Math 511: Linear Algebra

6.3 Linear Transformations as Matrices

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6.3.1 Linear Transformation and Their Matrices

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Dr. Gilbert Strang presents an excellent lecture on representing linear transformations as matrices in his video Linear Transformations and their Matrices

Theorem 6.3.1

The Standard Matrix Representation

If $L$ is a linear transformation $L:\mathbb{R}^n \rightarrow \mathbb{R}^m$, then there is a standard matrix $A\in\mathbb{R}^{m\times n}$ such that

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

for each $\mathbf{x}\in\mathbb{R}^n$. The $j^\text{th}$ column of this matrix is given by

$$ \mathbf{a}_j = L(\mathbf{e}_j)\qquad j=1,2,\ldots,n $$

Proof:

Let

$$ \mathbf{a}_j = L(\mathbf{e}_j) $$

for each $j=1,2,\ldots,n$. Define the matrix $A$ as

$$ A = \left[\mathbf{a}_1\ \mathbf{a}_2\ \ldots\ \mathbf{a}_n\right] $$

If each vector $\mathbf{x}\in\mathbb{R}^n$ is written as a linear combination of standard basis vectors in $\mathbb{R}^n$

$$ \mathbf{x} = x_1\mathbf{e}_1 + x_2\mathbf{e}_2 + \ldots + x_n\mathbf{e}_n $$

then

$$ \begin{align*} L(\mathbf{x}) &= x_1 L(\mathbf{e}_1) + x_2 L(\mathbf{e}_2) + \ldots + x_n L(\mathbf{e}_n) \\ \\ &= x_1\mathbf{a}_1 + x_2\mathbf{a}_2 + \ldots + x_n\mathbf{a}_n \\ \\ &= \left[\mathbf{a}_1\ \mathbf{a}_2\ \ldots\ \mathbf{a}_n\right]\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \\ \\ &= A\mathbf{x} \end{align*} $$ $\tombstone$

Note!

This proof is constructive. Not only does it tell us that there is an $m\times n$ matrix for every linear transformation between $\mathbb{R}^n$ and $\mathbb{R}^m$, it provides us with a procedure (an algorithm) to building the matrix. Namely, we set each $\mathbf{a}_j = L(\mathbf{e}_j)$ so that the $j^\text{th}$ column of the transformation matrix $A$ shows the location in $\mathbb{R}^m$ that the $j^\text{th}$ standard basis vector in $\mathbb{R}^n$ is mapped to.

We call this matrix the standard matrix representation because we are using the canonical or standard basis for both the domain and the codomain. For now we will always use the standard basis.

Also notice that our matrices are not square if the domain and codomain have different dimensions. Everyone should view Grant Sanderson's video

Nonsquare matrices as transformations between dimensions

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6.3.2 Examples

Example 1

Let $L:\mathbb{R}^3 \rightarrow \mathbb{R}^2$ be given by

$$ L(\mathbf{x}) = \left[\begin{array}{c} x_1 + x_2 \\ -x_3 \end{array}\right] $$

for each $\mathbf{x}\in\mathbb{R}^3$. To construct the matrix $A$ for the linear transformation, we determine which vectors in $\mathbb{R}^2$ each $\mathbf{e}_j\in\mathbb{R}^3$ is mapped to:

$$ \begin{align*} \mathbf{a}_1 = L(\mathbf{e}_1) &= L\left(\,\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\,\right) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ \\ \mathbf{a}_2 = L(\mathbf{e}_2) &= L\left(\,\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\,\right) = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\ \\ \mathbf{a}_3 = L(\mathbf{e}_3) &= L\left(\,\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\,\right) = \left[\begin{array}{r} 0 \\ -1 \end{array}\right] \end{align*} $$

Hence, the matrix representation of this linear transformation is

$$ A = \left[\begin{array}{rr} 1 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] $$

which we check via matrix multiplication to confirm its equivalence to $L$:

$$ A\mathbf{x} = \left[\begin{array}{rr} 1 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \left[\begin{array}{c} x_1 + x_2 \\ -x_3 \end{array}\right] = L(\mathbf{x}) $$

Example 2

One of the most useful linear transformations in $\mathbb{R}^2$ is a rotation. Let matrix $R$ be the standard matrix representation of a rotation.
$$ R = \left[\begin{array}{rr} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right] $$
rotates any vector $\mathbf{x}\in\mathbb{R}^2$ counterclockwise by the angle $\theta$. If $\theta = 0$ (no rotation), we see that $R$ is the identity matrix. When $\theta\neq 0$, the columns of $R$ show where $\mathbf{e}_1$ and $\mathbf{e}_2$ are mapped to, respectively.

