bmatrix} = \alpha L(\mathbf{x}) $$

and

$$ \begin{align*} L(\mathbf{x} + \mathbf{y}) &= L\left(\begin{bmatrix} x_1 + y_1 \\ x_2 + y_2 \end{bmatrix}\right) \\ \\ &= \begin{bmatrix} x_2 + y_2 \\ (x_1 + y_1) + (x_2 + y_2) \\ x_1 + y_1 \end{bmatrix} \\ \\ &= \begin{bmatrix} x_2 \\ x_1 + x_2 \\ x_1 \end{bmatrix} + \begin{bmatrix} y_2 \\ y_1 + y_2 \\ y_1 \end{bmatrix} \\ \\ &= L(\mathbf{x}) + L(\mathbf{y}) \end{align*} $$

---

6.1.4 Properties of Linear Transformations

Theorem 6.1.3

Let $T$ be a linear transformation from vector space $V$ to vector space $W$. If $\mathbf{u}$ and $\mathbf{v}$ are vectors in $V$, that is $\mathbf{u},\ \mathbf{v}\in V$, then the following are true

1. $T(\mathbf{0}) = \mathbf{0}$
2. $T(-\mathbf{v}) = -T(\mathbf{v})$
3. $T(\mathbf{u} - \mathbf{v}) = T(\mathbf{u}) - T(\mathbf{v})$

Proof:

1.

Since $\mathbf{u},\ \mathbf{v}\in V$ are any vectors in the domain, we have from the definition of a linear transformation, Theorem 6.1.1 and the properties of a vector space that

$$ T(\mathbf{0}) = T(0\,\mathbf{u}) = 0\,T(\mathbf{u}) = \mathbf{0} $$

2. Likewise

$$ T(-\mathbf{v}) = T((-1)\mathbf{v}) = (-1)T(\mathbf{v}) = -T(\mathbf{v}) $$

3. Finally

$$ T(\mathbf{u}-\mathbf{v}) = T((1)\mathbf{u} + (-1)\mathbf{v}) = 1\,T(\mathbf{u}) + (-1)\,T(\mathbf{v}) = T(\mathbf{u}) - T(\mathbf{v}) $$ ∎

---

6.1.5 Linear Transformations Between Abstract Vector Spaces

Example 7

Suppose $L:C[a,b]\rightarrow\mathbb{R}$ is the definite integral from $a$ to $b$ which maps $C[a,b]$ into $\mathbb{R}$.

$$ L(f) = \int_a^b f(x)\, dx $$

This is a linear functional, since for any vectors $f,g\in C[a,b]$ and scalars $\alpha,\beta\in\mathbb{R}$

$$ \begin{align*} L(\alpha f + \beta g) &= \int_a^b \alpha f(x) + \beta g(x)\, dx \\ \\ &= \alpha \int_a^b f(x)\, dx + \beta \int_a^b g(x)\, dx \\ \\ &= \alpha L(f) + \beta L(g) \end{align*} $$

Example 8

Let $D$ be the derivative of $f$ which maps $C^1[a,b]$ (the set of functions on $[a,b]$ with one continuous derivative) into $C[a,b]$, given by

$$ D(f) = f' $$

This is a linear transformation, since for any vectors $f,g\in C^1[a,b]$

$$ D\left(\alpha f + \beta g \right) = \left(\alpha f + \beta g \right)' = \alpha f' + \beta g' = \alpha D(f) + \beta D(g) $$

Example 9

Let $D$ be the linear transformation from $C^1[a,b]$ into $C[a,b]$ given by

$$ L(y) = py' + qy $$

where $y\in C^1[a,b]$ and $p,q\in C[a,b]$. This is an example of a linear first-order differential operator. It is linear since

$$ \begin{align*} L(\alpha y_1 + \beta y_2) &= p(\alpha y_1 + \beta y_2)' + q(\alpha y_1 + \beta y_2) \\ \\ &= p(\alpha y_1' + \beta y_2') + q(\alpha y_1 + \beta y_2) \\ \\ &= \alpha(p y_1' + q y_1) + \beta(p y_2' + q y_2) \\ \\ &= \alpha L(y_1) + \beta L(y_2) \end{align*} $$