The following interactive demonstration how the vectors $\mathbf{e}_1$ and $\mathbf{e}_2$ move as $\theta$ is changed.

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6.3.3 Graph, Networks, and Incidence Matrices

A Matrix is a two-dimensional array of numbers. There are several uses for matrices. For example

• A linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$.
• A transition from one basis in a vector space to another basis.
• A two-dimensional array of numbers
• A Markov process, or Markov chain
• A Stochastic process
• A directed graph
• An undirected graph

We discussed the directed graph of a Stochastic process in problem set 5. Directed and undirected graphs can be also be used in many applications

• find a minimum path or route
• analyze social or population relationships
• analyze networks
• model electrical circuits
• develop dependency schedules

The matrices used to represent directed and undirected graphs play an important role in analyzing and solving these problems. Everyone should study Dr. Strang's lecture

Graphs, Networks, Incidence Matrices

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6.3.4 Transformations and Basis

We now have several examples for creating

the standard matrix representation of a linear transformation.

However it is often more useful to represent a linear transformation

with respect to a different set of basis vectors.

In this way a linear transformation or function is an imaginary object like numbers or vectors.

Like vectors, the matrix that represents the linear transformation depends on the choice of basis. It is more complicated because we now need to

• choose a basis for the domain and
• choose a basis for the codomain

We must choose a basis for the domain so we can represent the input vectors to our linear transformation from the domain, and we must choose a basis for the codomain so we know how to represent the output vectors in the codomain.

Only after choosing these two bases can we create a matrix representation of a linear transformation

Example 3

Let us choose two nonstandard unit vectors (vectors on the unit circle) for our basis of $\mathbb{R}^2$.

Let use choose as our basis for $\mathbb{R}^2$,

$$ U = \left\{ \begin{bmatrix} \cos\left(\frac{\pi}{4}\right) \\ \sin\left(\frac{\pi}{4}\right) \end{bmatrix},\ \begin{bmatrix} \cos\left(\frac{3\pi}{4}\right) \\ \sin\left(\frac{3\pi}{4}\right) \end{bmatrix} \right\} = \left\{ \begin{bmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{bmatrix},\ \begin{bmatrix} -\frac{1}{\sqrt{2}} \\ \ \ \frac{1}{\sqrt{2}} \end{bmatrix} \right\} = \left\{ \mathbf{u}_1,\ \mathbf{u}_2 \right\} $$

We know from our previous work that the transition matrix on $\mathbb{R}^2$ from this new basis $U$ to the standard basis is given by

$$ U = \begin{bmatrix} \mathbf{u}_1 & \mathbf{u}_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \end{bmatrix} $$

The linear transformation $S_U$ on $\mathbb{R}^2$ is represented by matrix $U$,

$$ S_U:\mathbb{R}^2\rightarrow\mathbb{R}^2 $$

where the choice of basis in the domain $\mathbb{R}^2$ is the basis $U$. The basis for the codomain is the standard basis.

One can easily determine that the transition matrix from the standard basis to the new basis $U$ is given by

$$ U^{-1} = \begin{bmatrix}\ \ \frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \end{bmatrix} $$

This matrix also represents a linear transformation on $\mathbb{R}^2$,

$$ S_U^{-1}:\mathbb{R}^2\rightarrow\mathbb{R}^2 $$

This time the choice of basis for the domain is the standard basis $\left\{\ihat,\ \jhat\right\}$, and the choice of basis for the codomain is the ordered basis $U$.

Now we see that transition matrices were our first matrix representations of linear transformations with respect to nonstandard bases.

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6.3.5 Matrix Representations and Basis

Let $L$ be a linear transformation (operator) such that

$$ L:\mathbb{R}^2\rightarrow\mathbb{R^2}\text{ that rotates all vectors in standard position counterclockwise $\frac{\pi}{2}$.} $$

Example 4

Determine the standard matrix representation of linear operator $L$.

If we want the standard matrix representation of our rotation we simply use trigonometry of the unit circle to find where the linear transformation sends $\ihat$ and $\jhat$ to determine the columns of our matrix.

$$ \begin{align*} A_L &= \begin{bmatrix} L\left(\ihat\right)\ L\left(\jhat\right) \end{bmatrix} \\ \\ &= \begin{bmatrix} L\left(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right)\ L\left(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\right) \end{bmatrix} \\ \\ &= \begin{bmatrix} \begin{bmatrix} \cos\left(0 + \frac{\pi}{2}\right) \\ \sin\left(0 + \frac{\pi}{2}\right) \end{bmatrix}\ \begin{bmatrix} \cos\left(\frac{\pi}{2} + \frac{\pi}{2}\right) \\ \sin\left(\frac{\pi}{2} + \frac{\pi}{2}\right) \end{bmatrix} \end{bmatrix} \\ \\ &= \begin{bmatrix}\ \ 0 & -1 \\ \ \ 1 &\ \ 0 \end{bmatrix} \end{align*} $$

Example 5

Determine the matrix representation of $L$ with respect to the basis $U$ from Example 3.

There are two very different ways to consider the linear operator with respect to the standard basis and the $U$ basis.

1. Linear transformations as functions

In this view we see that the linear transformation $L$ can map vectors

• from $\mathbb{R}^2$ with the standard basis to $\mathbb{R}^2$ with the standard basis

or we can consider the linear transformation $L$ mapping

• from $\mathbb{R}^2$ with basis $U$ to $\mathbb{R}^2$ with basis $U$.

We also have transition operator $S_U$

• from $\mathbb{R}^2$ with basis $U$ to $\mathbb{R}^2$ with the standard basis,

and its inverse $S_U^{-1}$, the transition operator

• form $\mathbb{R}^2$ with the standard basis to $\mathbb{R}^2$ with basis $U$.

From these linear transformations we can create a commutative diagram

We call this a commutative diagram because

• every arrow represents the direction of a function from domain to codomain
• the codomain of one function is the domain of the next function
• function composition holds for the connected paths

That is

$$ L = S_U^{-1}\circ L\circ S_U $$

This is why we see the function composition operator in the middle of the diagram.

2. Linear Transformations as Matrices

We can also replace every function in our commutative diagram with its matrix representation:

• $A_L$ is the standard matrix representation of linear transformation $L$.
• $A_U$ is the matrix representation of linear operator $L$ choosing basis $U$ for both domain and codomain.
• $U$ is the matrix representation of transition operator $S_U$ from basis $U$ to the standard basis
• $U^{-1}$ is the matrix representation of transition operator $S_U^{-1}$ from the standard basis to basis $U$.

In this case

$$ A_U = U^{-1}A_LU, $$

where on the right-hand side of the equation we replaced function composition with matrix multiplication.

Finally we are prepared to create matrix $A_U$, the matrix representation of $L$ with respect to basis $U$.

$$ \begin{align*} A_U &= U^{-1}A_LU \\ \\ &= \begin{bmatrix}\ \ \frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \end{bmatrix}\begin{bmatrix}\ \ 0 & -1 \\ \ \ 1 &\ \ 0 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} &\ \ \frac{1}{\sqrt{2}} \end{bmatrix} \\ \\ &= \dfrac{1}{\sqrt{2}}\begin{bmatrix}\ \ 1 &\ \ 1 \\ -1 &\ \ 1 \end{bmatrix}\begin{bmatrix}\ \ 0 & -1 \\ \ \ 1 &\ \ 0 \end{bmatrix}\dfrac{1}{\sqrt{2}}\begin{bmatrix}\ \ 1 & -1 \\ \ \ 1 &\ \ 1 \end{bmatrix} \\ \\ &= \frac{1}{2}\begin{bmatrix}\ \ 1 & -1 \\ \ \ 1 &\ \ 1 \end{bmatrix}\begin{bmatrix}\ \ 1 & -1 \\ \ \ 1 &\ \ 1 \end{bmatrix} \\ \\ &= \frac{1}{2}\begin{bmatrix}\ \ 0 & -2 \\ \ \ 2 &\ \ 0 \end{bmatrix} = \begin{bmatrix}\ \ 0 & -1 \\ \ \ 1 &\ \ 0 \end{bmatrix} \end{align*} $$

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6.3.6 Matrix Representation Theorem

Any linear transformation $L: V\rightarrow W$ between vector spaces $V,W$ with dimension $n$ and $m$ respectively may be expressed as a matrix. To prove this, we let $E = \left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ and $F = \left\{\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_m\right\}$ be ordered bases on $V$ and $W$. Here we prove that it is possible to construct a transition matrix from $V$ to $W$ and hence represent the linear transformation.