---

6.1.6 Linear Transformation and their Matrices

We have known for some time that if we have a matrix like

$$ A = \begin{bmatrix}\ \ 2\ &\ \ 1\ \\ -3\ &\ \ 0\ \\ \ \ 1\ & -2\ \end{bmatrix} $$

then it defines a linear transformation $L$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ where

$$ L(\mathbf{x}) = A\mathbf{x} $$

This transformation is linear because for any vectors $\mathbf{x},\ \mathbf{y}\in\mathbb{R}^2$ and scalars $\alpha, \beta\in\mathbb{R}$ we have

$$ \begin{align*} L\left(\alpha\mathbf{x} + \beta\mathbf{y}\right) &= A\left(\alpha\mathbf{x} + \beta\mathbf{y}\right) \\ \\ &= \alpha A\mathbf{x} + \beta A\mathbf{y} = \alpha L(\mathbf{x}) + \beta L(\mathbf{y}) \end{align*} $$

from chapter one.

Example 10

If $A\in\mathbb{R}^{3\times 2}$ is given by

$$ A = \begin{bmatrix} 2 & 1 \\ 1 & 5 \\ 0 & 3 \end{bmatrix} $$

Define the linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^3$ by

$$ T(\mathbf{v}) := A\mathbf{v} = \begin{bmatrix} 2 & 1 \\ 1 & 5 \\ 0 & 3 \end{bmatrix}\mathbf{v} $$

Find $T\left(\mathbf{v}\right)$, where $\mathbf{v} = \langle 2, -1 \rangle$.

$$ T\left(\langle 2, -1 \rangle\right) = A\mathbf{v} = \begin{bmatrix} 2 & 1 \\ 1 & 5 \\ 0 & 3 \end{bmatrix}\begin{bmatrix}\ \ 2 \\ -1 \end{bmatrix} = \begin{bmatrix}\ \ 3 \\ -3 \\ -3 \end{bmatrix} $$

---

6.1.7 Linear Transformation Given by a Matrix

Theorem 6.1.4

Let $A\in\mathbb{R}^{m\times n}$. The function $T:\mathbb{R}^n\rightarrow\mathbb{R}^m$ defined by

$$ T(\mathbf{v}) = A\mathbf{v} $$

is a linear transformation.

Proof:

Given two arbitrary vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{R}^n$, and a scalar $\alpha$ in $\mathbb{R}$, then

$$ \begin{align*} L\left(\mathbf{u} + \mathbf{v}\right) &= A\left(\mathbf{u} + \mathbf{v}\right) \\ &= A\mathbf{u} + A\mathbf{v} = L(\mathbf{u}) + L(\mathbf{v}) \\ \\ L\left(\alpha\mathbf{u}\right) &= A\left(\alpha\mathbf{u}\right) = \alpha\,A\mathbf{u} = \alpha\,L(\mathbf{u}) \end{align*} $$ ∎

Exercise 1

Let $T:\mathbb{R}^{m\times n}\rightarrow\mathbb{R}^{m\times n}$ be the function that maps an $m\times n$ matrix $A$ to its transpose.

$$ L(A) = A^T $$

Show that $T$ is a linear transformation.

Check Your Work

Let $A,\ B\in\mathbb{R}^{m\times n}$ and let $c$ be a scalar in $\mathbb{R}$. Then
$$ \begin{align*} T(A + B) &= (A + B)^T = A^T + B^T = T(A) + T(B) \\ \\ T(cA) &= (cA)^T = cA^T = cL(A) \end{align*} $$
Hence $T$ is a linear transformation

---

Your use of this self-initiated mediated course material is subject to our Creative Commons License 4.0

---