Theorem 6.3.2

Matrix Representation Theorem for Linear Transformations Between Finite Vector Spaces

Let $E = \left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ and $F = \left\{\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_m\right\}$ be ordered bases on vector spaces $V$ and $W$, respectively, then for each linear transformation $L: V\rightarrow W$ there is a matrix $A\in\mathbb{R}^{m\times n}$ such that

$$ \left[\,L\left([\mathbf{v}]_E\right)\,\right]_F = A\left[\mathbf{v}\right]_E $$

for each $\mathbf{v}\in V$.

Additionally, we have that each column of the matrix $A$ has the form

$$ \mathbf{a}_j = \left[L(\mathbf{v}_j)\right]_F $$

for each $j=1,2,\ldots,n$.

Proof:

Let $L: V\rightarrow W$ be a linear transformation between the vector spaces $V$ and $W$ with ordered bases $E = \left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ on $V$ and $F = \left\{\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_m\right\}$ on $W$. Then, for any vector $\mathbf{v}\in V$ we may express $\mathbf{v}$ as a linear combination with respect to the basis $E$:

$$ \mathbf{x} = \left[\mathbf{v}\right]_E = x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + \ldots + x_n\mathbf{v}_n $$

The vector $\mathbf{x}$ is the coordinate vector of $\mathbf{v}$ with respect to $E$ and it is our goal to so that there is a matrix $A\in\mathbb{R}^{m\times n}$ which represents $L$ so that

$$ A\mathbf{x} = \mathbf{y}\qquad \Leftrightarrow \qquad \mathbf{y} = \left[\,L(\mathbf{v})\,\right]_F = y_1\mathbf{w}_1 + y_2\mathbf{w}_2 + \ldots + y_m\mathbf{w}_m, $$

where vector $\mathbf{y}$ is the coordinate vector of $L\left(\mathbf{v}\right)$ with respect to $F$. For this matrix $A$, the coordinate vector of $L(\mathbf{v})$ with respect to $F$ is would be given by

$$ \mathbf{y} = \left[\,L\left(\left[\mathbf{v}\right]_E\right)\,\right]_F = A\mathbf{x} $$

We show the construction of $A$ explicitly using a similar procedure to the proof for Theorem 6.3.1.

Define

$$ \mathbf{a}_j = \begin{bmatrix} a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix} = \left[L(\mathbf{v}_j)\right]_F $$

to be the coordinate vector of $L(\mathbf{v})$ with respect to the basis $F = \left\{\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_m\right\}$, so

$$ \left[\,L(\mathbf{v}_j)\,\right]_F = a_{1j}\mathbf{w}_1 + a_{2j}\mathbf{w}_2 + \ldots + a_{mj}\mathbf{w}_m $$

for $j=1,2,\ldots,n$. If $\mathbf{v}$ is given as a linear combination with respect to the basis $E$

$$ \left[\mathbf{v}\right]_E = x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + \ldots + x_n\mathbf{v}_n $$

and the matrix $A$ is defined as

$$ A = \left[a_{ij}\right] = \left[\mathbf{a}_1\ \mathbf{a}_2\ \ldots\ \mathbf{a}_n\right] $$

then

$$ \begin{align*} L(\mathbf{v}) &= L\left(\sum_{j=1}^n x_j\mathbf{v}_j\right) \\ \\ &= \sum_{j=1}^n x_j L(\mathbf{v}_j) \\ \\ &= \sum_{j=1}^n x_j \left(\sum_{i=1}^m a_{ij}\mathbf{w}_i\right) \\ \\ &= \sum_{j=1}^n \left( \sum_{i=1}^m a_{ij}x_j \right)\mathbf{w}_i \end{align*} $$

From here, we set

$$ y_i = \sum_{i=1}^m a_{ij}x_j $$

for each $i=1,2,\ldots,m$ and have that

$$ \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} = A\mathbf{x} $$

so $\mathbf{y}$ is the coordinate vector of $L(\mathbf{v})$ with respect to $F$. We have shown that $L$ may always be represented by a matrix and proven the theorem. The relationships between the different representations of $L$ are demonstrated by the following commutative diagram.

In this diagram

• $E$ is the transition matrix from basis $E$ to the standard basis in vector space $V$, and

• $F$ is the transition matrix from basis $F$ to the standard basis in vector space $W$. $\tombstone$

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6.3.7 More Linear Transformation Examples

Exercise 1

Let $L$ be the linear transformation mapping $\mathbb{R}^3$ into $\mathbb{R}^2$ where

$$ L(\mathbf{x}) = (x_1 + x_2)\mathbf{b}_1 + x_3\mathbf{b}_2 $$

for each $\mathbf{x}\in\mathbb{R}^3$ where

$$ \mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad \mathbf{b}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

Find the matrix $A$ representing $L$ with respect to the ordered bases $\left\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\right\}$ for the domain $\mathbb{R}^3$, and ordered basis $\left\{\mathbf{b}_1,\mathbf{b}_2\right\}$ for the codomain $\mathbb{R}^2$.

Follow Along

Apply the transformation to each basis element $\left\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\right\}$:

$$ \begin{align*} L(\mathbf{e}_1) &= (1+0)\mathbf{b}_1 + (0)\mathbf{b}_2 = 1\mathbf{b}_1 + 0\mathbf{b}_2 \\ \\ L(\mathbf{e}_2) &= (0+1)\mathbf{b}_1 + (0)\mathbf{b}_2 = 1\mathbf{b}_1 + 0\mathbf{b}_2 \\ \\ L(\mathbf{e}_3) &= (0+0)\mathbf{b}_1 + (1)\mathbf{b}_2 = 0\mathbf{b}_1 + 1\mathbf{b}_2 \end{align*} $$
The $i^\text{th}$ column of $A$ is the coordinates of $L(\mathbf{e}_i)$ with respect to $\left\{\mathbf{b}_1,\mathbf{b}_2\right\}$ for $i=1,2,3$. $A$ is given by

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

Exercise 2

Let $L$ be the linear transformation mapping on $\mathbb{R}^2$ defined by

$$ L(\alpha\mathbf{b}_1 + \beta\mathbf{b}_2) = (\alpha + \beta)\mathbf{b}_1 + 2\beta\mathbf{b}_2 $$
where $\left\{\mathbf{b}_1,\mathbf{b}_2\right\}$ are from Exercise 1. Find the matrix representation $A$ for $L$ with respect to $\left\{\mathbf{b}_1,\mathbf{b}_2\right\}$.

Follow Along

For this exercise, the linear transformation is defined as it is applied to a linear combination of the basis vectors $\left\{\mathbf{b}_1,\mathbf{b}_2\right\}$. To find $A$, we need to determine $L(\mathbf{b}_1)$ and $L(\mathbf{b}_2)$. This is done by choosing the appropriate values for $\alpha$ and $\beta$:

$$ \begin{align*} L(\mathbf{b}_1) &= L(1\mathbf{b}_1 + 0\mathbf{b}_2) = (1+0)\mathbf{b}_1 + 2(0)\mathbf{b}_2 = 1\mathbf{b}_1 + 0\mathbf{b}_2 \\ \\ L(\mathbf{b}_2) &= L(0\mathbf{b}_1 + 1\mathbf{b}_2) = (0+1)\mathbf{b}_1 + 2(1)\mathbf{b}_2 = 1\mathbf{b}_1 + 2\mathbf{b}_2 \end{align*} $$
Hence,

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

Exercise 3

The video below concerns a linear transformation $D$ defined by $D(p) = p'$ which maps $P$ (the set of all polynomials) to $P$ using a transformation from the basis $\left\{1,x,x^2,x^3,\ldots\right\}$ to $\left\{1,x,x^2,x^3,\ldots\right\}$. This technically is not an example of what we are discussing this section, since it requires an "infinite dimensional matrix" to define the transformation, but it is instructive in how one would build the transformation in the finite $P_n$ to $P_{n-1}$ case.

(Re)watch the video Abstract vector spaces, then write out the matrix for the linear transformation representing the derivative $D$ which maps $P_n$ to $P_{n-1}$.

Check Your Work

Since we know that given

$$ p = a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \ldots + a_2 x_2 + a_1 x + a_0 $$
we have

$$ p' = (n-1)a_{n-1}x^{n-2} + (n-2)a_{n-2}x^{n-2} + \ldots + 2 a_2 x_1 + a_1 $$
This tells us two important facts:

1. The coefficient that each value is multiplied by during the transformation.
2. That our matrix must be $n\times (n-1)$ since the derivative reduces the power of each $x^j$.

Hence,

$$ D = \begin{bmatrix} n-1 & 0 & \ldots & 0 & 0 & 0 \\ 0 & n-2 & \ldots & 0 & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & \ldots & 2 & 0 & 0 \\ 0 & 0 & \ldots & 0 & 1 & 0 \\ \end{bmatrix} $$

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6.3.8 Reduced Row Echelon Form and Transition Matrices

Another method for determining matrix representation of the linear transformation $L:\mathbb{R}^n\rightarrow\mathbb{R}^m$ with respect to the ordered bases $U = \left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ and $V = \left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_m\right\}$ involves taking advantage of row equivalency and thus augmented matrices. By finding a representation of each $L(\mathbf{u}_j)$ as a linear combination of $\mathbf{v}_1,\mathbf{v}_2,\ldots\mathbf{v}_m$, we can construct the appropriate matrix.

Theorem 6.3.3

Let $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ and $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_m\right\}$ be ordered bases for $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively. If $L:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear transformation between these bases and $A$ is its matrix representation, then

$$ \mathbf{a}_j = V^{-1}L(\mathbf{u}_j) $$

for each $j=1,2,\ldots,n$, where $V = \left[ \mathbf{v}_1\ \mathbf{v}_2\ \ldots\ \mathbf{v}_m \right]$, and $L(\mathbf{u_j})$ is the standard representation of $L$ applied to basis vector $\ \mathbf{u}_j$ in standard coordinates.

Proof:

If we describe both bases $U$ and $V$ in standard coordinates, then $L(\mathbf{u}_j)$ is the linear transformation applied to a vector in standard coordinates. The vector representation of vector $\mathbf{u}_j$

$$ \begin{bmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{bmatrix}_U = \left[\mathbf{u}_j\right]_U = \mathbf{u}_j = U\begin{bmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{bmatrix}_U $$

Thus when we write $L(\mathbf{u}_j)$ everything is written in standard coordinates.

$$ L(\mathbf{u}_j) = A_L\mathbf{u}_j = A_LU\left[\mathbf{u}_j\right]_U $$

We want our matrix $A = V^{-1}A_LU$

If $A$ represents $L$ with respect to the bases $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ and $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_m\right\}$, then for each $j=1,2,\ldots,n$ we may write $L(\mathbf{u}_j)$ as a linear combination with respect to $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_m\right\}$

$$ \begin{align*} L(\mathbf{u}_j) &= a_{1j}\mathbf{v}_1 + a_{2j}\mathbf{v}_2 + \ldots + a_{mj}\mathbf{v}_m \\ \\ &= V\mathbf{a}_j \end{align*} $$

We know that $V$ is nonsingular since its columns form a basis for $\mathbb{R}^m$, so

$$ \mathbf{a}_j = V^{-1}L(\mathbf{u}_j) = V^{-1}A_LU\left[\mathbf{u}_j\right]_U = V^{-1}L\left(\left[\mathbf{u}_j\right]_U\right) = \left[ L\left(\left[\mathbf{u}_j\right]_U\right) \right]_V $$

for each $j=1,2,\ldots,n$. $\tombstone$

We learn from this theorem that the definition in Linear Transformations and Matrices, is still true:

The column $\ \mathbf{a}_j$ of matrix representation $A$ is where basis vector $\ \mathbf{u}_j$ in domain coordinates lands in codomain coordinates.

The direct consequence of this theorem is the ability to find the matrix representation of the linear transformation using augmented matrices and reduced row echelon form.

Corollary 6.3.4

If $A$ is the matrix representation of the linear transformation $L:\mathbb{R}^n\rightarrow\mathbb{R}^m$ with respect to the bases $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ and $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_m\right\}$ then the reduced row echelon form of

$$ \left[\left.\ \mathbf{v}_1\ \mathbf{v}_2\ \ldots \mathbf{v}_m\ \right|\ L(\mathbf{u}_1)\ L(\mathbf{u}_2)\ \ldots\ L(\mathbf{u}_n)\ \right] $$

is $\ \left[\left.\ I\ \right|\ A\ \right]$.

Proof:

Let $V = \left[\ \mathbf{v}_1\ \mathbf{v}_2\ \ldots \mathbf{v}_m\ \right]$ be the transition matrix representation of the the transition operator $S_V$ from basis $V$ to the standard basis. Since $V$ is nonsingular, the augmented matrix

$$\left[\left.\ V\ \right|\ L(\mathbf{u}_1)\ L(\mathbf{u}_2)\ \ldots\ L(\mathbf{u}_n)\ \right]$$

is row equivalent to $\left[\left.\ I\ \right|\ A\ \right]$ because

$$ \begin{align*} V^{-1}\left[\left.\ V\ \right|\ L(\mathbf{u}_1)\ L(\mathbf{u}_2)\ \ldots\ L(\mathbf{u}_n)\ \right] &= \left[\left.\ V^{-1}V\ \right|\ V^{-1}L(\mathbf{u}_1)\ V^{-1}L(\mathbf{u}_2)\ \ldots\ V^{-1}L(\mathbf{u}_n)\ \right] \\ \\ &= \left[\left.\ I\ \right|\ \mathbf{a}_1\ \mathbf{a}_2\ \ldots\ \mathbf{a}_n\ \right] \\ \\ &= \left[\left.\ I\ \right|\ A\ \right] \end{align*} $$ $\tombstone$

Exercise 4

Let $L:\mathbb{R}^2\rightarrow\mathbb{R}^3$ be the linear transformation given by

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

with respect to the bases $\left\{\mathbf{u}_1,\mathbf{u}_2\right\}$ and $\left\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\right\}$ where

$$ \mathbf{u}_1 = \left[\begin{array}{r} 1 \\ -1 \end{array}\right]\qquad \mathbf{u}_2 = \left[\begin{array}{r} -3 \\ 1 \end{array}\right] $$

and

$$ \mathbf{v}_1 = \left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right] \qquad \mathbf{v}_2 = \left[\begin{array}{r} 2 \\ 1 \\ 0 \end{array}\right] \qquad \mathbf{v}_3 = \left[\begin{array}{r} 3 \\ 2 \\ 1 \end{array}\right]$$

Check Your Work

$$ A = \left[\begin{array}{rr} 0 & 0 \\ -3 & 5 \\ 2 & -4 \end{array} \right] $$

Follow Along

First, we compute $L(\mathbf{u}_1)$ and $L(\mathbf{u}_2)$:

$$ L(\mathbf{u}_1) = \left[\begin{array}{r} 0 \\ 1 \\ 2 \end{array}\right]\qquad L(\mathbf{u}_2) = \left[\begin{array}{r} -2 \\ -3 \\ -4 \end{array}\right] $$
Next, we construct the augmented matrix $\left[\left.\ V\ \right|\ L(\mathbf{u}_1)\ L(\mathbf{u}_2)\ \right]$

$$ \left[\ \begin{array}{rrr|rr} 1 & 2 & 3 & 0 & -2 \\ 0 & 1 & 2 & 1 & -3 \\ 0 & 0 & 1 & 2 & -4 \end{array}\ \right] $$
and place it into reduced row echelon form.

$$ \begin{align*} \left[\ \begin{array}{rrr|rr} 1 & 2 & 3 & 0 & -2 \\ 0 & 1 & 2 & \ \ 1 & -3 \\ 0 & 0 & 1 & 2 & -4 \end{array}\ \right]& \begin{array}{l} -3R_3 + R_1 \\ -2R_3 + R_2 \\ \ \end{array} \\ \\ \left[\ \begin{array}{rrr|rr} 1 & 2 & 0 & -6 & 10 \\ 0 & 1 & 0 & -3 & 5 \\ 0 & 0 & 1 & 2 & -4 \end{array}\ \right]& \begin{array}{l} -2R_2 + R_1 \\ \ \\ \ \end{array} \\ \\ \left[\ \begin{array}{rrr|rr} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -3 & 5 \\ 0 & 0 & 1 & 2 & -4 \end{array}\ \right]& \end{align*} $$
Hence, the matrix $A$ is given by

$$ A = \left[\begin{array}{rr} 0 & 0 \\ -3 & 5 \\ 2 & -4 \end{array} \right] $$
Techniques from Chapter One are still in practice.

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